Free Math Worksheets

- How does the sign of the derivative relate to whether a graph is increasing or decreasing? - Differentiate the polynomial using the power rule. - A positive derivative indicates increasing behavior; a negative derivative indicates decreasing behavior.

1. Differentiate: \(f'(x)=0.5x^4-6x^2+10\). 2. Evaluate the derivative: \(f'(2)=0.5(2)^4-6(2)^2+10=-6\). 3. Since \(f'(2)<0\), the graph is decreasing at \(x=2\).

\(f'(x)=0.5x^4-6x^2+10\); the graph is decreasing at \(x=2\).

- Determine the sign of each first derivative over all real numbers. - A constant function is both nondecreasing and nonincreasing. - A strictly increasing function is also nondecreasing, and a strictly decreasing function is also nonincreasing.

1. \(f_1'(x)=3x^2+5>0\) for every real \(x\). Thus, \(f_1\) is strictly increasing and also nondecreasing. 2. \(f_2'(x)=-2x\), which is positive for \(x<0\) and negative for \(x>0\). Thus, \(f_2\) is neither nondecreasing nor nonincreasing on all of \(\mathbb{R}\). 3. \(f_3'(x)=-2<0\) for every real \(x\). Thus, \(f_3\) is strictly decreasing and also nonincreasing. 4. \(f_4'(x)=0\) for every real \(x\). A constant function is both nondecreasing and nonincreasing, but it is not strictly monotonic.

\(f_1\): a), c) \(f_2\): none \(f_3\): b), d) \(f_4\): c), d)

- Find the zeros of the first derivative. - Determine where the derivative is positive. - Include the critical endpoints in the monotonicity interval.

1. Differentiate: \(f'(x)=-3x^2+12x+15=-3(x-5)(x+1)\). 2. The derivative is zero at \(x=-1\) and \(x=5\). 3. Because this quadratic opens downward, \(f'(x)>0\) for \(-1<x<5\) and equals zero at the endpoints. 4. Therefore, the largest interval on which \(f\) is strictly increasing is \([-1,5]\).

\([-1,5]\)

- A point on the y-axis has x-coordinate \(0\). - Use the sign of the first derivative to determine monotonicity. - Apply the chain rule to the exponential term. - Analyze the terms separately as \(x\to\infty\).

1. At \(x=0\), \(f_a(0)=8e^0+3=11\). This value is independent of \(a\), so every graph has y-intercept \((0, 11)\). 2. Differentiate: \(f_a'(x)=-4e^{-0.5x}-a\). Because \(e^{-0.5x}>0\) and \(a>0\), \(f_a'(x)<0\) for every real \(x\). Therefore, each function is strictly decreasing. 3. As \(x\to\infty\), \(8e^{-0.5x}\to0\) and \(-ax\to-\infty\). Hence, \(\lim_{x\to\infty}f_a(x)=-\infty\).

a) The common y-intercept is \((0, 11)\). b) \(f_a'(x)=-4e^{-0.5x}-a<0\), so \(f_a\) is strictly decreasing. c) \(\lim_{x\to\infty}f_a(x)=-\infty\)

- Relate the sign of the first derivative to monotonicity. - Rewrite the quadratic derivative by completing the square. - A strictly monotonic function is one-to-one.

1. Differentiate: \(f'(x)=x^2+2x+5\). 2. Complete the square: \(f'(x)=(x+1)^2+4\). 3. Since \((x+1)^2+4>0\) for every real \(x\), \(f\) is strictly increasing on \(\mathbb{R}\). 4. Therefore, \(f\) is one-to-one and has an inverse function on its entire domain.

Because \(f'(x)=(x+1)^2+4>0\) for every real \(x\), \(f\) is strictly increasing and has an inverse function on \(\mathbb{R}\).

- Determine the sign of the first derivative throughout the domain. - Remember that \(x\) is restricted to positive values. - A strictly monotonic function is one-to-one.

1. Differentiate: \(f'(x)=\frac{1}{x}+2\). 2. On the domain \(x>0\), \(\frac{1}{x}>0\). Therefore, \(f'(x)>2>0\). 3. Thus, \(f\) is strictly increasing on \((0,\infty)\), so it is one-to-one and has an inverse function on that interval.

\(f'(x)=\frac{1}{x}+2>0\) for every \(x>0\). Therefore, \(f\) is strictly increasing and has an inverse function on \((0,\infty)\).

- Relate the sign of the first derivative to monotonicity. - Recall the range of the exponential function \(e^x\). - A strictly monotonic function is one-to-one.

1. Differentiate: \(f'(x)=e^x+2\). 2. Since \(e^x>0\) for every real \(x\), \(f'(x)>2>0\). 3. Therefore, \(f\) is strictly increasing on \(\mathbb{R}\), so it is one-to-one and has an inverse function on its entire domain.

\(f'(x)=e^x+2>0\) for every real \(x\). Therefore, \(f\) is strictly increasing and has an inverse function on \(\mathbb{R}\).

- Use the quotient rule. - A nonzero square is always positive. - A strictly monotonic function is one-to-one.

1. Apply the quotient rule: \(f'(x)=\frac{(x-1)-(x+3)}{(x-1)^2}=-\frac{4}{(x-1)^2}\). 2. For every \(x>1\), the denominator is positive, so \(f'(x)<0\). 3. Therefore, \(f\) is strictly decreasing on \((1,\infty)\). It is one-to-one and has an inverse function on this domain.

\(f'(x)=-\frac{4}{(x-1)^2}<0\) for every \(x>1\). Thus, \(f\) is strictly decreasing and has an inverse function on \((1,\infty)\).

- Use the derivative to test monotonicity. - Determine the largest possible value of \(-2-3x^2\). - A strictly monotonic function is one-to-one.

1. Differentiate: \(f'(x)=-2-3x^2\). 2. Since \(x^2\ge0\), \(f'(x)\le-2<0\) for every real \(x\). 3. Therefore, \(f\) is strictly decreasing on \(\mathbb{R}\), so it is one-to-one and has an inverse function on its entire domain.

\(f'(x)=-2-3x^2<0\) for every real \(x\). Thus, \(f\) is strictly decreasing and has an inverse function on \(\mathbb{R}\).

- Identify the vertex and opening direction of the parabola. - Use the sign of the first derivative to describe where the function increases or decreases. - What is the slope of the line \(y=2x\)? - Find where the derivative has that slope. - Substitute the x-coordinate into the function to find the y-coordinate.

1. The function is defined for every real \(x\), so the domain is \((-\infty, \infty)\). Since \(k>0\), the parabola opens upward and has vertex \((0, k)\). Therefore, the range is \([k, \infty)\). 2. Differentiate: \(f_k'(x)=\frac{2x}{k}\). This derivative is negative for \(x<0\), zero at \(x=0\), and positive for \(x>0\). Thus \(f_k\) is decreasing on \((-\infty, 0]\) and increasing on \([0, \infty)\). 3. The line \(y=2x\) has slope \(2\). Set \(f_k'(x)=2\): \(\frac{2x}{k}=2\), so \(x=k\). 4. Then \(f_k(k)=\frac{k^2}{k}+k=2k\). Therefore, \(B_k=(k, 2k)\), which also lies on \(y=2x\).

a) Domain: \((-\infty, \infty)\); range: \([k, \infty)\) b) Decreasing on \((-\infty, 0]\); increasing on \([0, \infty)\) c) \(B_k=(k, 2k)\)

- Local extrema in the interior of the domain require a zero derivative. - Differentiate \(x^{-1}\) using the power rule. - Determine the sign of every term in the derivative. - Remember that \(x=0\) is not in the domain.

1. Differentiate: \(f'(x)=4+\frac{3}{x^2}\). 2. For every \(x\neq0\), \(\frac{3}{x^2}>0\). Therefore, \(f'(x)>4>0\) throughout the domain. 3. The derivative is never \(0\), so the function has no critical values in its domain and no local extrema.

Since \(f'(x)=4+\frac{3}{x^2}>0\) for every \(x\neq0\), the function has no local extrema.

- Find the first derivative and factor it. - Use the derivative's zeros to create a sign chart. - A positive derivative indicates increase; a negative derivative indicates decrease.

1. Differentiate: \(h'(x)=6x^2-30x+36=6(x-2)(x-3)\). 2. The critical numbers are \(x=2\) and \(x=3\). 3. The derivative is positive on \((-\infty,2)\), negative on \((2,3)\), and positive on \((3,\infty)\). 4. Therefore, \(h\) is strictly increasing on \((-\infty,2]\) and \([3,\infty)\), and strictly decreasing on \([2,3]\).

Strictly increasing on \((-\infty,2]\) and \([3,\infty)\); strictly decreasing on \([2,3]\)

- Factor the first derivative completely. - Use its zeros to divide the real line into intervals. - Test the sign of the derivative on each interval.

1. Differentiate: \(f'(x)=x^3-3x^2+2x=x(x-1)(x-2)\). 2. The critical numbers are \(x=0\), \(x=1\), and \(x=2\). 3. A sign chart gives \(f'(x)<0\) on \((-\infty,0)\), \(f'(x)>0\) on \((0,1)\), \(f'(x)<0\) on \((1,2)\), and \(f'(x)>0\) on \((2,\infty)\). 4. Therefore, \(f\) is strictly decreasing on \((-\infty,0]\) and \([1,2]\), and strictly increasing on \([0,1]\) and \([2,\infty)\).

Strictly increasing on \([0,1]\) and \([2,\infty)\); strictly decreasing on \((-\infty,0]\) and \([1,2]\)

- A positive derivative means the original function is increasing. - A negative derivative means the original function is decreasing. - A horizontal tangent occurs where the derivative is zero.

1. The graph of \(f'\) is above the x-axis on \((-2,1)\), so \(f'(x)>0\) there. It is below the x-axis on \((1,3)\), so \(f'(x)<0\) there. 2. On \([-1,0]\), the derivative is positive, so \(f\) is strictly increasing. Statement a) is true. 3. On \([1,2]\), the derivative is negative in the interior, so \(f\) is strictly decreasing, not increasing. Statement b) is false. 4. The graph of \(f'\) crosses the x-axis at \(x=1\), so \(f'(1)=0\). Therefore, \(f\) has a horizontal tangent at \(x=1\). Statement c) is true.

a) True. \(f'(x)>0\) on \((-1,0)\). b) False. \(f'(x)<0\) on \((1,2)\), so \(f\) is strictly decreasing there. c) True. \(f'(1)=0\).

- Check whether the derivative graph is above or below the x-axis on each interval. - A positive derivative means increasing; a negative derivative means decreasing.

1. On the interior of \(I_1=[-1,1]\), the graph of \(f'\) is above the x-axis, so \(f'(x)>0\). Therefore, \(f\) is strictly increasing on \(I_1\), which corresponds to A. 2. On the interior of \(I_2=[1,3]\), the graph of \(f'\) is below the x-axis, so \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \(I_2\), which corresponds to B. 3. On the interior of \(I_3=[3,4]\), the graph of \(f'\) is above the x-axis, so \(f'(x)>0\). Therefore, \(f\) is strictly increasing on \(I_3\), which corresponds to A.

I1: A, I2: B, I3: A

- Locate the turning points. - Read the graph from left to right. - Include the endpoints of the displayed domain.

1. The graph has a local maximum at \(x=-1\) and a local minimum at \(x=3\). 2. The graph rises from \(x=-3\) to \(x=-1\), falls from \(x=-1\) to \(x=3\), and rises again from \(x=3\) to \(x=5\). 3. Therefore, \(f\) is strictly increasing on \([-3,-1]\) and \([3,5]\), and strictly decreasing on \([-1,3]\).

Strictly increasing on \([-3,-1]\) and \([3,5]\); strictly decreasing on \([-1,3]\)

- Compare function values as you move from left to right within the red boundaries. - A constant segment prevents strict increase or strict decrease. - Use the order definition, not only the absence of horizontal segments.

1. In panel a), every larger x-value in \(I\) gives a smaller function value. The function is strictly decreasing, so a) matches 3. 2. In panel b), the graph rises, remains constant from \(x=2\) to \(x=4\), and then rises again. It never decreases, but it is not strictly increasing because of the constant segment. Thus, b) matches 2. 3. In panel c), every larger x-value in \(I\) gives a larger function value. The function is strictly increasing, so c) matches 1.

a) 3 b) 2 c) 1

- Check whether the graph changes direction within each interval. - An interval containing both rising and falling portions is not monotonic.

1. On \(I_1\), the graph rises throughout the interval, so \(f\) is strictly increasing. 2. On \(I_2\), the graph rises to a local maximum at \(x=0\) and then falls, so \(f\) is not monotonic. 3. On \(I_3\), the graph falls throughout the interval, so \(f\) is strictly decreasing.

\(I_1\): strictly increasing \(I_2\): not monotonic \(I_3\): strictly decreasing

- Locate each point where the graph changes direction. - Read the graph from left to right and record whether it rises or falls.

1. The graph decreases from \(x=-3\) to the local minimum at \(x=-2\). 2. It increases from \(x=-2\) to the local maximum at \(x=0\). 3. It decreases from \(x=0\) to the local minimum at \(x=2\). 4. It increases from \(x=2\) to \(x=3\). 5. Therefore, \(f\) is strictly decreasing on \([-3,-2]\) and \([0,2]\), and strictly increasing on \([-2,0]\) and \([2,3]\).

Strictly increasing on \([-2,0]\) and \([2,3]\); strictly decreasing on \([-3,-2]\) and \([0,2]\)

- Locate the valley and peak of the graph. - The graph is decreasing where it moves downward from left to right. - Estimate each turning point to the nearest tenth.

1. The graph has a local minimum near \(x=-1.4\) and a local maximum near \(x=1.4\). 2. The graph falls from \(x=-3\) to the local minimum, rises between the two extrema, and falls again from the local maximum to \(x=3\). 3. Therefore, the decreasing intervals are approximately \([-3,-1.4]\) and \([1.4,3]\).

Approximately \([-3,-1.4]\) and \([1.4,3]\)

- Find and factor the first derivative. - Use its zeros to make a sign chart. - Compare the sign chart with the turning points in the graph.

1. Differentiate: \(h'(x)=3x^2-3=3(x-1)(x+1)\). 2. The critical numbers are \(x=-1\) and \(x=1\). 3. The derivative is positive on \((-\infty,-1)\), negative on \((-1,1)\), and positive on \((1,\infty)\). 4. Therefore, \(h\) is strictly increasing on \((-\infty,-1]\) and \([1,\infty)\), and strictly decreasing on \([-1,1]\). This agrees with the graph.

Strictly increasing on \((-\infty,-1]\) and \([1,\infty)\); strictly decreasing on \([-1,1]\)

- Differentiate and factor the quadratic derivative. - Determine the sign of an upward-opening quadratic between its zeros. - It is enough for the derivative to be negative throughout the interval's interior.

1. Differentiate and factor: \(f'(x)=\frac{1}{2}x^2-x-\frac{3}{2}=\frac{1}{2}(x-3)(x+1)\). 2. The derivative is zero at \(x=-1\) and \(x=3\). 3. Because the quadratic opens upward, \(f'(x)<0\) for every \(x\in(-1,3)\). 4. By the monotonicity theorem, \(f\) is strictly decreasing on \([-1,3]\).

\(f'(x)=\frac{1}{2}(x-3)(x+1)<0\) for \(-1<x<3\), so \(f\) is strictly decreasing on \([-1,3]\).

- Locate the point where the graph changes from rising to falling. - Use the x-coordinates of the displayed endpoints and turning point.

1. The graph rises from \(x=-1\) to its maximum at \(x=2\). 2. It falls from \(x=2\) to \(x=5\). 3. Therefore, \(f\) is strictly increasing on \([-1,2]\) and strictly decreasing on \([2,5]\).

Strictly increasing on \([-1,2]\); strictly decreasing on \([2,5]\)

- Follow the profile from left to right. - The lowest point separates the downhill and uphill portions.

1. The graph descends from \(x=-2\) to its lowest point at \(x=2\). 2. It then rises from \(x=2\) to \(x=6\). 3. Therefore, the hiker travels downhill on \([-2,2]\) and uphill on \([2,6]\).

Downhill on \([-2,2]\); uphill on \([2,6]\)

- Locate the local minimum and local maximum. - Identify the portions that move downward from left to right.

1. The graph decreases from the left endpoint to the local minimum at \(x=-2\). 2. It increases from \(x=-2\) to the local maximum at \(x=4\). 3. It decreases again from \(x=4\) to the right endpoint. 4. Therefore, \(f\) is strictly decreasing on \([-4,-2]\) and \([4,6]\).

\([-4,-2]\) and \([4,6]\)

- Find each valley and peak. - Identify where the ramp rises as you move from left to right.

1. The graph has local minima at \(x=-2\) and \(x=2\), and a local maximum at \(x=0\). 2. It rises from \(x=-2\) to \(x=0\), and again from \(x=2\) to the right endpoint \(x=3\). 3. Therefore, \(k\) is strictly increasing on \([-2,0]\) and \([2,3]\).

\([-2,0]\) and \([2,3]\)

- Divide the domain at every turning point. - For each piece, decide whether the graph rises or falls from left to right.

1. The graph has local maxima at \(x=-2\) and \(x=2\), and a local minimum at \(x=0\). 2. The graph rises on \([-3,-2]\), falls on \([-2,0]\), rises on \([0,2]\), and falls on \([2,3]\). 3. Therefore, \(p\) is strictly increasing on \([-3,-2]\) and \([0,2]\), and strictly decreasing on \([-2,0]\) and \([2,3]\).

Strictly increasing on \([-3,-2]\) and \([0,2]\); strictly decreasing on \([-2,0]\) and \([2,3]\)

- A derivative below the x-axis is negative. - A derivative above the x-axis is positive. - Translate each derivative sign into the original function's monotonic behavior.

1. The graph of \(f'\) lies below the x-axis throughout \([-1,5]\), so \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \([-1,5]\). 2. The graph of \(g'\) lies above the x-axis throughout \([-1,5]\), so \(g'(x)>0\). Therefore, \(g\) is strictly increasing on \([-1,5]\).

\(f\): strictly decreasing on \([-1,5]\) \(g\): strictly increasing on \([-1,5]\)

- Find the zeros of the first derivative. - Compare both the monotonic intervals and the function values at the turning points.

1. Differentiate and factor: \(f'(x)=3x^2-6x=3x(x-2)\). 2. The derivative is positive for \(x<0\), negative for \(0<x<2\), and positive for \(x>2\). 3. Thus, \(f\) is strictly increasing on \((-\infty,0]\), strictly decreasing on \([0,2]\), and strictly increasing on \([2,\infty)\). 4. Also, \(f(0)=1\) and \(f(2)=-3\), matching the displayed local maximum and local minimum. The graph represents \(f\).

Yes. The function is strictly increasing on \((-\infty,0]\) and \([2,\infty)\), and strictly decreasing on \([0,2]\). Its turning points \((0,1)\) and \((2,-3)\) match the graph.

- Factor the first derivative completely. - Use a sign chart and compare its turning points with the graph.

1. Differentiate and factor: \(f'(x)=x^3-4x=x(x-2)(x+2)\). 2. The derivative is negative on \((-\infty,-2)\), positive on \((-2,0)\), negative on \((0,2)\), and positive on \((2,\infty)\). 3. Therefore, \(f\) is strictly decreasing on \((-\infty,-2]\) and \([0,2]\), and strictly increasing on \([-2,0]\) and \([2,\infty)\). 4. The displayed graph has local minima at \(x=-2\) and \(x=2\), and a local maximum at \(x=0\), so its monotonic behavior matches \(f\).

Yes. The graph matches the pattern decreasing, increasing, decreasing, increasing, with turning points at \(x=-2\), \(x=0\), and \(x=2\).

- Factor the first derivative. - Test its sign on each interval determined by the critical numbers. - Compare the sign chart with the graph's turning points.

1. Differentiate and factor: \(h'(x)=-3x^2+9x-6=-3(x-1)(x-2)\). 2. The critical numbers are \(x=1\) and \(x=2\). 3. The derivative is negative for \(x<1\), positive for \(1<x<2\), and negative for \(x>2\). 4. Therefore, \(h\) is strictly decreasing on \((-\infty,1]\) and \([2,\infty)\), and strictly increasing on \([1,2]\). This agrees with the graph.

Strictly increasing on \([1,2]\); strictly decreasing on \((-\infty,1]\) and \([2,\infty)\)

- Find where the derivative graph lies below the x-axis. - Zeros of the derivative form interval boundaries.

1. A function is strictly decreasing where its derivative is negative. 2. The graph of \(h'\) is below the x-axis on \((-6,-4)\) and \((1,5)\). 3. Therefore, \(h\) is strictly decreasing on \([-6,-4]\) and \([1,5]\).

\([-6,-4]\) and \([1,5]\)

- Locate the vertex. - A downward-opening parabola rises to its vertex and falls afterward.

1. The vertex is \((-2,3)\). 2. The graph rises to the vertex on \([-5,-2]\) and falls after the vertex on \([-2,1]\). 3. Therefore, \(f\) is strictly increasing on \([-5,-2]\) and strictly decreasing on \([-2,1]\).

Strictly increasing on \([-5,-2]\); strictly decreasing on \([-2,1]\)

- Locate the lowest point. - Read the graph from left to right.

1. The graph has its minimum at \(x=3\). 2. It falls from \(x=0\) to \(x=3\), then rises from \(x=3\) to \(x=6\). 3. Therefore, \(f\) is strictly decreasing on \([0,3]\) and strictly increasing on \([3,6]\).

Strictly decreasing on \([0,3]\); strictly increasing on \([3,6]\)

- Use the local maximum and local minimum as interval boundaries. - Determine whether each graph segment rises or falls.

1. The graph has a local maximum at \(x=-2\) and a local minimum at \(x=2\). 2. It rises on \([-4,-2]\), falls on \([-2,2]\), and rises on \([2,4]\). 3. Therefore, \(f\) is strictly increasing on \([-4,-2]\) and \([2,4]\), and strictly decreasing on \([-2,2]\).

Strictly increasing on \([-4,-2]\) and \([2,4]\); strictly decreasing on \([-2,2]\)

- Mark the local minimum and local maximum. - Decide whether each section rises or falls from left to right.

1. The graph falls from \(x=-1\) to the local minimum at \(x=0\). 2. It rises from \(x=0\) to the local maximum at \(x=2\). 3. It falls again from \(x=2\) to \(x=3\). 4. Therefore, \(f\) is strictly decreasing on \([-1,0]\) and \([2,3]\), and strictly increasing on \([0,2]\).

Strictly increasing on \([0,2]\); strictly decreasing on \([-1,0]\) and \([2,3]\)

- Locate the valley and peak. - Use their x-coordinates as interval boundaries.

1. The graph has a local minimum at \(x=-1\) and a local maximum at \(x=1\). 2. It falls on \([-2,-1]\), rises on \([-1,1]\), and falls on \([1,2]\). 3. Therefore, \(f\) is strictly decreasing on \([-2,-1]\) and \([1,2]\), and strictly increasing on \([-1,1]\).

Strictly increasing on \([-1,1]\); strictly decreasing on \([-2,-1]\) and \([1,2]\)

- Find where the graph changes direction. - Read each section from left to right.

1. The graph has a local minimum at \(x=-1\) and a local maximum at \(x=3\). 2. It decreases on \([-3,-1]\), increases on \([-1,3]\), and decreases on \([3,5]\). 3. Therefore, \(f\) is strictly increasing on \([-1,3]\), and strictly decreasing on \([-3,-1]\) and \([3,5]\).

Strictly increasing on \([-1,3]\); strictly decreasing on \([-3,-1]\) and \([3,5]\)

- Positive slope corresponds to increasing behavior. - Use each turning point as an interval boundary.

1. The graph has local maxima at \(x=-2\) and \(x=2\), and a local minimum at \(x=0\). 2. Its slope is positive on the rising sections \([-4,-2]\) and \([0,2]\), and negative on the falling sections \([-2,0]\) and \([2,4]\). 3. Therefore, \(g\) is strictly increasing on \([-4,-2]\) and \([0,2]\), and strictly decreasing on \([-2,0]\) and \([2,4]\).

Strictly increasing on \([-4,-2]\) and \([0,2]\); strictly decreasing on \([-2,0]\) and \([2,4]\)

- Factor the first derivative. - Use its zeros to create a sign chart. - Compare the sign chart with the graph.

1. Differentiate and factor: \(f'(x)=3x^2-12x+9=3(x-1)(x-3)\). 2. The derivative is positive for \(x<1\), negative for \(1<x<3\), and positive for \(x>3\). 3. Therefore, \(f\) is strictly increasing on \((-\infty,1]\) and \([3,\infty)\), and strictly decreasing on \([1,3]\). This agrees with the graph.

Strictly increasing on \((-\infty,1]\) and \([3,\infty)\); strictly decreasing on \([1,3]\)

- The original function increases where its derivative is above the x-axis. - It decreases where its derivative is below the x-axis.

1. The derivative is positive on \((-2,3)\), so \(f\) is strictly increasing on \([-2,3]\). 2. The derivative is negative on \([-4,-2)\) and \((3,6]\), so \(f\) is strictly decreasing on \([-4,-2]\) and \([3,6]\).

Strictly increasing on \([-2,3]\); strictly decreasing on \([-4,-2]\) and \([3,6]\)

- The first derivative is positive where the original graph rises from left to right. - Locate the vertex, where the slope changes sign.

1. The graph is an upward-opening parabola with vertex at \((2,-1)\). 2. The function rises to the right of the vertex, so its slope is positive for \(2<x\le6\). 3. Therefore, \(f'(x)>0\) on \((2,6]\).

\((2,6]\)

- Read the speed graph from left to right. - Turning points separate increasing and decreasing intervals.

1. The graph has a local maximum at \(t=2\) and a local minimum at \(t=6\). 2. The speed increases on \([0,2]\), decreases on \([2,6]\), and increases again on \([6,8]\).

Strictly increasing on \([0,2]\) and \([6,8]\); strictly decreasing on \([2,6]\)

- Analyze each graph segment separately. - Decide whether each segment rises, stays level, or falls.

1. The graph rises from \(x=0\) to \(x=3\), so \(g\) is strictly increasing on \([0,3]\). 2. It is horizontal from \(x=3\) to \(x=7\), so \(g\) is constant on \([3,7]\). 3. It falls from \(x=7\) to \(x=10\), so \(g\) is strictly decreasing on \([7,10]\).

Strictly increasing on \([0,3]\); constant on \([3,7]\); strictly decreasing on \([7,10]\)

- A positive derivative means the original function rises from left to right. - Mark the turning points, then identify the increasing intervals between them.

1. The derivative is positive where \(g\) is strictly increasing. 2. The graph decreases on \((-4, -2)\), increases on \((-2, 0)\), decreases on \((0, 2)\), and increases on \((2, 4)\). 3. Therefore, \(g^{\prime}(x)>0\) on \((-2, 0)\) and \((2, 4)\).

\((-2, 0)\) and \((2, 4)\)

- Connect the sign of the derivative to where the original function increases or decreases. - Check whether the derivative graph crosses the x-axis inside each interval. - A derivative sign change means the original function changes direction.

1. A differentiable function is invertible on an interval if it is strictly monotonic there. The sign of its derivative determines where it increases or decreases. 2. On \(I_1=[-1,1]\), \(h'(x)>0\) for \(-1\le x<0\) and \(h'(x)<0\) for \(0<x\le1\). 3. Because the derivative changes sign at \(x=0\), \(h\) increases and then decreases on \(I_1\). Therefore, it is not invertible on \(I_1\). 4. On \(I_2=[0,2]\), \(h'(x)<0\) for \(0<x<2\), with derivative zeros only at the endpoints. 5. Therefore, \(h\) is strictly decreasing on \([0,2]\) and is invertible on \(I_2\).

\(h\) is invertible on \(I_2=[0,2]\) because \(h'(x)<0\) throughout the interior of that interval, so \(h\) is strictly decreasing. It is not invertible on \(I_1=[-1,1]\) because \(h'\) changes sign at \(x=0\).

- A derivative of zero means the tangent line is horizontal. - Identify which marked points are smooth local maxima or minima.

1. The derivative equals the slope of the tangent line. 2. A derivative value of zero corresponds to a horizontal tangent, which occurs at the smooth local extrema. 3. Points \(A\), \(B\), and \(D\) are local extrema with horizontal tangents. Point \(C\) lies on a decreasing portion of the graph.

At points \(A\), \(B\), and \(D\).

- Recall how the sign of a derivative determines whether the original function increases or decreases. - Find where the derivative graph lies above the x-axis. - Use the x-intercepts as interval boundaries.

1. A function is strictly increasing where its derivative is positive. 2. On the displayed domain \([-3,3]\), the graph of \(f'\) is above the x-axis for \(-3\le x<-1\) and \(1<x\le3\). 3. The isolated zeros at \(x=-1\) and \(x=1\) may be included as interval endpoints. Therefore, \(f\) is strictly increasing on \([-3,-1]\) and \([1,3]\).

\(f\) is strictly increasing on \([-3,-1]\) and \([1,3]\).

- Recall how a negative derivative affects the original function. - Find where the derivative graph lies below the x-axis. - Use the x-intercepts as interval boundaries.

1. A function is strictly decreasing where its derivative is negative. 2. On the displayed domain \([-2,4]\), the graph of \(f'\) is below the x-axis for \(-2\le x<0\) and \(2<x\le4\). 3. The isolated zeros at \(x=0\) and \(x=2\) may be included as interval endpoints. Therefore, \(f\) is strictly decreasing on \([-2,0]\) and \([2,4]\).

\(f\) is strictly decreasing on \([-2,0]\) and \([2,4]\).

- Identify which derivative remains positive from \(x=0\) to \(x=2\). - A derivative graph above the x-axis corresponds to an increasing original function. - Check whether any derivative crosses the x-axis inside the interval.

1. A function is strictly increasing on the interval if its derivative is positive throughout the interior of the interval. 2. The graph of \(f'\) crosses the x-axis at \(x=1\), so \(f'\) is negative on part of \([0,2]\). 3. The graph of \(g'\) stays above the x-axis for every \(x\in[0,2]\), so \(g\) is strictly increasing there. 4. The graph of \(h'\) is below the x-axis for \(0<x\le2\), so \(h\) is strictly decreasing there. 5. Therefore, only \(g\) is strictly increasing on \([0,2]\).

Only \(g\) is strictly increasing on \([0,2]\).

- Find which derivative stays below the x-axis from \(x=-1\) to \(x=1\). - A negative derivative corresponds to a decreasing original function. - Check whether any derivative changes sign inside the interval.

1. A function is strictly decreasing on the interval if its derivative is negative throughout the interior of the interval. 2. The graph of \(f'\) stays below the x-axis on \([-1,1]\), so \(f\) is strictly decreasing there. 3. The graph of \(g'\) stays above the x-axis on \([-1,1]\), so \(g\) is strictly increasing there. 4. The graph of \(h'\) is negative for \(x<0\) and positive for \(x>0\), so \(h\) decreases and then increases. 5. Therefore, only \(f\) is strictly decreasing on \([-1,1]\).

Only \(f\) is strictly decreasing on \([-1,1]\).

- Recall what sign the derivative has when the original function is decreasing. - Find where the derivative graph lies below the x-axis.

1. A function is strictly decreasing where its derivative is negative. 2. The graph of \(f'\) crosses the x-axis at \(x=3\) and lies below the x-axis for \(3<x\le6\). 3. Therefore, \(f\) is strictly decreasing on \([3,6]\).

\(f\) is strictly decreasing on \([3,6]\).

- Use the sign of \(f'\) to determine whether \(f\) rises or falls. - A local extremum in the interior would require a change in monotonic behavior.

1. The graph of \(f'\) stays entirely above the x-axis, so \(f'(x)>0\) throughout the displayed interval. 2. Therefore, \(f\) is strictly increasing throughout the interval. 3. Since \(f'\) has no zeros or sign changes, \(f\) has no local extrema in the interior of the displayed interval.

No. The function is strictly increasing because \(f'(x)>0\) throughout the displayed interval.

- Rewrite the derivative as the negative of a square. - Determine where the derivative is negative and where it is zero. - An isolated zero of the derivative does not prevent strict decrease.

1. Differentiate: \(g'(x)=-x^2+2x-1=-(x-1)^2\). 2. Since \((x-1)^2\ge0\), \(g'(x)\le0\) for every real \(x\). 3. The derivative is zero only at the isolated value \(x=1\) and is negative on every interval around it except at that point. Therefore, \(g\) is strictly decreasing on all of \(\mathbb{R}\).

Yes. Since \(g'(x)=-(x-1)^2\le0\) for all real \(x\) and vanishes only at \(x=1\), \(g\) is strictly decreasing on \(\mathbb{R}\).

- Factor the first derivative. - A repeated zero does not necessarily change the derivative's sign. - Find where the derivative is positive.

1. Differentiate: \(f'(x)=x^3-2x^2=x^2(x-2)\). 2. The derivative is zero at \(x=0\) and \(x=2\). 3. Since \(x^2\ge0\), the sign is determined by \(x-2\). Thus, \(f'(x)<0\) for \(x<2\), except at \(x=0\), and \(f'(x)>0\) for \(x>2\). 4. Therefore, the largest interval on which \(f\) is strictly increasing is \([2,\infty)\).

\([2,\infty)\)

- Distinguish necessary conditions from sufficient conditions. - Consider \(f(x)=x^3\) as a possible counterexample. - Strict increase does not require the derivative to be positive at every single point. - A horizontal tangent does not always indicate an extremum.

1. Statement a) is true. A positive derivative throughout an interval guarantees that the function is strictly increasing there. 2. Statement b) is true because \(f\) is differentiable on \(\mathbb{R}\). Fermat's theorem gives \(f'(x_0)=0\) at an interior local extremum. 3. Statement c) is false. For example, \(f(x)=x^3\) has \(f'(0)=0\), but no local extremum at \(x=0\). 4. Statement d) is false. The function \(f(x)=x^3\) is strictly increasing on \(\mathbb{R}\), but \(f'(0)=0\).

a) True b) True c) False d) False

- Factor the first derivative. - Use its sign on the intervals separated by the critical values. - Evaluate the original function at each critical value.

1. Differentiate: \(f'(x)=2x^2-2x-12=2(x-3)(x+2)\). 2. The critical values are \(x=-2\) and \(x=3\). 3. The derivative is positive for \(x<-2\) and \(x>3\), and negative for \(-2<x<3\). Thus, \(f\) is strictly increasing on \((-\infty,-2]\) and \([3,\infty)\), and strictly decreasing on \([-2,3]\). 4. The derivative changes from positive to negative at \(x=-2\), so \(f\) has a local maximum there. Since \(f(-2)=\frac{56}{3}\), the point is \(\left(-2,\frac{56}{3}\right)\). 5. The derivative changes from negative to positive at \(x=3\), so \(f\) has a local minimum there. Since \(f(3)=-23\), the point is \((3,-23)\).

a) Strictly increasing on \((-\infty,-2]\) and \([3,\infty)\); strictly decreasing on \([-2,3]\) b) Local maximum at \(\left(-2,\frac{56}{3}\right)\); local minimum at \((3,-23)\)

- Find the derivative. - Use the range of the cosine function. - Horizontal tangent lines correspond to a derivative of zero. - A positive derivative implies increasing behavior.

1. Differentiate: \(f'(x)=1.5+\cos x\). 2. Since \(-1\le\cos x\le 1\), \(0.5\le f'(x)\le 2.5\). 3. The derivative is never zero, so the graph has no horizontal tangent lines. 4. Since \(f'(x)\ge 0.5>0\) for every real \(x\), \(f\) is strictly increasing on \(\mathbb{R}\). 5. The minimum possible slope is \(0.5\), and the maximum possible slope is \(2.5\).

a) \(f'(x)\ge 0.5\), so the derivative is never zero. b) \(f\) is strictly increasing on \(\mathbb{R}\). c) Minimum slope: \(0.5\); maximum slope: \(2.5\)

- Test whether \(f(-x)=f(x)\). - Apply the quotient rule. - The denominator of the derivative is positive on the stated interval. - Analyze the numerator separately on \((0, \frac{\pi}{2})\) and \([\frac{\pi}{2}, \pi)\).

1. Since sine is odd, \(f(-x)=\frac{\sin(-x)}{-x}=\frac{-\sin x}{-x}=f(x)\). Therefore, \(f\) is even, so its graph is symmetric about the y-axis. 2. By the quotient rule, \(f'(x)=\frac{x\cos x-\sin x}{x^2}\). 3. The denominator \(x^2\) is positive on \((0, \pi)\). For \(0<x<\frac{\pi}{2}\), \(x\cos x-\sin x=\cos x(x-\tan x)<0\) because \(\cos x>0\) and \(\tan x>x\). For \(\frac{\pi}{2}\le x<\pi\), \(x\cos x\le 0\) and \(\sin x>0\), so \(x\cos x-\sin x<0\). Thus, \(f'(x)<0\) throughout \((0, \pi)\), and \(f\) is strictly decreasing there.

1. \(f(-x)=f(x)\), so the graph is symmetric about the y-axis. 2. \(f'(x)=\frac{x\cos x-\sin x}{x^2}\) 3. \(f'(x)<0\) for all \(x\in(0, \pi)\), so \(f\) is strictly decreasing.

- Differentiate term by term. - Use the given inequality to determine the sign of the derivative. - Check the parity of each term in \(g\). - Compare every value with \(g(0)\).

1. Differentiate: \(g'(x)=x-\sin x\). For \(x\ge 0\), the given inequality implies \(g'(x)\ge 0\). Therefore, \(g\) is nondecreasing on \([0, \infty)\). 2. Since cosine and \(x^2\) are even, \(g(-x)=\cos(-x)+\frac{1}{2}(-x)^2-1=g(x)\). 3. Since \(g(0)=0\) and \(g\) is nondecreasing for nonnegative inputs, \(g(x)\ge 0\) for \(x\ge 0\). Because \(g\) is even, the same is true for negative inputs. Thus, \(\cos x+\frac{1}{2}x^2-1\ge 0\), which is equivalent to \(\cos x\ge 1-\frac{1}{2}x^2\).

1. \(g'(x)=x-\sin x\ge 0\) for \(x\ge 0\), so \(g\) is nondecreasing on \([0, \infty)\). 2. \(g(-x)=g(x)\). 3. \(\cos x\ge 1-\frac{1}{2}x^2\) for all \(x\in\mathbb{R}\).

- Test whether \(g(-x)=-g(x)\). - Identify the dominant exponential term at each end of the real line. - Use the sign of the first derivative to determine monotonicity. - Expand both squares and subtract corresponding terms.

1. \(g(-x)=\frac{1}{2}(e^{-x}-e^x)=-g(x)\), so \(g\) is odd. Also, \(g(0)=0\). The equation \(e^x=e^{-x}\) has only the solution \(x=0\), so the origin is the only intercept. 2. As \(x\to\infty\), the term \(e^x\) dominates, so \(g(x)\to\infty\). By odd symmetry, \(g(x)\to-\infty\) as \(x\to-\infty\). 3. The derivative is \(g'(x)=\frac{1}{2}(e^x+e^{-x})\). Since both terms are positive, \(g'(x)>0\) for all real \(x\). Therefore, \(g\) is strictly increasing. 4. Squaring gives \([g'(x)]^2=\frac{1}{4}(e^{2x}+2+e^{-2x})\) and \([g(x)]^2=\frac{1}{4}(e^{2x}-2+e^{-2x})\). Subtracting yields \([g'(x)]^2-[g(x)]^2=\frac{1}{4}\cdot4=1\).

a) \(g\) is odd, and the only intercept is \((0,0)\). b) \(\lim_{x\to\infty}g(x)=\infty\), and \(\lim_{x\to-\infty}g(x)=-\infty\). c) \(g'(x)=\frac{1}{2}(e^x+e^{-x})>0\), so \(g\) is strictly increasing on \(\mathbb{R}\). d) \([g'(x)]^2-[g(x)]^2=1\).

- Rewrite the derivative as a perfect square. - Determine where it is positive and where it is zero. - An isolated zero of the derivative does not prevent strict increase.

1. Differentiate: \(f'(x)=x^2+4x+4=(x+2)^2\). 2. The derivative satisfies \(f'(x)\ge0\) for every real \(x\), and it equals zero only at \(x=-2\). 3. Because the derivative is positive except at one isolated value, \(f\) is strictly increasing on all of \(\mathbb{R}\).

The function is strictly increasing on \(\mathbb{R}\).

- Differentiate the function. - Determine when the derivative is positive for every real \(x\). - Consider separately the cases \(a > 0\), \(a = 0\), and \(a < 0\).

1. Differentiate: \(f_a'(x) = 3ax^2 + 3\). 2. If \(a \ge 0\), then \(3ax^2 \ge 0\) for every real \(x\). Therefore, \(f_a'(x) \ge 3 > 0\), so \(f_a\) is strictly increasing on \(\mathbb{R}\). 3. If \(a < 0\), then \(3ax^2 + 3\) becomes negative when \(|x|\) is sufficiently large. Thus, the function decreases on part of its domain and is not strictly increasing on all of \(\mathbb{R}\). 4. Therefore, \(f_a\) is strictly increasing on \(\mathbb{R}\) exactly when \(a \ge 0\).

\(a \ge 0\)

- Differentiate the function. - Use the range of the sine function. - Evaluate both the function and derivative at \(x=\pi\). - Write the tangent line in point-slope form.

1. Differentiate: \(g'(x)=1-\sin x\). Therefore, \(g'\left(\frac{\pi}{2}\right)=0\). 2. Since \(\sin x\le 1\), \(g'(x)=1-\sin x\ge 0\) for every real \(x\). Thus, no tangent line has negative slope. 3. At \(x=\pi\), \(g(\pi)=\pi-1\) and \(g'(\pi)=1\). The tangent line is \(y-(\pi-1)=x-\pi\), or \(y=x-1\).

1. The slope is \(0\). 2. \(g'(x)\ge 0\) for all real \(x\), so negative slopes do not occur. 3. \(y=x-1\)

- Find where the first derivative is negative. - Use the zeros of the quadratic derivative to identify the interval. - Compute the distance between the interval's endpoints before and after replacing \(a\) with \(3a\).

1. Differentiate: \(f_a'(x) = x^2 - a^2 = (x - a)(x + a)\). 2. The derivative is negative between its zeros \(-a\) and \(a\). Therefore, \(f_a\) is strictly decreasing on \((-a, a)\). 3. The distance between the endpoints is \(a - (-a) = 2a\). 4. Replacing \(a\) with \(3a\) changes the interval to \((-3a, 3a)\), whose length is \(6a\). Since \(6a = 3(2a)\), the interval's length is tripled.

a) \((-a, a)\) b) Its length is \(2a\). When \(a\) is tripled, the length becomes \(6a\), so it is also tripled.

- Rewrite the derivative as a perfect square. - Check whether the derivative changes sign at its zero. - Use the second and third derivatives to classify the horizontal-tangent point.

1. Differentiate: \(g'(x)=3x^2-6x+3=3(x-1)^2\ge0\). 2. The derivative is positive for every \(x\ne1\) and zero only at \(x=1\). Therefore, \(g\) is strictly increasing on \(\mathbb{R}\). 3. The horizontal tangent occurs at \(x=1\). Since \(g'\) is positive on both sides, there is no local extremum. 4. Also, \(g''(1)=0\) and \(g'''(1)=6\ne0\), so the graph has a stationary inflection point at \((1,3)\).

a) \(g\) is strictly increasing on \(\mathbb{R}\). b) The horizontal tangent occurs at \((1,3)\). It is a stationary inflection point, not a local extremum.

- Relate the sign of the first derivative to increasing and decreasing behavior. - Find where the first derivative equals zero. - Use unit-circle values to solve the trigonometric equation. - Test the sign of the derivative between consecutive critical numbers.

1. Differentiate: \(f'(x)=-\sin(x)+\frac{1}{2}\). 2. Solve \(f'(x)=0\): \(\sin(x)=\frac{1}{2}\). In \([0,2\pi]\), the critical numbers are \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\). 3. On \(\left(0,\frac{\pi}{6}\right)\), \(\sin(x)<\frac{1}{2}\), so \(f'(x)>0\). On \(\left(\frac{\pi}{6},\frac{5\pi}{6}\right)\), \(\sin(x)>\frac{1}{2}\), so \(f'(x)<0\). On \(\left(\frac{5\pi}{6},2\pi\right)\), \(\sin(x)<\frac{1}{2}\), so \(f'(x)>0\). 4. Therefore, \(f\) is strictly increasing on \(\left[0,\frac{\pi}{6}\right]\) and \(\left[\frac{5\pi}{6},2\pi\right]\), and strictly decreasing on \(\left[\frac{\pi}{6},\frac{5\pi}{6}\right]\).

Strictly increasing on \(\left[0,\frac{\pi}{6}\right]\) and \(\left[\frac{5\pi}{6},2\pi\right]\). Strictly decreasing on \(\left[\frac{\pi}{6},\frac{5\pi}{6}\right]\).

- Differentiate \(\sin^2(x)\) using the chain rule. - Factor the derivative into a product. - Determine where each factor is positive, zero, or negative. - Use the restricted interval when solving the trigonometric equations.

1. Differentiate using the chain rule: \(g'(x)=2\sin(x)\cos(x)-\sin(x)\). 2. Factor: \(g'(x)=\sin(x)(2\cos(x)-1)\). 3. In \([0,\pi]\), the derivative is zero at \(x=0\), \(x=\frac{\pi}{3}\), and \(x=\pi\). 4. On \((0,\pi)\), \(\sin(x)>0\), so the sign of \(g'(x)\) is determined by \(2\cos(x)-1\). It is positive on \(\left(0,\frac{\pi}{3}\right)\) and negative on \(\left(\frac{\pi}{3},\pi\right)\). 5. Therefore, \(g\) is strictly increasing on \(\left[0,\frac{\pi}{3}\right]\) and strictly decreasing on \(\left[\frac{\pi}{3},\pi\right]\).

Strictly increasing on \(\left[0,\frac{\pi}{3}\right]\). Strictly decreasing on \(\left[\frac{\pi}{3},\pi\right]\).

- Use the product rule to find the first derivative. - Evaluate the derivative at each given input. - The exponential factor is positive for every real input. - Use the sign of the derivative to decide whether the graph is rising or falling.

1. Differentiate using the product rule: \(g'(x)=2xe^x+(x^2-3)e^x=(x^2+2x-3)e^x\). 2. At \(x=-4\), \(g'(-4)=5e^{-4}>0\), so the graph is rising. 3. At \(x=0\), \(g'(0)=-3<0\), so the graph is falling. 4. At \(x=2\), \(g'(2)=5e^2>0\), so the graph is rising.

At \(x=-4\), the graph is rising. At \(x=0\), the graph is falling. At \(x=2\), the graph is rising.

- Find the domain before analyzing the derivative. - Factor the numerator of the first derivative. - Make a sign chart that includes both critical numbers and the excluded value. - Compare the derivative's sign immediately to the left and right of each relevant x-value.

1. The domain is \(\mathbb{R}\setminus\{2\}\). 2. Differentiate using the quotient rule: \(f'(x)=\frac{2x(x-2)-(x^2-3)}{(x-2)^2}=\frac{x^2-4x+3}{(x-2)^2}=\frac{(x-1)(x-3)}{(x-2)^2}\). 3. The derivative is zero at \(x=1\) and \(x=3\). Since the denominator is positive for every domain value, the derivative's sign is determined by \((x-1)(x-3)\). 4. Thus, \(f'(x)>0\) on \((-\infty, 1)\) and \((3, \infty)\), while \(f'(x)<0\) on \((1, 2)\) and \((2, 3)\). 5. The function changes from increasing to decreasing at \(x=1\), giving a local maximum, and from decreasing to increasing at \(x=3\), giving a local minimum. 6. At the vertical asymptote \(x=2\), the derivative is negative on both sides. The domain is interrupted there, but the monotonic behavior does not switch.

Increasing on \((-\infty, 1)\) and \((3, \infty)\); decreasing on \((1, 2)\) and \((2, 3)\). The monotonic behavior changes at \(x=1\) and \(x=3\), but not across the vertical asymptote \(x=2\).

- State the domain before making a sign chart. - Simplify the derivative before analyzing its sign. - Treat the intervals on opposite sides of a domain break separately.

1. The domain is \(\mathbb{R}\setminus\{0\}\). 2. Differentiate: \(g'(x)=\frac{2-x}{x^3}\). 3. The derivative is negative for \(x<0\), positive for \(0<x<2\), and negative for \(x>2\). 4. Therefore, \(g\) is strictly decreasing on \((-\infty,0)\) and \((2,\infty)\), and strictly increasing on \((0,2)\). 5. The point \(x=0\) is not in the domain, so it is not an extremum or a monotonicity-change point of the graph. The left branch is decreasing as it approaches the break, while the right branch initially increases.

Strictly decreasing on \((-\infty,0)\) and \((2,\infty)\); strictly increasing on \((0,2)\). The domain break at \(x=0\) separates two branches and is not an extremum.

- Differentiate the reciprocal term. - Analyze each domain interval separately. - A strictly increasing function can still be bounded above or below.

1. Differentiate: \(f'(x)=\frac{2}{(x-1)^2}>0\) for every \(x\ne1\). Thus, \(f\) is strictly increasing on both \((-\infty,1)\) and \((1,\infty)\). 2. Since \(\frac{2}{x-1}\to0\) as \(x\to\pm\infty\), both limits equal \(3\). 3. The claim is false. A function can be strictly increasing on each branch while approaching a finite horizontal asymptote. Here, the horizontal asymptote is \(y=3\).

1) Strictly increasing on \((-\infty,1)\) and \((1,\infty)\) 2) \(\lim_{x\to\infty}f(x)=3\) and \(\lim_{x\to-\infty}f(x)=3\) 3) False; the graph approaches the horizontal asymptote \(y=3\).

- Factor the denominator and check the numerator at the excluded values. - Compare the degrees for end behavior. - Use the quotient rule and determine the derivative's sign. - A tangent line needs the point and the derivative value there.

1. Since \(x^2-9=(x-3)(x+3)\) and the numerator is nonzero at \(x=\pm 3\), the vertical asymptotes are \(x=-3\) and \(x=3\). 2. The denominator has greater degree than the numerator, so \(\lim_{x\to\pm\infty}f(x)=0\). Thus, the horizontal asymptote is \(y=0\). 3. By the quotient rule, \(f'(x)=\frac{5(x^2-9)-5x(2x)}{(x^2-9)^2}=-\frac{5(x^2+9)}{(x^2-9)^2}\). 4. For every domain value, \(x^2+9>0\) and \((x^2-9)^2>0\), so \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \((-\infty, -3)\), \((-3, 3)\), and \((3, \infty)\). 5. Since \(f(0)=0\) and \(f'(0)=-\frac{5}{9}\), the tangent line through the origin is \(y=-\frac{5}{9}x\).

a) Vertical asymptotes: \(x=-3\) and \(x=3\); horizontal asymptote: \(y=0\) b) \(f'(x)=-\frac{5(x^2+9)}{(x^2-9)^2}\); \(f\) is strictly decreasing on \((-\infty, -3)\), \((-3, 3)\), and \((3, \infty)\). c) \(y=-\frac{5}{9}x\)

- Use the quotient rule. - The denominator of the derivative is always positive. - Use sign changes in the numerator to determine monotonicity and extrema.

1. Differentiate using the quotient rule: \(g'(x)=\frac{3(x^2+9)-6x^2}{(x^2+9)^2}=\frac{27-3x^2}{(x^2+9)^2}\). 2. The derivative is zero at \(x=-3\) and \(x=3\). Its denominator is always positive, so its sign is determined by \(27-3x^2\). 3. The derivative is negative for \(x<-3\), positive for \(-3<x<3\), and negative for \(x>3\). Therefore, \(g\) is strictly decreasing on \((-\infty,-3]\) and \([3,\infty)\), and strictly increasing on \([-3,3]\). 4. At \(x=-3\), the derivative changes from negative to positive, so \(g\) has a local minimum at \((-3,-0.5)\). 5. At \(x=3\), the derivative changes from positive to negative, so \(g\) has a local maximum at \((3,0.5)\).

Strictly decreasing on \((-\infty,-3]\) and \([3,\infty)\); strictly increasing on \([-3,3]\) Local minimum: \((-3,-0.5)\) Local maximum: \((3,0.5)\)

- Simplify the rational expression before classifying the vertical asymptote. - How does the multiplicity of an uncanceled denominator zero affect sign changes? - What does the sign of the first derivative tell you about increasing and decreasing intervals? - How does differentiation change the denominator exponent near a vertical asymptote?

1. Simplify the function for \(x\neq -3\): \(f(x)=\frac{(x-3)(x+3)}{(x+3)^2}=\frac{x-3}{x+3}\). 2. One factor of \(x+3\) remains in the denominator, so \(f\) has a vertical asymptote of order \(1\) at \(x=-3\). Its graph changes sign across the asymptote. Lara’s claim is false because she did not simplify first. 3. Rewrite \(f(x)=1-\frac{6}{x+3}\). Then \(f'(x)=\frac{6}{(x+3)^2}\). 4. The derivative has a vertical asymptote of order \(2\) at \(x=-3\), and \(f'(x)>0\) for every \(x\) in the domain. 5. Therefore, \(f\) is strictly increasing on each interval \((-\infty, -3)\) and \((-3, \infty)\). Tom’s conclusion is correct, but his stated reason is incomplete: an even-order vertical asymptote of \(f'\) shows that \(f'\) has the same sign on both sides, not that the sign is positive. The explicit derivative \(f'(x)=\frac{6}{(x+3)^2}>0\) establishes that \(f\) is increasing on both intervals.

Lara’s claim is false. After simplification, \(f(x)=\frac{x-3}{x+3}\), so the vertical asymptote at \(x=-3\) has order \(1\) and the graph changes sign across it. Tom’s conclusion is correct, but his reasoning is incomplete. The order \(2\) of the vertical asymptote of \(f'\) only shows that \(f'\) has the same sign on both sides. Since \(f'(x)=\frac{6}{(x+3)^2}>0\) on the domain, \(f\) is increasing on both \((-\infty, -3)\) and \((-3, \infty)\).

- Find where the first derivative is positive and where it is negative. - Split the real line at the zeros of the derivative. - A function is invertible on any interval where it is strictly monotonic. - Make each interval as large as possible without crossing a change in monotonicity.

1. Differentiate: \(f'(x)=x^2+x-6=(x+3)(x-2)\). 2. The critical numbers are \(x=-3\) and \(x=2\). 3. A sign chart shows that \(f'(x)>0\) on \((-\infty,-3)\), \(f'(x)<0\) on \((-3,2)\), and \(f'(x)>0\) on \((2,\infty)\). 4. Therefore, \(f\) is strictly increasing on \((-\infty,-3]\), strictly decreasing on \([-3,2]\), and strictly increasing on \([2,\infty)\). 5. These are the maximal intervals on which \(f\) is one-to-one, so each corresponding restriction has an inverse function.

\((-\infty,-3]\), \([-3,2]\), and \([2,\infty)\)

- Use the product rule to find the derivative. - Determine where the derivative changes sign. - A strictly monotonic restriction is one-to-one. - An interval containing both sides of a local extremum is not one-to-one.

1. By the product rule, \(f'(x)=e^{-x}-(x+3)e^{-x}=(-x-2)e^{-x}\). 2. Since \(e^{-x}>0\), the derivative is positive for \(x<-2\), zero at \(x=-2\), and negative for \(x>-2\). 3. Thus, \(f\) is strictly increasing on \((-\infty,-2]\) and strictly decreasing on \([-2,\infty)\). For example, \([-2,\infty)\) is an interval on which \(f\) has an inverse function. 4. The interval \([-3,0]\) contains the local maximum at \(x=-2\) in its interior. The function increases and then decreases on this interval, so it is not one-to-one there and does not have an inverse function on that interval.

1) One possible interval is \([-2,\infty)\). 2) One possible interval is \([-3,0]\).

- Factor the derivative and use the fact that \(e^x>0\). - Identify the intervals on which the function is strictly monotonic. - For an interval of the form \([a,\infty)\), determine how far left the increasing portion can extend. - A local extremum inside an interval prevents the function from being one-to-one there.

1. By the product rule, \(g'(x)=2xe^x+x^2e^x=x(x+2)e^x\). 2. Because \(e^x>0\), the derivative is positive on \((-\infty,-2)\), negative on \((-2,0)\), and positive on \((0,\infty)\). 3. On \([0,\infty)\), \(g\) is strictly increasing. Any interval \([a,\infty)\) with \(a<0\) contains points from a decreasing portion and the increasing portion after \(0\), so it is not one-to-one. Therefore, the largest interval of the required form is \([0,\infty)\). 4. The interval \([-3,-1]\) contains \(x=-2\), where the derivative changes from positive to negative. Thus, \(g\) has a local maximum there and takes some output values twice on \([-3,-1]\). It is not one-to-one on that interval.

1) \([0,\infty)\) 2) The derivative changes from positive to negative at \(x=-2\), so \(g\) is not one-to-one on \([-3,-1]\).

- Find the zeros of the first derivative. - Use a sign chart to identify intervals of increase and decrease. - A function has an inverse on any interval where it is strictly monotonic. - Choose the largest monotonic interval containing \(x=1\).

1. Differentiate: \(f'(x)=x^2-4=(x-2)(x+2)\). 2. The derivative is positive on \((-\infty,-2)\), negative on \((-2,2)\), and positive on \((2,\infty)\). 3. Therefore, \(f\) changes monotonicity and is not one-to-one on all of \(\mathbb{R}\), so it does not have an inverse function on its entire domain. 4. Since \(1\in[-2,2]\) and \(f\) is strictly decreasing on \([-2,2]\), the restriction to this interval is one-to-one. 5. The derivative changes sign at both endpoints, so \([-2,2]\) is the largest such interval containing \(x=1\).

\(f\) is not invertible on \(\mathbb{R}\). The largest interval containing \(x=1\) on which it is invertible is \([-2,2]\).

- Find where the first derivative is zero. - Determine the derivative's sign on each side of the critical number. - A change in monotonicity prevents global invertibility. - Locate \(x=0\) relative to the critical number.

1. Differentiate: \(f'(x)=e^x-2\). 2. The derivative is zero when \(e^x=2\), so the critical number is \(x=\ln(2)\). 3. For \(x<\ln(2)\), \(f'(x)<0\), and for \(x>\ln(2)\), \(f'(x)>0\). Thus, \(f\) decreases and then increases. 4. Therefore, \(f\) is not one-to-one on \(\mathbb{R}\) and does not have an inverse function on its entire domain. 5. Since \(0<\ln(2)\), the largest monotonic interval containing \(0\) is \((-\infty,\ln(2)]\). On this interval, \(f\) is strictly decreasing and has an inverse.

\(f\) is not invertible on \(\mathbb{R}\). The largest interval containing \(x=0\) on which it is invertible is \((-\infty,\ln(2)]\).

- Find the zeros of each first derivative. - Use derivative sign charts to identify intervals of strict monotonicity. - A restriction has an inverse wherever the function is strictly monotonic. - Include each turning point as an endpoint of the adjacent maximal intervals.

1. a) \(f'(x)=3x^2+6x-9=3(x+3)(x-1)\). The derivative is positive on \((-\infty,-3)\), negative on \((-3,1)\), and positive on \((1,\infty)\). 2. Thus, \(f\) is not one-to-one on \(\mathbb{R}\). Its maximal invertibility intervals are \((-\infty,-3]\), \([-3,1]\), and \([1,\infty)\). 3. b) \(g'(x)=2e^{2x}-4e^x=2e^x(e^x-2)\). Since \(e^x>0\), the derivative changes from negative to positive at \(x=\ln(2)\). 4. Therefore, \(g\) is not one-to-one on \(\mathbb{R}\). Its maximal invertibility intervals are \((-\infty,\ln(2)]\) and \([\ln(2),\infty)\). 5. c) \(h'(x)=\cos(x)-\frac{1}{2}\). On \([0,2\pi]\), the derivative is zero at \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\). 6. The derivative is positive on \(\left(0,\frac{\pi}{3}\right)\), negative on \(\left(\frac{\pi}{3},\frac{5\pi}{3}\right)\), and positive on \(\left(\frac{5\pi}{3},2\pi\right)\). 7. Thus, \(h\) is not one-to-one on \([0,2\pi]\). Its maximal invertibility intervals are \(\left[0,\frac{\pi}{3}\right]\), \(\left[\frac{\pi}{3},\frac{5\pi}{3}\right]\), and \(\left[\frac{5\pi}{3},2\pi\right]\).

a) Not invertible on \(\mathbb{R}\); maximal intervals: \((-\infty,-3]\), \([-3,1]\), and \([1,\infty)\) b) Not invertible on \(\mathbb{R}\); maximal intervals: \((-\infty,\ln(2)]\) and \([\ln(2),\infty)\) c) Not invertible on \([0,2\pi]\); maximal intervals: \(\left[0,\frac{\pi}{3}\right]\), \(\left[\frac{\pi}{3},\frac{5\pi}{3}\right]\), and \(\left[\frac{5\pi}{3},2\pi\right]\)

- Apply the quotient rule and simplify the numerator. - The denominator of the derivative is always positive. - Relate the derivative's sign to strict monotonicity and invertibility.

1. By the quotient rule, \(f'(x)=\frac{2(x^2+1)-4x^2}{(x^2+1)^2}=\frac{2(1-x^2)}{(x^2+1)^2}\). 2. The denominator is always positive. The derivative is positive for \(|x|<1\), zero at \(x=\pm1\), and negative for \(|x|>1\). 3. a) Since the derivative changes sign, \(f\) changes monotonicity and is not one-to-one on \(\mathbb{R}\). 4. b) The maximal interval on which \(f\) is strictly increasing is \([-1,1]\). 5. c) Because \(f\) is strictly increasing on \([-1,1]\), its restriction to this interval is one-to-one and has an inverse function.

a) \(f\) is not invertible on \(\mathbb{R}\) because its derivative changes sign at \(x=-1\) and \(x=1\). b) \(I=[-1,1]\) c) The restriction is strictly increasing, so it has an inverse function.

- Use the product rule and factor the derivative. - Determine where the function decreases and where it increases. - A maximal interval of the form \([a,\infty)\) must begin at the turning point. - Choose a noninvertible interval that contains both sides of the turning point.

1. By the product rule, \(f'(x)=e^x+(x-4)e^x=(x-3)e^x\). 2. Since \(e^x>0\), the derivative is negative for \(x<3\), zero at \(x=3\), and positive for \(x>3\). 3. Therefore, \(f\) is strictly increasing on \([3,\infty)\). This is the largest interval of the required form on which \(f\) is one-to-one. 4. For example, \([2,4]\) contains the local minimum at \(x=3\) in its interior. The function decreases and then increases there, so it is not one-to-one on that interval.

Largest interval of the required form: \([3,\infty)\) One interval on which \(f\) is not invertible: \([2,4]\)

- Factor the derivative after applying the product and chain rules. - Check whether the derivative changes sign inside \([1,3]\). - A strictly monotonic restriction is invertible. - Use the critical number as an endpoint of an invertible interval.

1. By the product and chain rules, \(f'(x)=2xe^{-x}-x^2e^{-x}=x(2-x)e^{-x}\). 2. For \(x>0\), the exponential factor and \(x\) are positive. Thus, the derivative is positive on \((0,2)\) and negative on \((2,\infty)\). 3. On \([1,3]\), the function increases up to \(x=2\) and then decreases. Therefore, it is not one-to-one and does not have an inverse function on that interval. 4. The function is strictly increasing on \([0,2]\) and strictly decreasing on \([2,\infty)\). Either interval gives an invertible restriction.

\(f\) is not invertible on \([1,3]\) because it has a local maximum at \(x=2\). It is invertible, for example, on \([0,2]\) or on \([2,\infty)\).

- Factor signs separately in the derivative. - A derivative can be zero at an isolated point without changing the function's monotonicity. - Split the domain only where the derivative changes sign. - A strictly monotonic restriction has an inverse function.

1. The derivative is zero at \(x=0\) and \(x=3\). 2. Since \((x-3)^2\ge0\), the sign of \(f'(x)\) is determined by the sign of \(x\), except at the isolated zero \(x=3\). 3. Thus, \(f'(x)<0\) for \(x<0\), and \(f'(x)>0\) for \(x>0\) except that \(f'(3)=0\). 4. Therefore, \(f\) is strictly decreasing on \((-\infty,0]\) and strictly increasing on \([0,\infty)\). The zero at \(x=3\) does not split the increasing interval because the derivative does not change sign there. 5. These are the maximal intervals on which \(f\) is one-to-one and therefore invertible.

\((-\infty,0]\) and \([0,\infty)\)

- Find the zeros of the derivative. - Use a sign chart for \(x^2-1\). - A function that changes direction on an interval is not strictly monotonic.

1. The derivative factors as \(g'(x)=(x-1)(x+1)\), so its zeros are \(x=-1\) and \(x=1\). 2. The derivative is positive on \([-2,-1)\) and \((1,2]\), and negative on \((-1,1)\). 3. a) Therefore, \(g\) is strictly increasing on \([-2,-1]\) and \([1,2]\), and strictly decreasing on \([-1,1]\). 4. b) Because the derivative changes sign twice, \(g\) changes direction from increasing to decreasing and then back to increasing. A continuous one-to-one function on an interval must be strictly monotonic, so \(g\) is not one-to-one on \([-2,2]\).

a) Increasing on \([-2,-1]\) and \([1,2]\); decreasing on \([-1,1]\) b) \(g\) is not invertible on \([-2,2]\) because it is not strictly monotonic there.

- Use strict monotonicity to test invertibility. - Analyze the zeros and sign of the derivative. - For part b, test for even symmetry. - A one-to-one function cannot assign the same output to two different inputs.

1. Differentiate: \(f_k'(x)=x^2+k\). 2. If \(k\geq0\), then \(f_k'(x)\geq0\) for every \(x\), with at most one isolated zero. Thus, \(f_k\) is strictly increasing and invertible on \(\mathbb{R}\). 3. If \(k<0\), the derivative has two real zeros and changes sign, so \(f_k\) is not monotonic on all of \(\mathbb{R}\). Therefore, part a gives \(k\geq0\). 4. For part b, \(g(-x)=a(-x)^4+b(-x)^2+c=g(x)\). 5. Thus, \(g\) is even. For every \(x\neq0\), the distinct inputs \(x\) and \(-x\) have the same output, so \(g\) is not one-to-one and cannot be invertible on \(\mathbb{R}\).

a) \(k\geq0\) b) Since \(g(-x)=g(x)\), the function is not one-to-one and is not invertible on \(\mathbb{R}\).

- Use the quotient rule and determine the derivative's sign. - Strict monotonicity guarantees an inverse function. - Interchange input and output, then solve for the new output. - The domain and range switch when a function is inverted.

1. a) By the quotient rule, \(f'(x)=\frac{2(x-4)-2x}{(x-4)^2}=-\frac{8}{(x-4)^2}\). 2. For every \(x>4\), \(f'(x)<0\). Thus, \(f\) is strictly decreasing and one-to-one on its domain. 3. As \(x\to4^+\), \(f(x)\to\infty\), and as \(x\to\infty\), \(f(x)\to2\) from above. Therefore, the range of \(f\) is \((2,\infty)\). 4. b) Start with \(y=\frac{2x}{x-4}\). Then \(y(x-4)=2x\), so \(x(y-2)=4y\) and \(x=\frac{4y}{y-2}\). 5. Therefore, \(f^{-1}(x)=\frac{4x}{x-2}\). Its domain is \((2,\infty)\), and its range is \((4,\infty)\).

a) \(f'(x)=-\frac{8}{(x-4)^2}<0\), so \(f\) is invertible. b) \(f^{-1}(x)=\frac{4x}{x-2}\), with domain \((2,\infty)\) and range \((4,\infty)\)

- Use the derivative's sign to establish strict monotonicity. - Inverse graphs are reflections across \(y=x\). - For a strictly increasing function, a common point with its inverse must be a fixed point. - Determine whether \(e^x+2=0\) can have a real solution.

1. Since \(g'(x)=e^x+1>0\) for every real \(x\), \(g\) is strictly increasing and has an inverse function. 2. For a strictly increasing function, any common point of the graphs of \(g\) and \(g^{-1}\) must lie on \(y=x\). Thus, solve \(g(x)=x\). 3. The equation \(e^x+x+2=x\) simplifies to \(e^x+2=0\). 4. Since \(e^x>0\) for every real \(x\), this equation has no real solution. Therefore, the graphs have no common point.

1) \(g'(x)=e^x+1>0\), so \(g\) is strictly increasing and invertible. 2) The equation \(g(x)=x\) becomes \(e^x=-2\), which has no real solution. Thus, the graphs have no common point.

- A square-root radicand must be nonnegative. - Use the endpoint and end behavior to determine the range. - A positive derivative establishes strict increase. - Interchange input and output, then isolate the new output. - The inverse domain is the original range.

1. a) Require \(4x-8\ge0\), so the domain is \([2,\infty)\). 2. The minimum value occurs at \(x=2\), where \(h(2)=-2\). The function is unbounded above, so its range is \([-2,\infty)\). 3. b) For \(x>2\), \(h'(x)=\frac{2}{\sqrt{4x-8}}>0\). Together with continuity at the endpoint, this shows that \(h\) is strictly increasing on \([2,\infty)\) and therefore invertible. 4. c) Start with \(y=\sqrt{4x-8}-2\). Then \(y+2=\sqrt{4x-8}\), so \((y+2)^2=4x-8\). 5. Solving gives \(x=\frac{1}{4}(y+2)^2+2\). Therefore, \(h^{-1}(x)=\frac{1}{4}(x+2)^2+2\), with domain \([-2,\infty)\).

a) Domain: \([2,\infty)\); range: \([-2,\infty)\) b) \(h'(x)>0\) for \(x>2\), so \(h\) is strictly increasing and invertible. c) \(h^{-1}(x)=\frac{1}{4}(x+2)^2+2\), with domain \([-2,\infty)\)

- Account for the negative sign in front of the square root. - A negative derivative establishes strict decrease. - When isolating the square root, note the restriction \(3-y\ge0\). - The domain and range switch for an inverse function.

1. a) Require \(x+5\ge0\), so the domain is \([-5,\infty)\). 2. At \(x=-5\), \(k(-5)=3\). As \(x\to\infty\), \(k(x)\to-\infty\), so the range is \((-\infty,3]\). 3. b) For \(x>-5\), \(k'(x)=-\frac{1}{2\sqrt{x+5}}<0\). Together with continuity at the endpoint, this shows that \(k\) is strictly decreasing on \([-5,\infty)\), so it has an inverse. 4. c) Start with \(y=3-\sqrt{x+5}\). Then \(\sqrt{x+5}=3-y\), and \(x+5=(3-y)^2\). 5. Therefore, \(x=(3-y)^2-5\), so \(k^{-1}(x)=(3-x)^2-5\). 6. The domain of \(k^{-1}\) is \((-\infty,3]\), and its range is \([-5,\infty)\).

a) Domain: \([-5,\infty)\); range: \((-\infty,3]\) b) \(k'(x)<0\) for \(x>-5\), so \(k\) is strictly decreasing and invertible. c) \(k^{-1}(x)=(3-x)^2-5\), with domain \((-\infty,3]\) and range \([-5,\infty)\)

- Transplantation corresponds to \(t=0\). - Use the derivative's sign to establish strict monotonicity. - Isolate the square root before squaring. - An inverse function interchanges the input and output quantities.

1. a) \(f(0)=5\sqrt{1}=5\), so the tree is \(5\,\text{ft}\) tall when transplanted. 2. Solve \(25=5\sqrt{0.2t+1}\). Then \(5=\sqrt{0.2t+1}\), so \(25=0.2t+1\) and \(t=120\). The tree reaches \(25\,\text{ft}\) after \(120\) years. 3. b) \(f'(t)=\frac{0.5}{\sqrt{0.2t+1}}>0\) on \([0,120]\). Therefore, \(f\) is strictly increasing and invertible. 4. From \(y=5\sqrt{0.2t+1}\), obtain \(\frac{y^2}{25}=0.2t+1\). Thus, \(t=0.2y^2-5\). 5. Therefore, \(f^{-1}(x)=0.2x^2-5\). Since \(f(0)=5\) and \(f(120)=25\), its domain is \([5,25]\). 6. c) The inverse takes a tree height in feet and returns the number of years after transplantation when the tree reaches that height.

a) Initial height: \(5\,\text{ft}\); time to reach \(25\,\text{ft}\): \(120\) years b) \(f^{-1}(x)=0.2x^2-5\), with domain \([5,25]\) c) \(f^{-1}\) gives the number of years after transplantation when the tree reaches a specified height.

- Substitute \(x=0\) for the no-fertilizer case. - Use the derivative's sign to establish strict increase. - Solve the model equation for the fertilizer variable. - Track the units when the input and output switch.

1. a) \(g(0)=0\), so the modeled yield increase is \(0\) bushels per acre with no fertilizer. 2. Solve \(10=\frac{12x}{x+4}\). Then \(10x+40=12x\), so \(x=20\). This represents \(20\cdot10=200\,\text{lb/acre}\) of fertilizer. 3. b) \(g'(x)=\frac{48}{(x+4)^2}>0\) on \([0,20]\). Thus, \(g\) is strictly increasing and invertible. 4. From \(y=\frac{12x}{x+4}\), obtain \(yx+4y=12x\), so \(x=\frac{4y}{12-y}\). 5. Therefore, \(g^{-1}(x)=\frac{4x}{12-x}\). Since \(g(0)=0\) and \(g(20)=10\), the inverse domain is \([0,10]\). 6. c) \(g^{-1}(8)\) is the fertilizer amount, in units of \(10\,\text{lb/acre}\), needed for an \(8\)-bushel-per-acre yield increase. Since \(g^{-1}(8)=8\), this is \(80\,\text{lb/acre}\).

a) Yield increase with no fertilizer: \(0\) bushels per acre; fertilizer for a \(10\)-bushel-per-acre increase: \(200\,\text{lb/acre}\) b) \(g^{-1}(x)=\frac{4x}{12-x}\), with domain \([0,10]\) c) \(g^{-1}(8)=8\), meaning \(80\,\text{lb/acre}\) of fertilizer is needed for an \(8\)-bushel-per-acre increase.

- Require the radicand to be nonnegative. - Use the derivative's sign to determine monotonicity. - Isolate the square root before squaring. - The domain and range switch for an inverse function.

1. a) Require \(12-4x\ge0\), so \(x\le3\). The domain is \((-\infty,3]\). 2. The square root is nonnegative and can grow without bound as \(x\to-\infty\). Therefore, the range is \([-1,\infty)\). 3. b) For \(x<3\), \(f'(x)=-\frac{4}{\sqrt{12-4x}}<0\). Together with continuity at \(x=3\), this shows that \(f\) is strictly decreasing on its domain and is invertible. 4. c) Start with \(y=2\sqrt{12-4x}-1\). Then \(\frac{y+1}{2}=\sqrt{12-4x}\). 5. Squaring gives \(\frac{(y+1)^2}{4}=12-4x\), so \(x=3-\frac{1}{16}(y+1)^2\). 6. Thus, \(f^{-1}(x)=3-\frac{1}{16}(x+1)^2\). Its domain is \([-1,\infty)\), and its range is \((-\infty,3]\).

a) Domain: \((-\infty,3]\); range: \([-1,\infty)\) b) \(f'(x)<0\) for \(x<3\), so \(f\) is strictly decreasing and invertible. c) \(f^{-1}(x)=3-\frac{1}{16}(x+1)^2\), with domain \([-1,\infty)\) and range \((-\infty,3]\)

- Examine the behavior near the vertical and horizontal asymptotes. - The derivative's sign determines monotonicity. - Interchange input and output, then solve for the new output. - The domain and range switch for an inverse function.

1. a) As \(x\to-2^+\), \(g(x)\to\infty\). As \(x\to\infty\), \(g(x)\to-1\) from above. Therefore, the range is \((-1,\infty)\). 2. b) \(g'(x)=-\frac{5}{(x+2)^2}<0\) for every \(x>-2\). Thus, \(g\) is strictly decreasing and invertible. 3. c) Start with \(y=\frac{5}{x+2}-1\). Then \(y+1=\frac{5}{x+2}\), so \(x=\frac{5}{y+1}-2\). 4. Therefore, \(g^{-1}(x)=\frac{5}{x+1}-2\). Its domain is \((-1,\infty)\), and its range is \((-2,\infty)\).

a) \((-1,\infty)\) b) \(g'(x)<0\) on its domain, so \(g\) is strictly decreasing and invertible. c) \(g^{-1}(x)=\frac{5}{x+1}-2\), with domain \((-1,\infty)\) and range \((-2,\infty)\)

- Replace \(x\) with \(-x\) and compare the result with the original function. - The sign of the first derivative determines increasing and decreasing behavior. - Apply the chain rule to differentiate the square root. - A tangent line requires the graph point and local slope.

1. a) Since \(h(-x)=\sqrt{x^2+9}=h(x)\), the graph is symmetric about the y-axis. 2. Differentiate: \(h'(x)=\frac{x}{\sqrt{x^2+9}}\). 3. b) The denominator is always positive, so the derivative has the sign of \(x\). Thus, \(h\) is strictly decreasing on \((-\infty,0]\) and strictly increasing on \([0,\infty)\). 4. c) At \(x=4\), \(h(4)=5\) and \(h'(4)=\frac{4}{5}\). 5. The tangent line is \(y-5=\frac{4}{5}(x-4)\), or \(y=\frac{4}{5}x+\frac{9}{5}\).

a) Symmetric about the y-axis b) Strictly decreasing on \((-\infty,0]\); strictly increasing on \([0,\infty)\) c) \(y=\frac{4}{5}x+\frac{9}{5}\)

- Require the square-root radicand to be nonnegative. - Apply the chain rule when differentiating. - Use the point and derivative value to write the tangent line. - The sign of the derivative determines monotonicity and helps identify the range.

1. a) Require \(4x+8\ge0\), so \(D_f=[-2,\infty)\). By the chain rule, \(f'(x)=\frac{2}{\sqrt{4x+8}}\), for \(x>-2\). 2. b) At \(x_0=\frac{1}{4}\), \(f\left(\frac{1}{4}\right)=4\) and \(f'\left(\frac{1}{4}\right)=\frac{2}{3}\). 3. The tangent line is \(y-4=\frac{2}{3}\left(x-\frac{1}{4}\right)\), or \(y=\frac{2}{3}x+\frac{23}{6}\). 4. c) Since \(f'(x)>0\) for every \(x>-2\), \(f\) is strictly increasing on its domain. Its minimum value is \(f(-2)=1\), and it is unbounded above. Thus, the range is \([1,\infty)\).

a) \(D_f=[-2,\infty)\); \(f'(x)=\frac{2}{\sqrt{4x+8}}\) for \(x>-2\) b) \(y=\frac{2}{3}x+\frac{23}{6}\) c) \([1,\infty)\)

- Use the sign of the first derivative. - Find the endpoint value and the end behavior to determine the range. - Strict monotonicity guarantees that a function is one-to-one. - Solve the equation for the original input, then interchange variables.

1. For \(x>-4\), \(g'(x)=\frac{1}{4\sqrt{0.5x+2}}>0\). Since \(g\) is continuous at the endpoint and increasing on the interior, it is strictly increasing on \([-4,\infty)\). The minimum value is \(g(-4)=2\), and \(g(x)\to\infty\), so the range is \([2,\infty)\). 2. A strictly increasing function is one-to-one, so it has an inverse. Solve \(y=2+\sqrt{0.5x+2}\) for \(x\): \(y-2=\sqrt{0.5x+2}\), so \((y-2)^2=0.5x+2\), and \(x=2(y-2)^2-4\). Therefore, \(g^{-1}(x)=2(x-2)^2-4\). 3. The domain of the inverse is the range of the original function, so the domain of \(g^{-1}\) is \([2,\infty)\).

1. \(g'(x)=\frac{1}{4\sqrt{0.5x+2}}>0\) for \(x>-4\), so \(g\) is strictly increasing. Range: \([2,\infty)\). 2. \(g^{-1}(x)=2(x-2)^2-4\). 3. Domain of \(g^{-1}\): \([2,\infty)\).

- Show that the radicand is always positive. - Test for even or odd symmetry. - Use the quotient and chain rules. - Remember that \(\sqrt{x^2}=|x|\) when evaluating limits.

1. Since \(k\neq0\), \(x^2+k^2>0\) for every real \(x\). The denominator is always defined and nonzero, so the domain is \(\mathbb{R}\). 2. Because \(f_k(-x)=-f_k(x)\), each graph is symmetric about the origin. 3. Differentiate and simplify: \(f_k'(x)=\frac{k^3}{(x^2+k^2)^{3/2}}\). 4. The denominator is positive. If \(k>0\), the derivative is positive and the function is strictly increasing. If \(k<0\), the derivative is negative and the function is strictly decreasing. 5. Write \(\sqrt{x^2+k^2}=|x|\sqrt{1+k^2/x^2}\). Then \(\lim_{x\to\infty}f_k(x)=k\) and \(\lim_{x\to-\infty}f_k(x)=-k\). 6. Therefore, the horizontal asymptotes are \(y=k\) as \(x\to\infty\) and \(y=-k\) as \(x\to-\infty\).

a) Domain: \(\mathbb{R}\); symmetry about the origin. b) \(f_k'(x)=\frac{k^3}{(x^2+k^2)^{3/2}}\); increasing for \(k>0\), decreasing for \(k<0\). c) \(\lim_{x\to\infty}f_k(x)=k\), \(\lim_{x\to-\infty}f_k(x)=-k\); horizontal asymptotes \(y=k\) and \(y=-k\), respectively.

- Require the square-root radicand to be nonnegative. - Use the graph point and derivative value to write the tangent line. - The sign of the first derivative determines increasing behavior. - Analyze the signs of the numerator and denominator separately.

1. Require \(4x-8\ge0\), so \(D_h=[2,\infty)\). By the chain rule, \(h'(x)=\frac{2}{\sqrt{4x-8}}\), for \(x>2\). 2. At \(x=3\), \(h(3)=2\) and \(h'(3)=1\). Thus, \(y-2=x-3\), so the tangent line is \(y=x-1\). 3. For every \(x>2\), both the numerator and denominator of \(h'(x)\) are positive. Therefore, \(h'(x)>0\), so \(h\) is strictly increasing.

1. \(D_h=[2,\infty)\); \(h'(x)=\frac{2}{\sqrt{4x-8}}\) for \(x>2\) 2. \(y=x-1\) 3. Since \(h'(x)>0\) for \(x>2\), \(h\) is strictly increasing there.

- A positive first derivative guarantees strict increase. - Use the natural logarithm to undo the exponential. - Isolate the exponential expression before taking logarithms. - Composing a function with its inverse returns the original input.

1. a) By the chain rule, \(g'(x)=\frac{1}{2}e^{\frac{1}{2}x}>0\) for every real \(x\). Therefore, \(g\) is strictly increasing and invertible. 2. b) Start with \(y=e^{\frac{1}{2}x}+1\). Then \(y-1=e^{\frac{1}{2}x}\), so \(\ln(y-1)=\frac{1}{2}x\). 3. Thus, \(x=2\ln(y-1)\), and \(g^{-1}(x)=2\ln(x-1)\), for \(x>1\). 4. c) \(g(2)=e^1+1=e+1\). Then \(g^{-1}(e+1)=2\ln(e)=2\).

a) \(g'(x)>0\) for all real \(x\), so \(g\) is strictly increasing and invertible. b) \(g^{-1}(x)=2\ln(x-1)\) c) \(g(2)=e+1\), and \(g^{-1}(e+1)=2\)

- Use the chain rule and determine the derivative's sign. - Exponentiation reverses the natural logarithm. - The restricted domain determines the square-root sign. - The inverse domain is the original range.

1. By the chain rule, \(f'(x)=\frac{2x}{x^2+1}\). 2. For \(x>0\), the derivative is positive. Together with continuity at \(x=0\), this shows that \(f\) is strictly increasing on \([0,\infty)\) and therefore invertible. 3. Start with \(y=\ln(x^2+1)\). Exponentiating gives \(e^y=x^2+1\), so \(x^2=e^y-1\). 4. Because the original domain requires \(x\ge0\), take the nonnegative square root: \(x=\sqrt{e^y-1}\). 5. Therefore, \(f^{-1}(x)=\sqrt{e^x-1}\), with domain \([0,\infty)\).

\(f\) is strictly increasing on \([0,\infty)\), so it is invertible. \(f^{-1}(x)=\sqrt{e^x-1}\)

- Apply the chain rule to the exponential composition. - Use the derivative's sign to establish strict increase. - Take the natural logarithm after isolating the exponential. - The restricted domain determines the square-root sign.

1. By the chain rule, \(g'(x)=xe^{x^2}\). 2. For \(x>0\), \(g'(x)>0\). Together with continuity at \(x=0\), this shows that \(g\) is strictly increasing on \([0,\infty)\) and therefore invertible. 3. Start with \(y=\frac{1}{2}e^{x^2}\). Then \(2y=e^{x^2}\), so \(\ln(2y)=x^2\). 4. Because the original domain requires \(x\ge0\), \(x=\sqrt{\ln(2y)}\). 5. Therefore, \(g^{-1}(x)=\sqrt{\ln(2x)}\), with domain \(\left[\frac{1}{2},\infty\right)\).

\(g\) is strictly increasing on \([0,\infty)\), so it is invertible. \(g^{-1}(x)=\sqrt{\ln(2x)}\)

- Determine where the logarithm argument is positive. - A logarithm is zero when its argument equals \(1\). - Analyze the sign of the first derivative on each domain interval. - Examine the logarithm near the domain boundaries.

1. The logarithm requires \(x(x-4)>0\), which holds for \(x<0\) or \(x>4\). Thus the domain is \((-\infty,0)\cup(4,\infty)\). 2. A zero satisfies \(x^2-4x=1\). Solving \(x^2-4x-1=0\) gives \(x=2\pm\sqrt{5}\), and both values lie in the domain. 3. The derivative is \(g'(x)=\frac{2x-4}{x^2-4x}\). For \(x<0\), the numerator is negative and the denominator is positive, so \(g'(x)<0\). For \(x>4\), both are positive, so \(g'(x)>0\). 4. Therefore, \(g\) is strictly decreasing on \((-\infty,0)\) and strictly increasing on \((4,\infty)\). 5. As \(x\to0^-\) or \(x\to4^+\), the logarithm argument approaches \(0^+\), so \(g(x)\to-\infty\). The vertical asymptotes are \(x=0\) and \(x=4\).

a) Domain: \((-\infty,0)\cup(4,\infty)\). Zeros: \(x=2-\sqrt{5}\) and \(x=2+\sqrt{5}\). b) Strictly decreasing on \((-\infty,0)\); strictly increasing on \((4,\infty)\). c) Vertical asymptotes: \(x=0\) and \(x=4\).

- Simplify the difference before differentiating. - A positive derivative implies strict increase. - Intersections correspond to zeros of the difference function. - Differentiate the proposed antiderivative and compare.

1. The difference function is \(h(x)=e^{x-2}-\frac{1}{x-1}\). 2. Differentiate: \(h'(x)=e^{x-2}+\frac{1}{(x-1)^2}>0\) for every \(x>1\). Therefore, \(h\) is strictly increasing. 3. At \(x=2\), \(f(2)=2\) and \(g(2)=2\), so the graphs intersect there. Equivalently, \(h(2)=0\). Since a strictly increasing function can have at most one zero, this is the only intersection. 4. Differentiate \(F\): \(F'(x)=e^{x-2}+1=f(x)\). Therefore, \(F\) is an antiderivative of \(f\).

a) \(h'(x)=e^{x-2}+\frac{1}{(x-1)^2}>0\), so \(h\) is strictly increasing. b) The unique intersection has x-coordinate \(x=2\); the intersection point is \((2,2)\). c) \(F'(x)=f(x)\).

- Use the sign of the first derivative. - A strictly monotonic function is one-to-one. - The inverse swaps the original domain and range. - Points on \(y=x\) satisfy \(k^*(x)=x\).

1. Differentiate: \(k'(x)=\frac{1}{2}(\ln(x)-1)\). For \(x\ge e\), \(\ln(x)\ge1\), so \(k'(x)\ge0\), with equality only at the endpoint. Therefore, \(k\) is strictly increasing on \([e,\infty)\). 2. Because \(k^*\) is strictly increasing, it is one-to-one and invertible. Its minimum value is \(k(e)=-\frac{e}{2}\), and \(k(x)\to\infty\). Thus the range of \(k^*\) is \(\left[-\frac{e}{2},\infty\right)\). 3. Therefore, the inverse has domain \(\left[-\frac{e}{2},\infty\right)\) and range \([e,\infty)\). 4. For an intersection with \(y=x\), solve \(\frac{1}{2}x(\ln(x)-2)=x\). Since \(x\ge e>0\), divide by \(x\): \(\frac{1}{2}(\ln(x)-2)=1\), so \(\ln(x)=4\) and \(x=e^4\). The intersection is \((e^4,e^4)\).

a) \(k'(x)=\frac{1}{2}(\ln(x)-1)\ge0\) for \(x\ge e\), so \(k\) is strictly increasing. b) Domain of \((k^*)^{-1}\): \(\left[-\frac{e}{2},\infty\right)\). Range: \([e,\infty)\). c) \((e^4,e^4)\).

- The logarithm argument must be positive. - Examine the argument near domain boundaries and at infinity. - Apply the chain rule to the logarithm. - Use a sign chart for the derivative denominator.

1. The logarithm argument must be positive: \(\frac{2x-4}{x+1}>0\). A sign chart gives \(x<-1\) or \(x>2\), so the domain is \((-\infty,-1)\cup(2,\infty)\). 2. As \(x\to\pm\infty\), the rational expression approaches \(2\), so \(g(x)\to\ln(2)\). The horizontal asymptote is \(y=\ln(2)\). 3. As \(x\to-1^-\), the logarithm argument approaches \(\infty\), so \(g(x)\to\infty\). As \(x\to2^+\), the argument approaches \(0^+\), so \(g(x)\to-\infty\). Thus the vertical asymptotes are \(x=-1\) and \(x=2\). 4. Differentiate using the chain and quotient rules: \(g'(x)=\frac{6}{(2x-4)(x+1)}\). On both domain intervals, the two factors in the denominator have the same sign, so their product is positive. Therefore, \(g'(x)>0\), and \(g\) is strictly increasing on each interval of its domain.

a) \((-\infty,-1)\cup(2,\infty)\). b) Vertical asymptotes: \(x=-1\) and \(x=2\). Horizontal asymptote: \(y=\ln(2)\). c) \(g'(x)=\frac{6}{(2x-4)(x+1)}>0\) throughout the domain, so \(g\) is strictly increasing on both domain intervals.

- Interpret changes inside and outside the power separately. - Find the sign of the first derivative on each side of its zero.

1. Shift the graph of \(x^4\) left \(1\) unit, stretch it vertically by a factor of \(2\), and shift it down \(8\) units. 2. Differentiate: \(f'(x)=8(x+1)^3\). The derivative is zero at \(x=-1\), negative for \(x<-1\), and positive for \(x>-1\). 3. Therefore, \(f\) is strictly decreasing on \((-\infty,-1]\) and strictly increasing on \([-1,\infty)\).

1) Left \(1\), vertical stretch by \(2\), down \(8\) 2) Strictly decreasing on \((-\infty,-1]\); strictly increasing on \([-1,\infty)\)

- A negative outside factor reflects a graph across the x-axis. - Rewrite the derivative as a negative multiple of a square. - An isolated zero does not prevent strict decrease.

1. Shift the graph of \(x^3\) right \(2\) units, reflect it across the x-axis, compress it vertically by a factor of \(\frac{1}{4}\), and shift it up \(3\) units. 2. Differentiate: \(f'(x)=-\frac{3}{4}(x-2)^2\le0\) for every real \(x\). 3. The derivative is negative except at the isolated value \(x=2\), so \(f\) is strictly decreasing on all of \(\mathbb{R}\).

1) Right \(2\), reflect across the x-axis, vertical compression by \(\frac{1}{4}\), up \(3\) 2) Strictly decreasing on \(\mathbb{R}\)

- Find and factor the first derivative. - Use the zeros of the derivative to divide the real line into intervals. - Test the sign of the derivative on each interval.

1. Differentiate: \(g'(x)=x^3-3x^2-4x+12\). 2. Factor the derivative: \(g'(x)=(x-2)(x-3)(x+2)\). Its zeros are \(x=-2\), \(x=2\), and \(x=3\). 3. A sign chart gives \(g'(x)<0\) on \((-\infty,-2)\), \(g'(x)>0\) on \((-2,2)\), \(g'(x)<0\) on \((2,3)\), and \(g'(x)>0\) on \((3,\infty)\). 4. Therefore, \(g\) is strictly decreasing on \((-\infty,-2]\) and \([2,3]\), and strictly increasing on \([-2,2]\) and \([3,\infty)\).

Strictly increasing on \([-2,2]\) and \([3,\infty)\); strictly decreasing on \((-\infty,-2]\) and \([2,3]\)

- Find the zeros of the derivative. - Use those values to divide the real line into intervals. - Test the sign of the derivative on each interval.

1. Factor the derivative: \(f'(x)=-\frac{1}{2}x(x-2)(x+2)\). Its zeros are \(x=-2\), \(x=0\), and \(x=2\). 2. A sign chart gives \(f'(x)>0\) on \((-\infty,-2)\), \(f'(x)<0\) on \((-2,0)\), \(f'(x)>0\) on \((0,2)\), and \(f'(x)<0\) on \((2,\infty)\). 3. Therefore, \(f\) is strictly increasing on \((-\infty,-2]\) and \([0,2]\), and strictly decreasing on \([-2,0]\) and \([2,\infty)\).

Strictly increasing on \((-\infty,-2]\) and \([0,2]\); strictly decreasing on \([-2,0]\) and \([2,\infty)\)

- Factor the derivative completely. - A factor with even multiplicity does not change sign at its zero. - An isolated zero of the derivative does not necessarily interrupt strict monotonicity.

1. Factor the derivative: \(f'(x)=(x-2)(x+2)(x-1)^2\). Its zeros are \(x=-2\), \(x=1\), and \(x=2\). 2. Because \((x-1)^2\ge0\), the derivative is positive for \(x<-2\), negative for \(-2<x<2\) except at \(x=1\), and positive for \(x>2\). 3. The derivative does not change sign at the double zero \(x=1\), so \(f\) remains strictly decreasing across that point. 4. Therefore, \(f\) is strictly increasing on \((-\infty,-2]\) and \([2,\infty)\), and strictly decreasing on \([-2,2]\).

Strictly increasing on \((-\infty,-2]\) and \([2,\infty)\); strictly decreasing on \([-2,2]\)

- Find and factor the first derivative. - Identify the derivative's zeros. - Test the sign of the derivative on each interval determined by those zeros.

1. Differentiate: \(g_a'(x) = 2ax - x^3 = x(2a - x^2)\). 2. The critical numbers are \(x = -\sqrt{2a}\), \(x = 0\), and \(x = \sqrt{2a}\). 3. A sign analysis of the derivative gives \(g_a'(x) > 0\) on \((-\infty, -\sqrt{2a})\) and \((0, \sqrt{2a})\). 4. Therefore, \(g_a\) is strictly increasing on \((-\infty, -\sqrt{2a})\) and \((0, \sqrt{2a})\).

\((-\infty, -\sqrt{2a})\) and \((0, \sqrt{2a})\)

- Factor the first derivative. - Order the critical numbers separately for \(a > 0\), \(a = 0\), and \(a < 0\). - Determine the derivative's sign on each resulting interval.

1. Differentiate: \(f_a'(x) = 3x^2 - 2ax = x(3x - 2a)\). The critical numbers are \(x = 0\) and \(x = \frac{2a}{3}\). 2. If \(a = 0\), then \(f_0'(x) = 3x^2 \ge 0\), and \(f_0\) is strictly increasing on \(\mathbb{R}\). 3. If \(a > 0\), the critical numbers are ordered \(0 < \frac{2a}{3}\). The derivative is positive on \((-\infty, 0)\) and \(\left(\frac{2a}{3}, \infty\right)\), and negative on \(\left(0, \frac{2a}{3}\right)\). 4. If \(a < 0\), the critical numbers are ordered \(\frac{2a}{3} < 0\). The derivative is positive on \(\left(-\infty, \frac{2a}{3}\right)\) and \((0, \infty)\), and negative on \(\left(\frac{2a}{3}, 0\right)\).

For \(a = 0\): strictly increasing on \(\mathbb{R}\) For \(a > 0\): strictly increasing on \((-\infty, 0)\) and \(\left(\frac{2a}{3}, \infty\right)\); strictly decreasing on \(\left(0, \frac{2a}{3}\right)\) For \(a < 0\): strictly increasing on \(\left(-\infty, \frac{2a}{3}\right)\) and \((0, \infty)\); strictly decreasing on \(\left(\frac{2a}{3}, 0\right)\)

- Factor the first derivative. - Determine when \(x^2 = k\) has real solutions. - Analyze the sign of the derivative between its critical numbers.

1. Differentiate: \(g_k'(x) = x^3 - kx = x(x^2 - k)\). 2. If \(k \le 0\), then \(x^2 - k > 0\) for \(k < 0\), and \(x^2 - k = x^2\) for \(k = 0\). The derivative has the sign of \(x\), except that it is zero at \(x = 0\). Thus, the function is strictly decreasing on \((-\infty, 0)\) and strictly increasing on \((0, \infty)\). 3. If \(k > 0\), the critical numbers are \(-\sqrt{k}\), \(0\), and \(\sqrt{k}\). The derivative is negative on \((-\infty, -\sqrt{k})\), positive on \(( -\sqrt{k}, 0)\), negative on \((0, \sqrt{k})\), and positive on \((\sqrt{k}, \infty)\). 4. These signs give the required intervals of increase and decrease.

For \(k \le 0\): strictly decreasing on \((-\infty, 0)\); strictly increasing on \((0, \infty)\) For \(k > 0\): strictly decreasing on \((-\infty, -\sqrt{k})\) and \((0, \sqrt{k})\); strictly increasing on \(( -\sqrt{k}, 0)\) and \((\sqrt{k}, \infty)\)

- Differentiate first. - Use the fact that even powers are nonnegative. - Connect the sign of the derivative to monotonicity.

1. Differentiate: \(g'(x)=x^4+2x^2+3\). 2. Since \(x^4\ge0\) and \(2x^2\ge0\), \(g'(x)\ge3>0\) for every real \(x\). Therefore, \(g'\) has no real zeros. 3. Because \(g'(x)>0\) everywhere, \(g\) is strictly increasing on \(\mathbb{R}\). 4. A differentiable function can have a local extremum at an interior point only if its derivative is zero there. Since \(g'\) is never zero, \(g\) has no local extrema.

a) \(g'(x)=x^4+2x^2+3\ge3\), so \(g'\) has no real zeros. b) \(g\) is strictly increasing on \(\mathbb{R}\) and has no local extrema.

- A function increases where its first derivative is positive. - First find where the derivative equals zero. - Test the intervals between the critical values. - A local maximum occurs where the derivative changes from positive to negative.

1. Differentiate: \(P'(x)=-3x^2+18x-15=-3(x-1)(x-5)\). 2. The critical values are \(x=1\) and \(x=5\). 3. The derivative is positive on \((1,5)\), so profit is increasing on that interval. 4. The second derivative is \(P''(x)=-6x+18\). Since \(P''(5)=-12<0\), \(x=5\) is a local maximum. 5. The local maximum value is \(P(5)=5\), corresponding to \(\$5000\).

a) Profit is increasing for \(1<x<5\). b) A local maximum occurs at \(x=5\) batches, with profit \(P(5)=5\), or \(\$5000\).

- Factor the first derivative completely. - A zero of even multiplicity does not force the derivative to change sign. - Use the sign of the first derivative to determine both extrema and monotonicity. - Not every horizontal tangent is a local extremum.

1. Differentiate: \(f'(x) = 2x^3 - 12x^2 + 18x = 2x(x - 3)^2\). 2. The critical numbers are \(x = 0\) and \(x = 3\). 3. Because \((x - 3)^2 \ge 0\), the sign of \(f'(x)\) is determined by \(x\). Thus, \(f'(x) < 0\) for \(x < 0\) and \(f'(x) > 0\) for \(x > 0\), except that \(f'(3) = 0\). 4. The derivative changes from negative to positive at \(x = 0\), so \((0, 0)\) is a local minimum. The derivative does not change sign at \(x = 3\), so \((3, 13.5)\) is a stationary inflection point, not an extremum. 5. Since \(f'(x) < 0\) for every \(x < 0\), the largest interval on which \(f\) is strictly decreasing is \((-\infty, 0)\).

a) The only local extremum is the local minimum \((0, 0)\). The point \((3, 13.5)\) is a stationary inflection point. b) \((-\infty, 0)\)

- A quotient is zero when its numerator is zero and its denominator is not. - Vertical asymptotes can occur where the denominator approaches zero. - Use the quotient rule and the identity \(\sin^2x+\cos^2x=1\). - Connect the sign of the first derivative to increasing or decreasing behavior.

1. A quotient is zero when its numerator is zero and its denominator is nonzero. On \((0, \pi)\), \(\cos x=0\) at \(x=\frac{\pi}{2}\), so this is the only zero. The denominator approaches \(0\) at the endpoints \(x=0\) and \(x=\pi\). Since the numerator is nonzero there, the graph has vertical asymptotes \(x=0\) and \(x=\pi\). 2. By the quotient rule, \(f'(x)=\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{\sin^2x}\) \(=-\frac{\sin^2x+\cos^2x}{\sin^2x}=-\frac{1}{\sin^2x}\). 3. For every \(x\in(0, \pi)\), \(\sin^2x>0\). Therefore, \(f'(x)=-\frac{1}{\sin^2x}<0\). Since the derivative is negative throughout the interval, \(f\) is strictly decreasing on \((0, \pi)\).

1. Zero: \(x=\frac{\pi}{2}\); vertical asymptotes: \(x=0\) and \(x=\pi\). 2. \(f'(x)=-\frac{1}{\sin^2x}\) 3. Because \(f'(x)<0\) for all \(x\in(0, \pi)\), \(f\) is strictly decreasing.

- Require the logarithm’s argument to be positive. - Factor the quadratic and use a sign chart. - Apply the chain rule. - Use the sign of the first derivative to determine monotonicity.

1. Factor the logarithm’s argument: \(x^2+2x-3=(x+3)(x-1)\). It is positive when \(x<-3\) or \(x>1\). Thus, \(D_f=(-\infty, -3)\cup(1, \infty)\). 2. Differentiate: \(f'(x)=\frac{2x+2}{x^2+2x-3}\). The denominator is positive throughout the domain. On \((-\infty, -3)\), the numerator is negative, so \(f\) is strictly decreasing. On \((1, \infty)\), the numerator is positive, so \(f\) is strictly increasing.

a) \(D_f=(-\infty, -3)\cup(1, \infty)\) b) Strictly decreasing on \((-\infty, -3)\); strictly increasing on \((1, \infty)\)

- Find where the quadratic inside the logarithm is positive. - Apply the chain rule. - The denominator of the derivative is positive on the domain. - Determine where the numerator changes sign.

1. Require \(2x-x^2=x(2-x)>0\). This holds for \(0<x<2\), so \(D_g=(0, 2)\). 2. Differentiate: \(g'(x)=\frac{2-2x}{2x-x^2}\). The denominator is positive on the domain. The numerator is positive for \(0<x<1\), zero at \(x=1\), and negative for \(1<x<2\). Therefore, \(g\) is strictly increasing on \((0, 1]\) and strictly decreasing on \([1, 2)\).

a) \(D_g=(0, 2)\) b) \(g'(x)=\frac{2-2x}{2x-x^2}\); increasing on \((0, 1]\) and decreasing on \([1, 2)\)

- Read the graph from left to right and locate its turning points. - Differentiate and factor the derivative. - Determine the sign of the derivative between its zeros.

1. From the graph, \(f\) decreases from \(x=-3\) to \(x=-1\), increases from \(x=-1\) to \(x=2\), and decreases from \(x=2\) to \(x=4\). 2. Differentiate: \(f'(x)=-x^2+x+2=-(x-2)(x+1)\). 3. The derivative is positive for \(-1<x<2\) and equals zero at \(x=-1\) and \(x=2\). 4. Therefore, \(f\) is strictly increasing on \([-1,2]\), confirming the graph-based conclusion.

a) Strictly increasing on \([-1,2]\); strictly decreasing on \([-3,-1]\) and \([2,4]\) b) \(f'(x)=-(x-2)(x+1)>0\) for \(-1<x<2\), so \(f\) is strictly increasing on \([-1,2]\).

- The sign of \(f'\) determines whether \(f\) increases or decreases. - A sign change in \(f'\) classifies a local extremum. - Local extrema of \(f'\) correspond to zeros and sign changes of \(f''\). - Count the turning points of the derivative graph.

1. The derivative is positive on \((-4,-1)\) and \((3,4)\), so \(f\) is increasing on \([-4,-1]\) and \([3,4]\). It is negative on \((-5,-4)\) and \((-1,3)\), so \(f\) is decreasing on \([-5,-4]\) and \([-1,3]\). 2. At \(x=-4\), \(f'\) changes from negative to positive, giving a local minimum. At \(x=-1\), it changes from positive to negative, giving a local maximum. At \(x=3\), it changes from negative to positive, giving a local minimum. 3. Inflection points of \(f\) occur at local extrema of \(f'\). The derivative graph has two local extrema, so \(f\) has two inflection points.

a) Increasing on \([-4,-1]\) and \([3,4]\); decreasing on \([-5,-4]\) and \([-1,3]\) b) Local minima at \(x=-4\) and \(x=3\); local maximum at \(x=-1\) c) Two inflection points

- The original function increases where its derivative is positive. - It decreases where its derivative is negative. - An isolated zero of the derivative does not necessarily interrupt strict increase.

1. In panel a), \(f'(x)<0\) on \([-1,1)\), \(f'(x)>0\) on \((1,4)\), and \(f'(x)<0\) on \((4,6]\). Therefore, \(f\) is strictly decreasing on \([-1,1]\) and \([4,6]\), and strictly increasing on \([1,4]\). 2. In panel b), \(f'(x)\ge0\) throughout \([-1,5]\), and the derivative is zero only at \(x=2\). Because the derivative is positive except at this isolated point, \(f\) is strictly increasing on \([-1,5]\). There is no strictly decreasing interval in the displayed domain.

a) Strictly increasing on \([1,4]\); strictly decreasing on \([-1,1]\) and \([4,6]\) b) Strictly increasing on \([-1,5]\); no strictly decreasing interval

- Differentiate and factor the derivative. - Use each zero of the derivative as a boundary in a sign chart. - Compare the size of the small decrease with the graph's scale.

1. Differentiate: \(f'(x)=-x^3+4.2x^2-0.8x=-x(x-0.2)(x-4)\). 2. The zeros of the derivative are \(x=0\), \(x=0.2\), and \(x=4\). 3. A sign chart gives \(f'(x)>0\) on \((-\infty,0)\), \(f'(x)<0\) on \((0,0.2)\), \(f'(x)>0\) on \((0.2,4)\), and \(f'(x)<0\) on \((4,\infty)\). 4. Therefore, \(f\) is strictly increasing on \((-\infty,0]\) and \([0.2,4]\), and strictly decreasing on \([0,0.2]\) and \([4,\infty)\). 5. The visual conjecture is false because \(f\) decreases on the very short interval \([0,0.2]\). The change is only \(f(0.2)-f(0)=-0.0052\), which is too small to see at the displayed scale.

a) The decrease occurs over only \(0.2\) unit in \(x\) and \(0.0052\) unit in \(y\), so it is not visually resolved at this scale. b) \(f'(x)=-x(x-0.2)(x-4)\). The function is strictly increasing on \((-\infty,0]\) and \([0.2,4]\), and strictly decreasing on \([0,0.2]\) and \([4,\infty)\). The conjecture is false.

- Use the sign of \(f'\) to determine where \(f\) increases or decreases. - Look for zeros of \(f'\) where its sign changes. - A change from negative to positive indicates a local minimum; a change from positive to negative indicates a local maximum. - An extremum of \(f'\) can indicate an inflection point of \(f\).

1. The graph of \(f'\) is above the x-axis for \(-3<x<1\) and below the x-axis for \(-5<x<-3\) and \(1<x<3\). 2. Therefore, \(f\) is strictly increasing on \([-3,1]\) and strictly decreasing on \([-5,-3]\) and \([1,3]\). 3. At \(x=-3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 4. The derivative \(f'\) has a local maximum at \(x=-1\). It changes from increasing to decreasing there, so \(f''\) changes sign. Thus, \(f\) has an inflection point at \(x=-1\).

a) Strictly increasing on \([-3,1]\); strictly decreasing on \([-5,-3]\) and \([1,3]\) b) Local minimum at \(x=-3\); local maximum at \(x=1\) c) Inflection point at \(x=-1\)

- Differentiate the given function and find its critical numbers. - Use a sign chart to determine its monotonic intervals. - Compare those boundaries with the turning points shown in the graph.

1. Differentiate and factor: \(f'(x)=-x^2+2x+3=-(x-3)(x+1)\). 2. The critical numbers are \(x=-1\) and \(x=3\). The derivative is negative on \((-\infty,-1)\), positive on \((-1,3)\), and negative on \((3,\infty)\). 3. Thus, the given function is strictly decreasing on \((-\infty,-1]\), strictly increasing on \([-1,3]\), and strictly decreasing on \([3,\infty)\). 4. The displayed graph instead changes from decreasing to increasing at \(x=-2\) and from increasing to decreasing at \(x=2\). Its monotonic intervals do not match those of \(f\), so the student's claim is false.

The claim is false. The given function is strictly increasing on \([-1,3]\) and strictly decreasing on \((-\infty,-1]\) and \([3,\infty)\), while the displayed graph turns at \(x=-2\) and \(x=2\).

- Use the local maximum and local minimum to divide the displayed domain. - Distinguish a function value from the slope of the graph.

1. The graph has a local maximum at \((-2,5)\) and a local minimum at \((2,-3)\). 2. On the displayed domain, the graph rises on \([-4,-2]\), falls on \([-2,2]\), and rises on \([2,4]\). 3. The student's claim is false. A positive function value describes the graph's height, not whether it is rising or falling. Near \(x=0\), the graph slopes downward, so \(f\) is strictly decreasing there.

a) Strictly increasing on \([-4,-2]\) and \([2,4]\); strictly decreasing on \([-2,2]\) b) The claim is false. The value \(f(0)=1\) gives the graph's height, while monotonicity depends on its slope. The graph is decreasing near \(x=0\).

- Factor the derivative as a perfect square. - Use the order definition: \(x_1<x_2\) must imply \(f(x_1)<f(x_2)\). - An isolated horizontal tangent is different from a horizontal segment.

1. Differentiate and factor: \(f'(x)=3x^2-6x+3=3(x-1)^2\). Thus, \(f'(x)\ge0\) for every real \(x\), and \(f'(x)=0\) only at \(x=1\). 2. Rewrite the function as \(f(x)=(x-1)^3+2\). If \(x_1<x_2\), then \(x_1-1<x_2-1\), and cubing preserves the inequality. Therefore, \((x_1-1)^3+2<(x_2-1)^3+2\), so \(f(x_1)<f(x_2)\). 3. Tim's statement is false. The isolated horizontal tangent at \(x=1\) does not create a constant interval, and \(f\) is strictly increasing on \(\mathbb{R}\).

a) \(f'(x)=3(x-1)^2\ge0\), with slope zero only at \(x=1\). b) Tim's statement is false. For every \(x_1<x_2\), \(f(x_1)<f(x_2)\), so \(f\) is strictly increasing on \(\mathbb{R}\).

- Use the sign of \(f'\) to determine where \(f\) increases or decreases. - A sign change in \(f'\) identifies and classifies a local extremum of \(f\). - A local extremum of \(f'\) can indicate an inflection point of \(f\). - Check the entire displayed interval when counting inflection points.

1. Statement (1) is false. At \(x=-2\), \(f'\) changes from positive to negative, so \(f\) has a local maximum, not a local minimum. 2. Statement (2) is true. The graph of \(f'\) is above the x-axis for \(1<x<3\), so \(f\) is strictly increasing on \([1,3]\). 3. Statement (3) is true. At \(x=1\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 4. Statement (4) is true. The graph of \(f'\) has two local extrema, near \(x=-0.8\) and \(x=2.1\). At each one, \(f'\) changes from increasing to decreasing or from decreasing to increasing, so \(f''\) changes sign. Therefore, \(f\) has exactly two inflection points on the interval.

(1) False (2) True (3) True (4) True

- Connect the sign of a derivative to whether the original function is increasing or decreasing. - Look for where each derivative graph is above or below the x-axis. - A strictly monotonic function is one-to-one. - For part b, locate the derivative zeros surrounding \(x=0\).

1. a) The graph of \(f'\) lies entirely above the x-axis. In fact, \(f'(x)\ge0.5>0\) for every real \(x\). 2. Therefore, \(f\) is strictly increasing on \(\mathbb{R}\), so it is one-to-one and invertible on \(\mathbb{R}\). 3. b) The graph of \(g'\) is on or above the x-axis for \(-2\le x\le2\), and it is below the x-axis outside this interval. 4. Thus, \(g\) is strictly increasing on \([-2,2]\). Its monotonicity changes at \(x=-2\) and \(x=2\), so this is the largest interval containing \(0\) on which the restriction of \(g\) is invertible.

a) Since \(f'(x)>0\) for all real \(x\), \(f\) is strictly increasing and therefore invertible on \(\mathbb{R}\). b) \([-2,2]\), because \(g'(x)\ge0\) on this interval and changes sign at its endpoints.

- Local extrema of a differentiable function have horizontal tangents. - Where \(g\) decreases, \(g^{\prime}\) is negative. - The derivative of a cubic polynomial is a quadratic polynomial.

1. (A) is true. Function \(g\) has a local maximum at \(x=-1\) and a local minimum at \(x=1\), so the tangent is horizontal at both points. 2. (B) is true. Function \(g\) decreases for \(-1<x<1\), so \(g^{\prime}(x)<0\) there. 3. (C) is false. At \(x=0\), the graph is decreasing with slope about \(-1.5\). 4. (D) is true. The derivative of a cubic polynomial is quadratic, and here its leading coefficient is positive.

(A), (B), and (D) are true. (C) is false.

- Locate the peak and valley in the temperature graph. - Each unit on the horizontal axis represents one hour.

1. The graph has a local maximum at \(t=-2\) and a local minimum at \(t=3\). 2. It increases on \([-4,-2]\), decreases on \([-2,3]\), and increases on \([3,5]\). 3. Since each increase of \(1\) in \(t\) represents one hour, these intervals correspond to increasing from 8:00 a.m. to 10:00 a.m., decreasing from 10:00 a.m. to 3:00 p.m., and increasing from 3:00 p.m. to 5:00 p.m.

Increasing on \([-4,-2]\) from 8:00 a.m. to 10:00 a.m., and on \([3,5]\) from 3:00 p.m. to 5:00 p.m.; decreasing on \([-2,3]\) from 10:00 a.m. to 3:00 p.m.

- Use the sign of \(f'\) to determine increasing and decreasing behavior. - At a zero of \(f'\), check whether the sign changes. - Local extrema of \(f'\) correspond to possible inflection points of \(f\). - The value of \(f''(1)\) is the slope of the graph of \(f'\) at \(x=1\).

1. Statement a) is true. At \(x=-3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 2. Statement b) is false. For \(1<x<4\), the graph of \(f'\) is below the x-axis, so \(f\) is strictly decreasing there. 3. Statement c) is true. The graph of \(f'\) has two local extrema, near \(x=-1.4\) and \(x=2.7\). At each one, the slope of \(f'\), which is \(f''\), changes sign. Therefore, \(f\) has exactly two inflection points over the displayed domain. 4. Statement d) is true. The graph of \(f'\) is decreasing at \(x=1\), so its slope satisfies \(f''(1)<0\).

a) True b) False c) True d) True

- A positive derivative gives strict increase, and a negative derivative gives strict decrease. - A derivative equal to zero throughout an interval means the original function is constant there. - Distinguish nonincreasing from strictly decreasing.

1. For \(h\), the derivative is negative on \([-1,1)\), positive on \((1,3)\), and negative on \((3,5]\). Therefore, \(h\) is strictly decreasing on \([-1,1]\) and \([3,5]\), and strictly increasing on \([1,3]\). 2. For \(k\), the derivative is nonpositive on \([-2,4]\), so \(k\) is nonincreasing on that entire interval. The derivative is negative on \((-2,0)\) and \((2,4)\), so \(k\) is strictly decreasing on \([-2,0]\) and \([2,4]\). Since \(k'(x)=0\) for \(0\le x\le2\), \(k\) is constant on \([0,2]\).

\(h\): strictly increasing on \([1,3]\); strictly decreasing on \([-1,1]\) and \([3,5]\) \(k\): nonincreasing on \([-2,4]\); strictly decreasing on \([-2,0]\) and \([2,4]\); constant on \([0,2]\)

- Determine the sign of the first derivative on the intervals separated by \(-1\) and \(1\). - Compare the resulting pattern with the graph from left to right.

1. Differentiate: \(f'(x)=-3x^2+3=-3(x-1)(x+1)\). 2. The derivative is negative for \(x<-1\), positive for \(-1<x<1\), and negative for \(x>1\). 3. Therefore, \(f\) is strictly decreasing on \((-\infty,-1]\), strictly increasing on \([-1,1]\), and strictly decreasing on \([1,\infty)\). 4. The displayed graph has the opposite pattern: it increases, then decreases, then increases. Therefore, it does not represent \(f\).

No. The given function decreases on \((-\infty,-1]\), increases on \([-1,1]\), and decreases on \([1,\infty)\), while the displayed graph has the opposite monotonic behavior.

- Factor the first derivative and count its real zeros. - Compare the number of derivative sign changes with the number of turning points in the graph.

1. Differentiate and factor: \(f'(x)=4x^3+4x=4x(x^2+1)\). 2. Since \(x^2+1>0\) for every real \(x\), the only critical number is \(x=0\). 3. The derivative is negative for \(x<0\) and positive for \(x>0\). Therefore, \(f\) is strictly decreasing on \((-\infty,0]\) and strictly increasing on \([0,\infty)\), with one local minimum at \(x=0\). 4. The displayed graph has three local extrema and four monotonic intervals. Therefore, it cannot represent \(f\).

The claim is false. The function \(f(x)=x^4+2x^2\) has only one local extremum and changes from decreasing to increasing at \(x=0\), while the displayed graph has three local extrema.

- Factor the first derivative completely. - Compare both the derivative sign chart and the function values at critical numbers with the graph.

1. Differentiate and factor: \(f'(x)=4x^3-12x^2+8x=4x(x-1)(x-2)\). 2. The derivative is negative on \((-\infty,0)\), positive on \((0,1)\), negative on \((1,2)\), and positive on \((2,\infty)\). 3. Thus, \(f\) decreases on \((-\infty,0]\), increases on \([0,1]\), decreases on \([1,2]\), and increases on \([2,\infty)\). 4. The critical-point values are \(f(0)=-1\), \(f(1)=0\), and \(f(2)=-1\). Both the monotonic pattern and these points match the displayed graph, so the graph represents \(f\).

Yes. The derivative \(f'(x)=4x(x-1)(x-2)\) gives the pattern decreasing, increasing, decreasing, increasing, with turning points \((0,-1)\), \((1,0)\), and \((2,-1)\), matching the graph.

- Factor the first derivative. - Compare the sequence of increasing and decreasing intervals with the graph.

1. Differentiate and factor: \(f'(x)=x^2-x-2=(x-2)(x+1)\). 2. The derivative is positive for \(x<-1\), negative for \(-1<x<2\), and positive for \(x>2\). 3. Therefore, \(f\) is strictly increasing on \((-\infty,-1]\), strictly decreasing on \([-1,2]\), and strictly increasing on \([2,\infty)\). 4. The displayed graph decreases, then increases, then decreases, which is the opposite pattern. Therefore, it does not represent \(f\).

No. The function increases on \((-\infty,-1]\), decreases on \([-1,2]\), and increases on \([2,\infty)\), while the displayed graph shows the opposite pattern.

- Factor the first derivative completely. - A double zero does not change the derivative's sign. - To classify the horizontal-tangent point further, check whether concavity changes.

1. Differentiate and factor: \(k'(x)=0.5x^4-2x^2=0.5x^2(x-2)(x+2)\). 2. The derivative is zero at \(x=-2\), \(x=0\), and \(x=2\), with \(x=0\) a double zero. 3. The derivative is positive for \(x<-2\), negative for \(-2<x<2\) except at \(x=0\), and positive for \(x>2\). 4. Therefore, \(k\) is strictly decreasing on \([-2,2]\). At \(x=0\), the graph has a horizontal tangent, but the derivative does not change sign, so there is no local extremum. 5. Also, \(k''(x)=2x^3-4x\) changes sign at \(x=0\), so the point is a stationary inflection point.

a) \(k'(x)=0.5x^2(x-2)(x+2)\), with zeros \(-2\), \(0\), and \(2\) b) \(k\) is strictly decreasing on \([-2,2]\). At \(x=0\), it has a horizontal tangent but no local extremum; the point is a stationary inflection point.

- Apply the order definition directly to \(x^3\). - Distinguish a sufficient derivative condition from the definition of strict increase.

1. For any real numbers \(x_1<x_2\), cubing preserves order, so \(x_1^3<x_2^3\). Therefore, \(f(x)=x^3\) is strictly increasing on \(\mathbb{R}\). 2. Also, \(g'(x)=3x^2+2>0\) for every real \(x\), so \(g\) is strictly increasing on \(\mathbb{R}\). 3. Lukas confuses a sufficient derivative condition with the definition. An isolated point where the derivative is zero does not prevent strict increase. His statement is false.

Lukas is incorrect. Both \(f\) and \(g\) are strictly increasing on \(\mathbb{R}\). The horizontal tangent of \(f\) at \(x=0\) does not violate the order definition.

- Read the graph from left to right. - Apply the order definition at interval endpoints.

1. The graph decreases on \([-3,-2]\), increases on \([-2,0]\), decreases on \([0,2]\), and increases on \([2,3]\). 2. The student's claim is false. Strict monotonicity compares function values at distinct x-values. Including a turning point as an endpoint does not violate the required inequalities, even though the derivative is zero there.

1. Strictly increasing on \([-2,0]\) and \([2,3]\); strictly decreasing on \([-3,-2]\) and \([0,2]\) 2. The claim is false. Horizontal-tangent turning points may be included as endpoints of strict-monotonicity intervals.

- A zero of the derivative corresponds to a horizontal tangent. - The sign of the derivative indicates whether the original function increases or decreases. - Estimate the slope of a tangent line at \(x=0\).

1. Zeros of \(f^{\prime}\) occur where \(f\) has horizontal tangents. The graph has local extrema at \(x=-1\), \(x=1\), and \(x=3\). 2. A tangent at \((0, 1)\) has slope approximately \(1.2\), so \(f^{\prime}(0)\approx1.2\). 3. The derivative is negative where \(f\) decreases: on \((-2, -1)\) and \((1, 3)\).

1. \(x=-1, 1, 3\) 2. \(f^{\prime}(0)\approx1.2\) 3. \((-2, -1)\) and \((1, 3)\)

- Use the asymptotes to determine \(b\) and \(c\). - Substitute a marked point to find \(a\). - Differentiate the function and analyze the sign of the derivative. - Treat the two domain intervals separately.

1. The vertical asymptote \(x=2\) gives \(b=2\), and the horizontal asymptote \(y=1\) gives \(c=1\). 2. Using the point \(P=(3, 2)\), \(2=\frac{a}{3-2}+1\), so \(a=1\). Therefore, \(f(x)=\frac{1}{x-2}+1\). 3. The domain is \((-\infty, 2)\cup(2, \infty)\). The vertical asymptote is \(x=2\), and the horizontal asymptote is \(y=1\). 4. The derivative is \(f'(x)=-\frac{1}{(x-2)^2}\), which is negative for every \(x\) in the domain. Therefore, \(f\) is decreasing on \((-\infty, 2)\) and on \((2, \infty)\), with no intervals of increase.

a) \(f(x)=\frac{1}{x-2}+1\) b) Domain: \((-\infty, 2)\cup(2, \infty)\); vertical asymptote: \(x=2\); horizontal asymptote: \(y=1\) c) \(f\) is decreasing on \((-\infty, 2)\) and \((2, \infty)\). It is not increasing on any interval.

- Use the quotient rule. - Simplify the numerator of the derivative completely. - The denominator is always positive, so focus on the sign of the numerator.

1. Apply the quotient rule: \(g'(x)=\frac{2x(x^2+3)-(x^2-9)(2x)}{(x^2+3)^2}=\frac{24x}{(x^2+3)^2}\). 2. Since \((x^2+3)^2>0\) for every real \(x\), the sign of \(g'(x)\) is determined by \(24x\). 3. Thus, \(g'(x)<0\) for \(x<0\), \(g'(0)=0\), and \(g'(x)>0\) for \(x>0\). 4. Therefore, \(g\) is strictly increasing on \([0,\infty)\).

\(g\) is strictly increasing on \([0,\infty)\).

- Identify where the function is strictly increasing or strictly decreasing. - Find the turning points by setting the derivative equal to zero. - The interval cannot extend across a point where the graph reverses direction.

1. A continuous function has an inverse on an interval when it is one-to-one there. For this graph, use intervals of strict monotonicity. 2. Differentiate: \(f^{\prime}(x)=-0.3x^2+1.5\). 3. The critical numbers satisfy \(-0.3x^2+1.5=0\), so \(x^2=5\) and \(x=\pm\sqrt{5}\). 4. The derivative is positive on \((-\sqrt{5},\sqrt{5})\), so \(f\) is strictly increasing between the local minimum and local maximum. 5. Because this interval contains \(x=1\), and extending past either critical number would cross a turning point, the largest interval is \([-\sqrt{5},\sqrt{5}]\).

\([-\sqrt{5}, \sqrt{5}]\), approximately \([-2.24, 2.24]\)

- What does a horizontal tangent tell you about the first derivative? - What does a quadratic graph look like when it has exactly one zero? - Check both the sign of the first derivative and the change in concavity at the horizontal tangent. - Rewrite the derivative in vertex form to determine when it is nonnegative for every real \(x\).

1. Differentiate: \(g_c'(x) = 3x^2 - 6x + c\). Horizontal tangents occur at the real zeros of \(g_c'\). 2. The derivative has two distinct real zeros when its discriminant is positive: \(D = (-6)^2 - 4 \cdot 3 \cdot c = 36 - 12c > 0\). Therefore, \(c < 3\). 3. Exactly one horizontal tangent occurs when \(D = 0\), so \(c = 3\). Then \(g_3'(x) = 3(x - 1)^2\), which has a double zero at \(x = 1\) and does not change sign there. Thus, the point is not a local extremum. Also, \(g_3''(x) = 6x - 6\) changes sign at \(x = 1\), so the graph changes concavity. Therefore, the point is a stationary inflection point. 4. Rewrite the derivative as \(g_c'(x) = 3(x - 1)^2 + c - 3\). If \(c > 3\), then \(g_c'(x) > 0\) for every \(x\), so \(g_c\) is strictly increasing. 5. If \(c = 3\), then \(g_3(x) = (x - 1)^3 + 1\), which is also strictly increasing. If \(c < 3\), the derivative is negative between its two real zeros, so the function is not increasing on its entire domain. Therefore, \(c \ge 3\).

a) \(c < 3\) b) When \(c = 3\), the derivative has a double zero at \(x = 1\) and does not change sign, while the second derivative changes sign there. Therefore, the graph has a stationary inflection point. c) \(c \ge 3\)

- Set the functions equal and clear the denominator. - Use the derivative of the resulting cubic to determine whether it is monotonic. - A strictly increasing cubic with opposite end behavior has exactly one real zero. - Check that the zero is not the excluded x-value.

1. Set the functions equal and clear the denominator: \(x+2=(x^2+2)(x-1)\). 2. Simplify to \(q(x)=x^3-x^2+x-4=0\). 3. Differentiate: \(q'(x)=3x^2-2x+1\). Its discriminant is \((-2)^2-4\cdot 3\cdot 1=-8<0\), and its leading coefficient is positive. Therefore, \(q'(x)>0\) for all real \(x\), so \(q\) is strictly increasing. 4. Since \(q(x)\to-\infty\) as \(x\to-\infty\) and \(q(x)\to\infty\) as \(x\to\infty\), the strictly increasing function has exactly one real zero. 5. Since \(q(1)=-3\neq 0\), that zero is not the excluded value \(x=1\). Therefore, the graphs have exactly one intersection point.

The graphs have exactly \(1\) intersection point.

- Differentiate with respect to \(x\), treating \(a\) as a constant. - Translate “nonnegative slope” into an inequality for the derivative. - Find the minimum possible value of \(a\cos x\). - Solve the resulting absolute-value inequality.

1. Differentiate: \(g_a'(x)=1+a\cos x\). 2. The expression \(a\cos x\) ranges from \(-|a|\) to \(|a|\). Therefore, the minimum value of the derivative is \(1-|a|\). 3. For the derivative to be nonnegative for every real \(x\), \(1-|a|\ge 0\). Thus, \(|a|\le 1\), so \(a\in[-1, 1]\).

\(a\in[-1, 1]\)

- Find the critical point in terms of \(k\). - Substitute that critical point into the original function. - Use the sign of the first derivative to determine where the hill descends. - Substitute the given location into the slope function and solve for \(k\).

1. Differentiate: \(f_k'(x)=10k^2x(2-kx)e^{2-kx}\). For \(x>0\), the critical point is \(x_H=\frac{2}{k}\). 2. \(f_k\left(\frac{2}{k}\right)=10k^2\left(\frac{2}{k}\right)^2e^0=40\). Thus, the maximum height is always \(40\,\text{m}\). 3. For \(x>\frac{2}{k}\), the factors \(x\) and \(e^{2-kx}\) are positive while \(2-kx<0\). Therefore, \(f_k'(x)<0\), so the hill descends. Also, \(f_k(x)=\frac{10k^2e^2x^2}{e^{kx}}\to0\) because exponential growth dominates polynomial growth. 4. At \(x=\frac{2+\sqrt2}{k}\), \(f_k'(x)=-10k(2\sqrt2+2)e^{-\sqrt2}\). Set its magnitude equal to \(0.25\): \(10k(2\sqrt2+2)e^{-\sqrt2}=0.25\). Thus, \(k=\frac{0.25}{10(2\sqrt2+2)e^{-\sqrt2}}\approx0.0213\,\text{m}^{-1}\).

a) \(40\,\text{m}\) b) \(f_k'(x)<0\) for \(x>\frac{2}{k}\), and \(\lim_{x\to\infty}f_k(x)=0\) c) \(k\approx0.0213\,\text{m}^{-1}\)

- Relate the sign of the derivative to increasing behavior. - Determine when an upward-opening quadratic is never below the \(x\)-axis. - Use the discriminant of the derivative.

1. Differentiate: \(f'(x) = 3x^2 + 2kx + 3\). 2. The function is strictly increasing on \(\mathbb{R}\) when the derivative is nonnegative everywhere and can equal zero only at isolated points. 3. Since \(f'\) is an upward-opening quadratic, it is nonnegative for every real \(x\) exactly when its discriminant is nonpositive. 4. Compute the discriminant: \(D = (2k)^2 - 4 \cdot 3 \cdot 3 = 4k^2 - 36\). 5. Solve \(4k^2 - 36 \le 0\): \(k^2 \le 9\), so \(-3 \le k \le 3\). 6. At \(k = \pm 3\), the derivative has one double zero but is otherwise positive, so the function remains strictly increasing. Therefore, the full parameter interval is \([-3, 3]\).

\(-3 \le k \le 3\)

- Use both the product and chain rules. - Analyze the discriminant of the quadratic factor in the derivative. - A local extremum requires a sign change in the first derivative. - The sign of the derivative determines increasing and decreasing intervals.

1. Differentiate: \(f_a'(x)=(-x^2+2x-a)e^{-x}\). 2. Since \(e^{-x}>0\), critical numbers satisfy \(x^2-2x+a=0\). Its discriminant is \(4-4a\). 3. If \(a<1\), there are two distinct critical numbers and the derivative changes sign at both, so there are two local extrema. If \(a=1\), there is a double zero at \(x=1\) with no sign change. If \(a>1\), there are no real critical numbers. Thus, there are no local extrema for \(a\geq1\). 4. For \(a=-3\), \(f_{-3}'(x)=(-x^2+2x+3)e^{-x}\), whose zeros are \(x=-1\) and \(x=3\). 5. The derivative is negative on \((-\infty, -1)\), positive on \((-1, 3)\), and negative on \((3, \infty)\). 6. Therefore, the local minimum is \((-1, -2e)\), and the local maximum is \((3, 6e^{-3})\).

a) Two local extrema for \(a<1\); no local extrema for \(a\geq1\). b) Decreasing on \((-\infty, -1)\) and \((3, \infty)\); increasing on \((-1, 3)\). Local minimum: \((-1, -2e)\); local maximum: \((3, 6e^{-3})\).

- Compare \(g(-x)\) with \(g(x)\). - Use the quotient rule and determine the derivative's sign. - Evaluate the derivative at \(x=0\). - Examine the one-sided limits at the endpoints of \((-1, 1)\) and use continuity.

1. Since \(g(-x)=\frac{-2x}{1-x^2}=-g(x)\), the graph is symmetric about the origin. 2. The denominator has greater degree than the numerator, so \(\lim_{x\to\infty}g(x)=0\). 3. By the quotient rule, \(g'(x)=\frac{2(1-x^2)-2x(-2x)}{(1-x^2)^2}=\frac{2x^2+2}{(1-x^2)^2}\). 4. Both the numerator and denominator are positive for every domain value, so \(g'(x)>0\). Therefore, \(g\) is strictly increasing on \((-\infty, -1)\), \((-1, 1)\), and \((1, \infty)\). 5. The tangent slope at the origin is \(g'(0)=2\). 6. On \((-1, 1)\), the function is continuous and strictly increasing, with \(\lim_{x\to-1^+}g(x)=-\infty\) and \(\lim_{x\to 1^-}g(x)=\infty\). By the Intermediate Value Theorem, it takes every real value on that interval.

a) \(g(-x)=-g(x)\), so the graph is symmetric about the origin; \(\lim_{x\to\infty}g(x)=0\) b) \(g'(x)=\frac{2x^2+2}{(1-x^2)^2}\); strictly increasing on \((-\infty, -1)\), \((-1, 1)\), and \((1, \infty)\) c) Tangent slope: \(2\); on \((-1, 1)\), the range is \(\mathbb{R}\).

- Connect invertibility on \(\mathbb{R}\) with strict monotonicity. - Analyze the sign of the first derivative. - Determine when an upward-opening quadratic never becomes negative. - Use the discriminant.

1. A function that is strictly monotonic on \(\mathbb{R}\) is one-to-one and therefore invertible on its range. 2. Differentiate: \(g_k'(x)=3x^2+2kx+12\). 3. This upward-opening quadratic is nonnegative for every real \(x\) when its discriminant is nonpositive. 4. Compute the discriminant: \((2k)^2-4\cdot3\cdot12=4k^2-144\). 5. Solve \(4k^2-144\leq0\): \(k^2\leq36\), so \(-6\leq k\leq6\). 6. At \(k=\pm6\), the derivative has one isolated double zero but remains nonnegative, so the cubic is still strictly increasing. 7. If \(|k|>6\), the derivative has two distinct zeros and is negative between them, so the cubic has a local maximum and a local minimum and is not one-to-one. Therefore, \(-6\leq k\leq6\).

\(-6\leq k\leq6\)

- Origin symmetry does not by itself guarantee strict monotonicity. - Compare the two end behaviors of even- and odd-degree polynomials. - For part c, determine the derivative's sign on both sides of its only zero. - A counterexample must satisfy the hypothesis but fail the conclusion.

1. a) False. The polynomial \(f(x)=x^3-x\) is odd, so its graph has origin symmetry. However, \(f'(x)=3x^2-1\) changes sign, so \(f\) is not strictly monotonic and is not one-to-one. 2. b) True. A continuous one-to-one function on \(\mathbb{R}\) must be strictly monotonic. An even-degree polynomial has the same end behavior as \(x\to-\infty\) and \(x\to\infty\), so it cannot be strictly monotonic on all of \(\mathbb{R}\). Therefore, an invertible polynomial must have odd degree. 3. c) True. Let the only zero of \(f'\) be \(x=a\). Since \(f'\) is continuous, nonzero away from \(a\), and does not change sign at \(a\), it is either positive for all \(x\ne a\) or negative for all \(x\ne a\). 4. Thus, \(f\) is strictly increasing or strictly decreasing on \(\mathbb{R}\), so it is one-to-one and has an inverse function.

a) False; for example, \(f(x)=x^3-x\) b) True c) True

- Use the zeros of the quadratic to determine where it is nonnegative. - The derivative is undefined when its radical denominator is zero. - The sign of the first derivative determines increasing and decreasing behavior. - Consider each connected interval of the original domain separately.

1. a) Require \(x(x+6)\ge0\). Thus, \(D_f=(-\infty,-6]\cup[0,\infty)\). 2. b) By the chain rule, \(f'(x)=\frac{2x+6}{2\sqrt{x^2+6x}}=\frac{x+3}{\sqrt{x^2+6x}}\). 3. The derivative requires \(x^2+6x>0\), so \(D_{f'}=(-\infty,-6)\cup(0,\infty)\). 4. c) On \((-\infty,-6)\), the denominator is positive and \(x+3<0\), so \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \((-\infty,-6]\). 5. On \((0,\infty)\), both the denominator and \(x+3\) are positive, so \(f'(x)>0\). Therefore, \(f\) is strictly increasing on \([0,\infty)\).

a) \(D_f=(-\infty,-6]\cup[0,\infty)\) b) \(f'(x)=\frac{x+3}{\sqrt{x^2+6x}}\), with \(D_{f'}=(-\infty,-6)\cup(0,\infty)\) c) Decreasing on \((-\infty,-6]\); increasing on \([0,\infty)\)

- Distinguish sufficient conditions from necessary conditions. - Test part b with an odd power function. - Use the fact that a continuous one-to-one function on an interval is strictly monotonic. - Track the order of inputs and outputs for a decreasing function.

1. a) True. If \(f'(x)>0\) throughout \(I\), then \(f\) is strictly increasing, so it is one-to-one and has an inverse on its range. 2. b) False. The function \(f(x)=x^3\) is strictly increasing and invertible on \(\mathbb{R}\), but \(f'(0)=0\). 3. c) True. A continuous one-to-one function on an interval must be strictly monotonic. An interior local extremum prevents strict monotonicity, so the function cannot be one-to-one on the entire interval. 4. d) True. If \(y_1<y_2\) and \(x_i=f^{-1}(y_i)\), then strict decrease of \(f\) implies \(x_1>x_2\). Hence, \(f^{-1}(y_1)>f^{-1}(y_2)\), so the inverse is strictly decreasing.

a) True b) False; for example, \(f(x)=x^3\) c) True d) True

- The derivative should be zero where the monotonic behavior changes. - Build a quadratic derivative that is negative between \(-2\) and \(4\) and positive outside that interval. - Find an antiderivative, then use the y-intercept to determine the constant.

1. The monotonic behavior changes at \(x=-2\) and \(x=4\), so choose a derivative with zeros there: \(g'(x)=a(x+2)(x-4)\), where \(a>0\). 2. One convenient choice is \(a=3\), giving \(g'(x)=3x^2-6x-24\). This derivative is positive outside \([-2,4]\) and negative inside it. 3. Find an antiderivative: \(g(x)=x^3-3x^2-24x+C\). 4. The y-intercept condition gives \(g(0)=C=5\). 5. Therefore, one possible function is \(g(x)=x^3-3x^2-24x+5\).

One possible function is \(g(x)=x^3-3x^2-24x+5\).

- Factor out \(\sin x\) when solving for zeros. - Use the range of cosine on the stated interval. - Differentiate tangent using \(\frac{d}{dx}(\tan x)=\sec^2x\). - Determine monotonicity from the sign of the first derivative.

1. Factor the function: \(g_k(x)=\sin x\left(1-\frac{k}{\cos x}\right)\). Therefore, \(g_k(x)=0\) when \(\sin x=0\) or \(\cos x=k\). On \(I\), \(\sin x=0\) gives \(x=0\). If \(0<k<1\), the equation \(\cos x=k\) gives one additional zero, \(x=\arccos k\). If \(k\ge1\), it gives no additional zero in \(I\). Thus there are two zeros for \(0<k<1\) and one zero for \(k\ge1\). 2. For \(k=0.5\), \(g_{0.5}'(x)=\cos x-\frac{0.5}{\cos^2x}=\frac{2\cos^3x-1}{2\cos^2x}\). Since \(\cos x>0\) on \(I\), the sign depends on \(2\cos^3x-1\). Setting the numerator equal to zero gives \(\cos x=\sqrt[3]{\frac{1}{2}}\), so \(x_c=\arccos\left(\sqrt[3]{\frac{1}{2}}\right)\approx0.65\). The derivative is positive before \(x_c\) and negative after \(x_c\). Therefore, \(g_{0.5}\) is increasing on \([0, x_c]\), decreasing on \(\left[x_c, \frac{\pi}{2}\right)\), and has a local maximum at \(x=x_c\).

a) For \(0<k<1\), \(g_k\) has two zeros. For \(k\ge1\), it has one zero, \(x=0\). b) Let \(x_c=\arccos\left(\sqrt[3]{\frac{1}{2}}\right)\approx0.65\). The function is increasing on \([0, x_c]\), decreasing on \(\left[x_c, \frac{\pi}{2}\right)\), and has a local maximum at \(x=x_c\).

- Use monotonicity between the critical numbers to establish the number of zeros. - The slopes of perpendicular nonvertical lines are negative reciprocals. - Use the normal line intercepts as the legs of the right triangle. - Check function values at the endpoints, the midpoint, and the critical numbers.

1. Differentiate: \(g'(x)=-\sin(x)+\frac{2}{\pi}\) and \(g''(x)=-\cos(x)\). Let \(\alpha=\arcsin\left(\frac{2}{\pi}\right)\) and \(M=\sqrt{1-\frac{4}{\pi^2}}+\frac{2\alpha}{\pi}-1\). The critical numbers are \(\alpha\) and \(\pi-\alpha\). Since \(g''(\alpha)<0\), there is a local maximum at \((\alpha,M)\). Also, \(g(\pi-x)=-g(x)\), so the graph has point symmetry about \(\left(\frac{\pi}{2},0\right)\); hence there is a local minimum at \((\pi-\alpha,-M)\). Numerically, the extrema are approximately \((0.69,0.21)\) and \((2.45,-0.21)\). 2. The function increases on \([0,\alpha]\), decreases on \([\alpha,\pi-\alpha]\), and increases on \([\pi-\alpha,\pi]\). Also, \(g(0)=0\), \(g\left(\frac{\pi}{2}\right)=0\), and \(g(\pi)=0\). The monotonicity on the three intervals shows that each of these zeros is unique there, so there are exactly three zeros. 3. At \(x=\frac{\pi}{2}\), the point is \(\left(\frac{\pi}{2},0\right)\), and the tangent slope is \(g'\left(\frac{\pi}{2}\right)=\frac{2-\pi}{\pi}\). Therefore, the normal slope is \(\frac{\pi}{\pi-2}\), and \(n(x)=\frac{\pi}{\pi-2}\left(x-\frac{\pi}{2}\right)\). 4. The x-intercept of the normal is \(\frac{\pi}{2}\), and the magnitude of its y-intercept is \(\frac{\pi^2}{2(\pi-2)}\). Thus the triangle area is \(\frac{1}{2}\cdot\frac{\pi}{2}\cdot\frac{\pi^2}{2(\pi-2)}=\frac{\pi^3}{8(\pi-2)}\approx3.395\).

a) Let \(\alpha=\arcsin\left(\frac{2}{\pi}\right)\) and \(M=\sqrt{1-\frac{4}{\pi^2}}+\frac{2\alpha}{\pi}-1\). Local maximum: \((\alpha,M)\approx(0.69,0.21)\). Local minimum: \((\pi-\alpha,-M)\approx(2.45,-0.21)\). b) The zeros are \(x=0,\frac{\pi}{2},\pi\). c) \(n(x)=\frac{\pi}{\pi-2}\left(x-\frac{\pi}{2}\right)\). d) \(\frac{\pi^3}{8(\pi-2)}\approx3.395\) square units.