Free math worksheets

- 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 \(D\).

- A local maximum is at least as large as all nearby function values. - Check the actual filled point when a graph is discontinuous. - Compare the graph on both sides of \(x_0\).

1. In graph a), \(x_0\) is the vertex of an upward-opening parabola, so it is a local minimum. 2. In graph b), \(x_0\) is the vertex of a downward-opening parabola, so nearby values are less than or equal to \(f(x_0)\). Therefore, \(x_0\) is a local maximum. 3. In graph c), the function is increasing through \(x_0\), so there is no local extremum. 4. In graph d), the defined point at \(x_0\) lies below nearby values, so it is a local minimum.

b)

- The theorem is about existence, not about locating the extrema. - Check the function's continuity on the entire stated interval. - What two properties of the interval matter?

1. The square-root expression is defined for every \(x\in[0,4]\) and is continuous there. 2. The interval \([0,4]\) is closed and bounded. 3. Therefore, the Extreme Value Theorem guarantees that \(f\) attains both an absolute maximum and an absolute minimum on \([0,4]\).

Yes. The function is continuous on the closed, bounded interval \([0,4]\), so the Extreme Value Theorem guarantees both absolute extrema.

- A critical number does not have to make the derivative equal to zero. - Compare the slopes immediately to the left and right of the corner. - Which interior inputs belong in an absolute-extrema candidate set?

1. For \(x<2\), \(f'(x)=-1\), and for \(x>2\), \(f'(x)=1\). 2. At \(x=2\), the one-sided derivatives are different, so \(f'(2)\) does not exist. 3. Since \(f\) is defined at \(x=2\), this derivative-undefined input is a critical number. It must be considered along with endpoints when checking candidates for absolute extrema on a closed interval.

The only interior critical number is \(x=2\), because \(f'(2)\) does not exist even though \(f(2)\) is defined.

- What must the slope be when a tangent line is horizontal? - Which function gives the slope of the tangent line? - After finding the \(x\)-coordinates, how will you find the corresponding \(y\)-coordinates?

1. A horizontal tangent has slope \(0\), so solve \(f'(x) = 0\). 2. Differentiate: \(f'(x) = 6x^2 + 6x - 12\). 3. Solve \(6x^2 + 6x - 12 = 0\), or \(x^2 + x - 2 = 0\). Factoring gives \((x + 2)(x - 1) = 0\), so \(x = -2\) or \(x = 1\). 4. Evaluate the function: \(f(-2) = 24\) and \(f(1) = -3\). 5. The points are \((-2, 24)\) and \((1, -3)\).

\((-2, 24)\) and \((1, -3)\)

- What condition does the derivative satisfy at a horizontal tangent? - First factor out the common factor \(x\). - How many real zeros can the cubic derivative have at most?

1. A horizontal tangent occurs where \(g'(x) = 0\). 2. Differentiate: \(g'(x) = x^3 - 3x^2 + 2x\). 3. Factor: \(x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x - 1)(x - 2)\). 4. Set each factor equal to zero. The solutions are \(x = 0\), \(x = 1\), and \(x = 2\).

\(x = 0\), \(x = 1\), and \(x = 2\)

- A horizontal tangent has slope zero. - Differentiate while treating \(k\) as a constant. - Substitute the given input into the derivative. - Use the exact cosine value at \(\frac{\pi}{3}\).

1. Differentiate: \(f_k'(x)=4\cos x-k\). 2. A horizontal tangent requires \(f_k'\left(\frac{\pi}{3}\right)=0\). Therefore, \(4\cos\left(\frac{\pi}{3}\right)-k=0\). Since \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\), \(2-k=0\), so \(k=2\).

\(k=2\)

- Use the product rule on \(x\sin x\). - Combine like trigonometric terms after differentiating. - A horizontal tangent has derivative zero. - Use the zero-product property.

1. Differentiate the product and the cosine term: \(f'(x)=\sin x+x\cos x-\sin x=x\cos x\). 2. A horizontal tangent requires \(x\cos x=0\). Thus, either \(x=0\) or \(\cos x=0\). In \([0, 2\pi]\), the solutions are \(x=0\), \(x=\frac{\pi}{2}\), and \(x=\frac{3\pi}{2}\).

\(f'(x)=x\cos x\); horizontal tangents at \(x=0, \frac{\pi}{2}, \frac{3\pi}{2}\)

- What is the slope of a line parallel to the x-axis? - Use the derivative to represent the tangent slope. - Set the derivative equal to \(0\). - After finding each x-coordinate, substitute it into \(f\) to find the corresponding y-coordinate.

1. Differentiate: \(f'(x)=x^2-4x+3\). 2. A horizontal tangent has slope \(0\), so solve \(f'(x)=0\): \(x^2-4x+3=(x-1)(x-3)=0\). Thus \(x=1\) or \(x=3\). 3. Evaluate the function: \(f(1)=\frac{19}{3}\) and \(f(3)=5\). 4. The points are \(\left(1,\frac{19}{3}\right)\) and \((3,5)\).

\(\left(1,\frac{19}{3}\right)\) and \((3,5)\)

- The necessary condition at an interior local extremum is a zero derivative. - Differentiate using the power rule. - Rewrite a reciprocal as a negative power if helpful. - Check all solutions against the domain.

1. For part a), \(f'(x)=x^2-2x-3=(x-3)(x+1)\). Setting \(f'(x)=0\) gives \(x=-1\) and \(x=3\). 2. For part b), \(g'(x)=1-\frac{4}{x^2}\). Setting \(g'(x)=0\) gives \(x^2=4\), so \(x=-2\) and \(x=2\). Both are in the domain.

a) \(x=-1\) and \(x=3\) b) \(x=-2\) and \(x=2\)

- Differentiate using the chain rule. - Set the first derivative equal to zero. - Use the zeros of cosine in the relevant angle interval. - Solve each resulting equation for \(x\).

1. Differentiate using the chain rule: \(g'(x)=6\cos(2x-\pi)\). 2. Set the derivative equal to zero: \(\cos(2x-\pi)=0\). 3. Since \(x\in[0, \pi]\), the angle \(2x-\pi\) ranges from \(-\pi\) to \(\pi\). In this interval, cosine is zero at \(2x-\pi=-\frac{\pi}{2}\) and \(2x-\pi=\frac{\pi}{2}\). 4. Solving gives \(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\). Both values lie in the interior of the stated interval.

\(x=\frac{\pi}{4}\) and \(x=\frac{3\pi}{4}\)

- Local extrema in the interior of the domain require a critical number. - 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 numbers 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.

- Compare \(f(0)\) with values immediately to the left and right. - A local maximum must be at least as large as all nearby values. - A local minimum must be at most as large as all nearby values. - Analyze the two formulas separately.

1. The function value is \(f(0)=1\). 2. For every \(x>0\), \(f(x)=x^2+1>1=f(0)\). 3. For every \(x<0\), \(f(x)=x<0<1=f(0)\). 4. Every neighborhood of \(0\) therefore contains function values both greater than and less than \(f(0)\). Thus \(f(0)\) is neither a local maximum nor a local minimum.

There is no local extremum at \(x=0\) because every neighborhood of \(0\) contains values above and below \(f(0)=1\).

- Use the definition of strict decrease at the left endpoint. - Local extrema compare only nearby values. - Recall the hypotheses of the extreme value theorem.

1. Statement (1) is true. Strict decrease gives \(f(a)>f(x)\) for every \(x>a\) in the interval. 2. Statement (2) is true. “Local” compares values only in a neighborhood, so a local maximum in one region can lie below a distant local minimum. 3. Statement (3) is true by the extreme value theorem.

(1) True (2) True (3) True

- State the necessary derivative condition for an interior local extremum. - Use \(f'(x)=\cos x\). - Solve \(\cos x=0\) on the given interval. - Check how the sign of cosine changes at each zero.

1. If a differentiable function has an interior local extremum at \(x_0\), then the necessary condition is \(f'(x_0)=0\). Since \(f'(x)=\cos x\), every interior local extremum of sine must occur at a zero of cosine. 2. Solve \(\cos x=0\) on \([0, 2\pi]\). The solutions are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). 3. Cosine changes from positive to negative at \(\frac{\pi}{2}\), so sine has a local maximum there. Cosine changes from negative to positive at \(\frac{3\pi}{2}\), so sine has a local minimum there.

The necessary condition shows that an interior local extremum of sine must be a zero of cosine. In \([0, 2\pi]\), the zeros are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). The first is a local maximum of sine, and the second is a local minimum.

- The derivative gives the slope of the graph. - Set the derivative equal to \(0\) and solve for \(x\). - Look for the point on the parabola where the graph changes from increasing to decreasing.

1. Differentiate: \(f'(x)=-2x+4\). 2. Set the slope equal to \(0\): \(-2x+4=0\), so \(x=2\). 3. The graph has a maximum at \(x=2\), where the tangent line is horizontal. This confirms that the slope is \(0\) there.

a) \(f'(x)=-2x+4\) b) \(x=2\) c) The graph has a horizontal tangent at its maximum when \(x=2\).

- What does differentiability look like on a graph? - Compare the slope immediately before and after \(x_e=2\). - Recall the derivative as a limit of difference quotients. - What assumption is required before setting a derivative equal to \(0\)?

1. At \(x_e=2\), the graph has a corner. The slope from the left is \(1\), while the slope from the right is \(-1\). Since the one-sided slopes are different, the derivative does not exist at \(x_e=2\). 2. The condition that the derivative equal \(0\) applies only at points where the derivative exists. Because \(f\) is not differentiable at \(x_e=2\), this local maximum cannot be found by solving an equation that sets the derivative equal to \(0\).

1. The function is not differentiable at \(x_e=2\) because the left-hand slope is \(1\) and the right-hand slope is \(-1\). 2. The derivative does not exist at the maximum, so the condition that the derivative equal \(0\) does not apply.

- A local minimum is no greater than all nearby function values. - At a discontinuity, use the filled point as the actual function value. - Compare values immediately to the left and right of \(x_0\).

1. In graph a), the function is decreasing through \(x_0\), so there is no local extremum. 2. In graph b), \(x_0\) is the vertex of a downward-opening parabola, so it is a local maximum. 3. In graph c), the filled point at \(x_0\) lies above the nearby curve values, so it is a local maximum, not a local minimum. 4. In graph d), the filled point at \(x_0\) lies below all nearby values. Therefore, \(x_0\) is a local minimum.

d)

- Look for peaks and valleys. - A local maximum occurs where the graph changes from increasing to decreasing. - A local minimum occurs where the graph changes from decreasing to increasing.

1. The graph changes from increasing to decreasing at \(x=-4\) and \(x=3\), so these are local-maximum locations. 2. The graph changes from decreasing to increasing at \(x=-1\) and \(x=5\), so these are local-minimum locations.

Local maxima at \(x=-4\) and \(x=3\); local minima at \(x=-1\) and \(x=5\)

- Separate continuity of the function from the shape of its domain interval. - Ask whether the endpoint values that would be extreme are actually in the domain. - A supremum or infimum need not be attained.

1. The function is continuous on \((0,1)\), but the interval is not closed, so the Extreme Value Theorem does not apply. 2. Values of \(f(x)\) can be made arbitrarily close to \(0\), but \(0\) is not attained because \(x=0\) is excluded. Thus, there is no absolute minimum. 3. Values can likewise be made arbitrarily close to \(1\), but \(1\) is not attained because \(x=1\) is excluded. Thus, there is no absolute maximum.

a) The domain interval is not closed. b) Neither an absolute maximum nor an absolute minimum is attained.

- Separate what the theorem guarantees from the later calculation that locates the extrema. - Which interior inputs must be added to the endpoints when forming the candidate set? - Compare function values, not only classifications of critical numbers.

1. Because \(f\) is a polynomial, it is continuous on the closed, bounded interval \([-2,2]\). The Extreme Value Theorem therefore guarantees that both absolute extrema are attained. 2. Differentiate: \(f'(x)=3x^2-3=3(x-1)(x+1)\). The interior critical numbers are \(x=-1\) and \(x=1\). 3. Evaluate the endpoints and critical numbers: \(f(-2)=-2\), \(f(-1)=2\), \(f(1)=-2\), and \(f(2)=2\). 4. The absolute maximum value is \(2\), attained at \(x=-1\) and \(x=2\). The absolute minimum value is \(-2\), attained at \(x=-2\) and \(x=1\).

a) Both an absolute maximum and an absolute minimum are guaranteed to be attained. b) Absolute maximum: \(2\) at \(x=-1\) and \(x=2\); absolute minimum: \(-2\) at \(x=-2\) and \(x=1\)

- A horizontal tangent has derivative \(0\). - Find the full coordinates of all three points. - Notice whether two points form a horizontal base. - Use \(\frac{1}{2}(\text{base})(\text{height})\).

1. Differentiate: \(f'(x)=x^3-4x=x(x^2-4)\). 2. Horizontal tangent lines occur where \(f'(x)=0\). Thus, \(x=-2\), \(x=0\), and \(x=2\). 3. Evaluate the function: \(f(-2)=-3\), \(f(0)=1\), and \(f(2)=-3\). The points are \((-2,-3)\), \((0,1)\), and \((2,-3)\). 4. The horizontal base from \((-2,-3)\) to \((2,-3)\) has length \(4\), and the vertical height to \((0,1)\) is \(4\). 5. The area is \(\frac{1}{2}\cdot 4\cdot 4=8\).

\(8\) square units

- What conditions must the first and second derivatives satisfy at a stationary inflection point? - How is the number of local extrema related to the zeros of the first derivative? - Use the discriminant to determine when a quadratic equation has two distinct real solutions.

1. Differentiate: \(f_t'(x) = x^2 + 2tx + 9\) and \(f_t''(x) = 2x + 2t\). 2. At a stationary inflection point, \(f_t'(x) = 0\) and \(f_t''(x) = 0\). From \(2x + 2t = 0\), we get \(x = -t\). 3. Substitute into the first derivative: \(f_t'(-t) = t^2 - 2t^2 + 9 = 9 - t^2\). Setting this equal to zero gives \(t = \pm 3\). Since \(t > 0\), \(t = 3\). 4. The corresponding input is \(x = -3\), and \(f_3(-3) = \frac{1}{3}(-3)^3 + 3(-3)^2 + 9(-3) = -9\). The stationary inflection point is \((-3, -9)\). 5. Two local extrema occur when \(f_t'(x) = x^2 + 2tx + 9\) has two distinct real zeros. Its discriminant is \(D = (2t)^2 - 4 \cdot 1 \cdot 9 = 4t^2 - 36\). 6. Solve \(4t^2 - 36 > 0\): \(t^2 > 9\), so \(t < -3\) or \(t > 3\). Each simple zero of the derivative produces a sign change, so these values give two local extrema.

a) \(t = 3\); the stationary inflection point is \((-3, -9)\). b) \(t < -3\) or \(t > 3\)

- Distinguish conditions that must be true from conditions that are only sufficient. - The growth rate is the first derivative. - At an interior local extremum of a differentiable function, its derivative is \(0\). - A negative second derivative can confirm a local maximum, but it is not required.

1. Situation (1) directly states the function value \(h(12)=80\), so card A must be true. 2. The growth rate is \(h'\). If \(h'\) has a local maximum at \(t=6\), Fermat's theorem applied to \(h'\) gives \(h''(6)=0\). The value \(h'(6)\) need not be positive, and \(h'''(6)<0\) is a sufficient but not necessary test for the maximum. Thus only card B must be true. 3. A local maximum height of \(160\,\text{cm}\) at \(t=25\) gives \(h(25)=160\). Since \(h\) is differentiable, Fermat's theorem gives \(h'(25)=0\). The condition \(h''(25)<0\) is sufficient but not necessary for a local maximum. Thus cards C and D must be true.

(1) A (2) B (3) C, D

- Distinguish conditions that must be true from conditions that are only sufficient. - The first derivative represents vertical velocity. - “Descending” determines the sign of the velocity. - At an interior local extremum of a differentiable function, its derivative is \(0\).

1. Descending at the greatest local rate means that the vertical velocity \(H'\) has a local minimum at \(t=10\) and is negative there. Fermat's theorem applied to \(H'\) gives \(H''(10)=0\), and descending gives \(H'(10)<0\). The condition \(H'''(10)>0\) is sufficient but not necessary for the local minimum. Thus cards D and E must be true. 2. A local maximum altitude of \(120\,\text{m}\) at \(t=30\) gives \(H(30)=120\). Since \(H\) is differentiable, Fermat's theorem gives \(H'(30)=0\). The condition \(H''(30)<0\) is sufficient but not necessary for a local maximum. Thus cards A and B must be true.

(1) D, E (2) A, B

- Find the first derivative and analyze how many real zeros it has. - Use the discriminant of the quadratic derivative. - A double zero of the derivative does not produce a local extremum. Check whether concavity changes there.

1. Differentiate: \(f_k'(x) = 3x^2 - 2kx + 3k\). 2. The number of critical points depends on the discriminant of \(3x^2 - 2kx + 3k = 0\): \(D = (-2k)^2 - 4 \cdot 3 \cdot 3k = 4k(k - 9)\). 3. If \(D > 0\), the derivative has two distinct real zeros. This occurs for \(k < 0\) or \(k > 9\). The derivative changes sign at each simple zero, so the graph has two local extrema. 4. If \(D < 0\), which occurs for \(0 < k < 9\), the derivative has no real zeros, so the graph has no local extrema. 5. If \(D = 0\), then \(k = 0\) or \(k = 9\). The derivative has a double zero and does not change sign, so there is no local extremum. Because \(f_k''(x) = 6x - 2k\) changes sign at the double zero, the graph has a stationary inflection point. 6. Therefore, the graph has no local extrema for every \(k\) in \([0, 9]\), with stationary inflection points occurring at the endpoint parameter values \(k = 0\) and \(k = 9\).

Two local extrema: \(k < 0\) or \(k > 9\) No local extrema: \(0 \le k \le 9\) Stationary inflection point: \(k = 0\) or \(k = 9\)

- What must the first derivative equal at a point with a horizontal tangent? - Substitute \(x = -3\) into the derivative. - Clear the denominator before solving the resulting quadratic equation.

1. Differentiate: \(g_a'(x) = \frac{3}{a}x^2 + 12x + 9a\). 2. A horizontal tangent at \(x = -3\) requires \(g_a'(-3) = 0\). Substitution gives \(\frac{27}{a} - 36 + 9a = 0\). 3. Because \(a \ne 0\), multiply by \(a\): \(9a^2 - 36a + 27 = 0\). 4. Divide by \(9\) and factor: \(a^2 - 4a + 3 = (a - 1)(a - 3) = 0\). 5. Therefore, \(a = 1\) or \(a = 3\). Both values satisfy the domain restriction.

\(a = 1\) or \(a = 3\)

- Find the critical numbers and identify the one that depends on \(k\). - Compute the corresponding function value in terms of \(k\). - Solve the \(x\)-coordinate equation for \(k\), then substitute into the \(y\)-coordinate.

1. Differentiate: \(f_k'(x) = x^2 - 2kx = x(x - 2k)\). The critical numbers are \(x = 0\) and \(x = 2k\). 2. The parameter-dependent critical number is \(x = 2k\). Its function value is \(f_k(2k) = \frac{8}{3}k^3 - 4k^3 + 4 = -\frac{4}{3}k^3 + 4\). 3. Solve \(x = 2k\) for the parameter: \(k = \frac{x}{2}\). 4. Substitute into the function value: \(y = -\frac{4}{3}\left(\frac{x}{2}\right)^3 + 4 = -\frac{1}{6}x^3 + 4\). 5. Because \(k \ne 0\), the parameter-dependent extrema have \(x \ne 0\). Therefore, the locus is \(y = -\frac{1}{6}x^3 + 4\), for \(x \ne 0\).

\(y = -\frac{1}{6}x^3 + 4\), for \(x \in \mathbb{R} \setminus \{0\}\)

- Factor the polynomial before finding its zeros. - Substitute the coordinates of the given point. - Set the derivative equal to zero and use \(k\ne0\).

1. Factor the function: \(f_k(x)=kx(x-3)^2\). Since \(k\ne0\), the zeros are \(x=0\) and \(x=3\), independent of \(k\). 2. Substituting \(x=1\) and \(f_k(1)=8\) gives \(8=k(1-6+9)=4k\), so \(k=2\). 3. The derivative is \(f_k'(x)=k(3x^2-12x+9)=3k(x-1)(x-3)\). Since \(k\ne0\), the critical points occur at \(x=1\) and \(x=3\), independent of \(k\).

a) \(x=0\) and \(x=3\) b) \(k=2\) c) Critical-point x-coordinates: \(1\) and \(3\)

- What do the function value and derivative equal at a local extremum on the x-axis? - What is the derivative of \(e^x\)? - Apply the product rule to the entire expression for \(g\). - Why does multiplying by the always-positive factor \(e^x\) preserve the sign of \(f\)?

1. Since \(f\) has a local extremum at \((0, 0)\), \(f(0) = 0\) and \(f'(0) = 0\). 2. Evaluate \(g\) at the origin: \(g(0) = e^0f(0) = 0\). 3. Apply the product rule: \(g'(x) = e^xf(x) + e^xf'(x)\). Therefore, \(g'(0) = e^0f(0) + e^0f'(0) = 0\). 4. Also, \(e^x > 0\) for every real \(x\), so \(g(x)\) has the same sign as \(f(x)\) near \(x = 0\). Because \(f\) has a local extremum with value \(0\), it does not change sign there. Thus \(g\) does not cross the x-axis at the origin and has a horizontal tangent there, so its graph touches the x-axis at \((0, 0)\).

Since \(g(0) = 0\), \(g'(0) = 0\), and \(g\) has the same sign as \(f\) near \(0\), the graph of \(g\) touches the x-axis at \((0, 0)\).

- Use the known zeros of cosine. - Include the derivative of the inner quadratic expression when applying the chain rule. - Check systematically which integer values produce solutions inside the interval. - The range of cosine can help classify the critical points.

1. For the zeros, solve \(\cos\left(\frac{\pi}{4}x^2\right)=0\). Thus \(\frac{\pi}{4}x^2=\frac{\pi}{2}+k\pi\), so \(x^2=2+4k\). Within \(0\le x^2\le 9\), this gives \(x^2=2\) and \(x^2=6\). Therefore, the zeros are \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{6}\). 2. Differentiate: \(g'(x)=-\frac{\pi}{2}x\sin\left(\frac{\pi}{4}x^2\right)\). Critical numbers occur when \(x=0\) or \(\sin\left(\frac{\pi}{4}x^2\right)=0\). Thus \(x^2=4k\), which gives \(x=\pm2\) and \(x=\pm\sqrt{8}\), in addition to \(x=0\). 3. Evaluate the function at the critical numbers. Since \(g(0)=1\), \(g(\pm2)=-1\), and \(g(\pm\sqrt{8})=1\), the points with value \(1\) are local maxima and the points with value \(-1\) are local minima.

a) \(x=\pm\sqrt{2}\approx\pm1.414\) and \(x=\pm\sqrt{6}\approx\pm2.449\). b) Local maxima: \((0,1)\), \(\left(-\sqrt{8},1\right)\), and \(\left(\sqrt{8},1\right)\), where \(\sqrt{8}\approx2.828\). Local minima: \((-2,-1)\) and \((2,-1)\).

- Use the standard zeros of cosine. - Apply the chain rule when differentiating \(\cos(x^2)\). - Set each factor of the first derivative equal to zero. - Use the maximum and minimum values of cosine to classify the critical points. - Keep only values inside the stated interval.

1. Solve \(\cos(x^2)=0\). This requires \(x^2=\frac{\pi}{2}+k\pi\). Because \(0\le x^2\le4\), only \(x^2=\frac{\pi}{2}\) is possible. Thus the zeros are \(x=\pm\sqrt{\frac{\pi}{2}}\). 2. Differentiate using the chain rule: \(f'(x)=-2x\sin(x^2)\). Critical numbers satisfy \(x=0\) or \(\sin(x^2)=0\). Within the interval, this gives \(x=0\) and \(x=\pm\sqrt{\pi}\). 3. Since \(f(0)=1\), the point \((0,1)\) is a local maximum. Since \(f(\pm\sqrt{\pi})=-1\), the points \(\left(-\sqrt{\pi},-1\right)\) and \(\left(\sqrt{\pi},-1\right)\) are local minima.

a) \(x=\pm\sqrt{\frac{\pi}{2}}\approx\pm1.253\). b) Local maximum: \((0,1)\). Local minima: \(\left(-\sqrt{\pi},-1\right)\) and \(\left(\sqrt{\pi},-1\right)\), with x-coordinates approximately \(\pm1.772\).

- Recall when sine equals zero. - Apply the chain rule to the quadratic argument. - Set each factor of the first derivative equal to zero. - Use the maximum and minimum values of sine to classify the critical points. - Check that \(\sqrt{6}\) lies inside the interval.

1. Solve \(\sin\left(\frac{\pi}{4}x^2\right)=0\). Then \(\frac{\pi}{4}x^2=k\pi\), so \(x^2=4k\). Since \(0\le x^2\le6.25\), the possible values are \(x^2=0\) and \(x^2=4\). Thus the zeros are \(x=-2,0,2\). 2. Differentiate: \(h'(x)=\frac{\pi}{2}x\cos\left(\frac{\pi}{4}x^2\right)\). Critical numbers occur when \(x=0\) or \(\cos\left(\frac{\pi}{4}x^2\right)=0\). This gives \(x^2=2+4k\), so the interior critical numbers are \(x=0\), \(x=\pm\sqrt{2}\), and \(x=\pm\sqrt{6}\). 3. At \(x=0\), the derivative changes from negative to positive, so \((0,0)\) is a local minimum. At \(x=\pm\sqrt{2}\), the function value is \(1\), so these are local maxima. At \(x=\pm\sqrt{6}\), the function value is \(-1\), so these are local minima.

a) \(x=-2,0,2\). b) Local maxima: \(\left(-\sqrt{2},1\right)\) and \(\left(\sqrt{2},1\right)\). Local minima: \((0,0)\), \(\left(-\sqrt{6},-1\right)\), and \(\left(\sqrt{6},-1\right)\).

- The exponential factor is never zero. - Use the derivative at the x-intercept to find the crossing angle. - Apply the product and chain rules. - Eliminate \(k\) from the extremum coordinates to find the locus.

1. Since the exponential factor is positive, the x-intercept satisfies \(x-k=0\), so \(x=k\). 2. Differentiate: \(f_k'(x)=\left(2-\frac{x}{k}\right)e^{1-x/k}\). At the intercept, \(f_k'(k)=1\). Thus, the tangent slope is always \(1\), and the angle with the positive x-axis is \(\arctan(1)=45^\circ\). 3. The critical point satisfies \(2-\frac{x}{k}=0\), so \(x=2k\). The second derivative is \(f_k''(x)=\frac1k e^{1-x/k}\left(\frac{x}{k}-3\right)\). Therefore, \(f_k''(2k)=-\frac1{ke}<0\), so the extremum is a local maximum. Its coordinates are \(E_k=\left(2k, \frac{k}{e}\right)\). 4. From \(x=2k\), \(k=x/2\). Substituting into \(y=k/e\) gives \(y=\frac{x}{2e}\). Thus, all extrema lie on this line through the origin.

a) The x-intercept is \(x=k\), and the crossing angle is \(45^\circ\). b) \(E_k=\left(2k, \frac{k}{e}\right)\), a local maximum. c) \(y=\frac{x}{2e}\)

- At an interior maximum, the first derivative is zero. - Use the product rule and chain rule. - After finding the parameter, check how the derivative changes sign across the whole modeled interval.

1. Differentiate using the product and chain rules: \(g'(t)=10e^{kt}(-1+60k-kt)\). 2. A maximum at \(t=20\) requires \(g'(20)=0\). Since \(10e^{20k}\neq0\), \(-1+40k=0\), so \(k=\frac{1}{40}=0.025\,\text{min}^{-1}\). 3. For \(k=\frac{1}{40}\), \(g'(t)=\frac14 e^{t/40}(20-t)\). Thus \(g'(t)>0\) for \(0\leq t<20\) and \(g'(t)<0\) for \(20<t\leq60\). 4. Therefore, \(g\) increases up to \(t=20\) and decreases afterward, so \(t=20\) is the absolute maximum on \([0,60]\).

\(k=0.025\,\text{min}^{-1}\)

- Find the critical time in terms of \(k\). - Substitute that time into the original function. - Evaluate the first derivative at \(t=0\). - Analyze how \(\frac{1}{k}\) changes as \(k\) increases.

1. Differentiate: \(f_k'(t)=ke^{-kt}(1-kt)\). The critical time is \(t=\frac{1}{k}\). Substitution gives \(f_k\left(\frac{1}{k}\right)=e^{-1}=\frac{1}{e}\), which is independent of \(k\). 2. \(f_k'(0)=k\). 3. The maximum occurs at \(t_{\max}=\frac{1}{k}\), so a larger \(k\) makes the maximum occur sooner. Since \(f_k'(0)=k\), a larger \(k\) also gives a steeper initial increase.

1. Maximum value: \(\frac{1}{e}\approx0.368\) 2. \(f_k'(0)=k\) 3. Larger \(k\) means a faster initial increase and an earlier maximum

- Solve \(f_k(x)=0\) and use the squared form. - Differentiate to count the critical points. - Compare the two ends of the graph when testing y-axis symmetry. - Evaluate the derivative at \(x=0\).

1. The function equals zero when \(e^x=k\), so \(x=\ln k\). Because the function is a square and is never negative, this zero is a tangency point. The statement is true for every \(k>0\). 2. Differentiate: \(f_k'(x)=2e^x(e^x-k)\). Since \(e^x>0\), the only critical point is \(x=\ln k\). The statement is true for every \(k>0\). 3. As \(x\to\infty\), \(f_k(x)\to\infty\), while as \(x\to-\infty\), \(f_k(x)\to k^2\). Therefore, the graph cannot be symmetric about the y-axis for any \(k>0\). The statement is false for every \(k\). 4. At \(x=0\), \(f_k'(0)=2(1-k)\). This value is positive when \(0<k<1\), zero when \(k=1\), and negative when \(k>1\). Therefore, the statement depends on \(k\).

1) True for every \(k>0\). 2) True for every \(k>0\). 3) False for every \(k>0\). 4) Depends on \(k\); the slope is positive exactly when \(0<k<1\).

- Factor the polynomial to find the intercepts. - Compare two different family members to locate their common points. - Use the derivative sign and second derivative to classify the nonzero critical number.

1. Factor: \(f_k(x)=kx^3(x-4)\). Because \(k\ne0\), every graph has intercepts \((0, 0)\) and \((4, 0)\). If \(k\ne m\) and \(f_k(x)=f_m(x)\), then \((k-m)(x^4-4x^3)=0\), so \(x=0\) or \(x=4\). Therefore, there are no other common points. 2. Differentiate: \(f_k'(x)=4kx^2(x-3)\). The critical numbers are \(0\) and \(3\). At \(x=0\), the derivative does not change sign, so there is no local extremum. At \(x=3\), \(f_k(3)=-27k\) and \(f_k''(3)=36k\). Thus, \((3,-27k)\) is a local minimum when \(k>0\) and a local maximum when \(k<0\). 3. Every local extremum has x-coordinate \(3\), so all local extrema lie on the vertical line \(x=3\).

a) The only common points are \((0, 0)\) and \((4, 0)\). b) The local extremum is \((3,-27k)\). It is a local minimum for \(k>0\) and a local maximum for \(k<0\). c) \(x=3\)

- Identify the domain before finding critical numbers. - Use the quotient rule to differentiate. - Analyze the discriminant of the numerator of the derivative. - Check that any derivative zeros lie in the function's domain and produce a sign change.

1. The domain is \(\mathbb{R} \setminus \{1\}\). 2. Apply the quotient rule: \(f_a'(x) = \frac{2x(x - 1) - (x^2 + a)}{(x - 1)^2} = \frac{x^2 - 2x - a}{(x - 1)^2}\). 3. Critical numbers must satisfy \(x^2 - 2x - a = 0\). The discriminant is \(D = 4 + 4a = 4(1 + a)\). 4. If \(a < -1\), then \(D < 0\), so the derivative has no real zeros and the function has no local extrema. 5. If \(a = -1\), the numerator of the derivative is \((x - 1)^2\), whose only zero is \(x = 1\). This value is not in the domain, so there are no local extrema. 6. If \(a > -1\), the derivative has two distinct zeros \(x = 1 \pm \sqrt{1 + a}\). Both are in the domain, and the numerator changes sign at each simple zero while the denominator remains positive. Therefore, the function has exactly two local extrema.

No local extrema for \(a \le -1\) Exactly two local extrema for \(a > -1\)

- Use the product and chain rules. - The exponential function has no zeros. - Identify which factor controls the zeros of the derivative. - A quadratic equation has at most two real solutions.

1. Apply the product and chain rules: \(f'(x)=(2ax+b)e^{-x}-(ax^2+bx+c)e^{-x}\). 2. Factor the exponential term: \(f'(x)=e^{-x}\bigl(-ax^2+(2a-b)x+b-c\bigr)\). 3. Since \(e^{-x}>0\), zeros of the derivative can occur only when the quadratic factor is zero. 4. Because \(a\neq0\), this factor is a quadratic polynomial and therefore has at most two real zeros. 5. Local extrema can occur only at zeros of the derivative, so \(f\) has at most two local extrema.

The derivative is \(f'(x)=e^{-x}Q(x)\), where \(Q\) is quadratic. Since \(e^{-x}\neq0\) and a quadratic has at most two real zeros, \(f\) can have at most two local extrema.

- Require the expression under the square root to be nonnegative. - Use the symmetry of a parabola or complete the square to find its vertex. - A strictly increasing outer function preserves the ordering of inner function values. - Apply the chain rule to verify the critical number.

1. a) The zeros of \(f\) are \(-1\) and \(7\). Since the parabola opens downward, \(f(x)\ge0\) on \([-1,7]\). Thus, \(D_g=[-1,7]\). 2. b) The vertex lies midway between the zeros at \(x=3\), and \(f(3)=16\). Thus, the vertex is \((3,16)\). 3. The square-root function is strictly increasing on \([0,\infty)\), so maximizing \(f(x)\) also maximizes \(\sqrt{f(x)}\). Therefore, \(g\) has its maximum at \(x=3\). 4. c) By the chain rule, \(g'(x)=\frac{-2x+6}{2\sqrt{-x^2+6x+7}}=\frac{3-x}{\sqrt{-x^2+6x+7}}\), for \(-1<x<7\). 5. The derivative is zero only at \(x=3\), confirming the maximum location. Also, \(g(3)=4\), so the maximum point of \(g\) is \((3,4)\).

a) \(D_g=[-1,7]\) b) The vertex of \(f\) is \((3,16)\); because the square-root function is strictly increasing, \(g\) is maximized at \(x=3\). c) \(g'(x)=\frac{3-x}{\sqrt{-x^2+6x+7}}\); the maximum point is \((3,4)\).

- Ground-level width is the distance between the x-intercepts. - Because the square root is increasing, maximize the radicand. - Compare critical points with domain endpoints. - The slope at a point is the derivative value there.

1. a) The graph meets the ground at \(x=-3\) and \(x=6\), so the width is \(6-(-3)=9\,\text{m}\). 2. Let \(u(x)=(x+3)^2(6-x)=-x^3+27x+54\). Because the square root is increasing, \(f\) is maximized where \(u\) is maximized. 3. Since \(u'(x)=-3x^2+27\), the interior critical number is \(x=3\). Comparing \(u(3)\) with the endpoint values at \(x=-3\) and \(x=6\) shows the maximum occurs at \(x=3\). 4. b) \(f(3)=\frac{1}{4}\sqrt{108}=\frac{3\sqrt{3}}{2}\approx2.60\,\text{m}\). 5. c) For interior points, \(f'(x)=\frac{-3x^2+27}{8\sqrt{-x^3+27x+54}}\). Thus, \(f'(0)=\frac{9}{8\sqrt{6}}\approx0.46\).

a) \(9\,\text{m}\) b) \(2.60\,\text{m}\) c) \(f'(0)=\frac{9}{8\sqrt{6}}\approx0.46\)

- Substitute \(z=x^2\) to solve the quartic intercept equation as a quadratic. - Apply the chain rule to the square-root function. - Compare all critical-point values and endpoint values. - A horizontal tangent has slope \(0\).

1. a) Set the radicand equal to \(0\). With \(z=x^2\), \(-z^2+8z+9=0\), so \(z=9\) or \(z=-1\). Thus, the real intercepts are \(x=\pm3\), and the base width is \(6\) units. 2. For interior points, \(f'(x)=\frac{-2x^3+8x}{\sqrt{-x^4+8x^2+9}}\). Critical numbers occur at \(x=0\) and \(x=\pm2\). 3. The function values are \(f(0)=3\) and \(f(\pm2)=5\), while the endpoint values are \(0\). Therefore, the maximum points are \((-2,5)\) and \((2,5)\). 4. c) Since \(f'(0)=0\), the tangent is horizontal. Because \(f(0)=3\), its equation is \(y=3\).

a) \(6\) units b) \((-2,5)\) and \((2,5)\) c) \(f'(0)=0\), so the tangent line is \(y=3\).

- Relate local extrema to the real zeros of the first derivative. - Use the discriminant of the quadratic derivative. - Determine what happens when the derivative has a double zero.

1. Differentiate: \(g_a'(x) = 3x^2 - 6ax + 6a - 3\). 2. The derivative has discriminant \(D = (-6a)^2 - 4 \cdot 3(6a - 3) = 36a^2 - 72a + 36 = 36(a - 1)^2\). 3. Since \(D \ge 0\) for every \(a\), the only way to avoid two distinct critical numbers is to have \(D = 0\). 4. Solve \(36(a - 1)^2 = 0\): \(a = 1\). 5. For \(a = 1\), \(g_1'(x) = 3(x - 1)^2\), so the derivative does not change sign at \(x = 1\). Therefore, the graph has no local extremum there. Also, \(g_1''(x) = 6x - 6\) changes sign at \(x = 1\), so the point is a stationary inflection point.

Yes. For \(a = 1\), the function has no local extrema.

- Determine which powers can appear in an even polynomial. - Differentiate and use the symmetry of the resulting odd function. - Focus on the sign of the first derivative immediately to the left and right of \(0\). - Apply the first derivative test to that sign change.

1. Y-axis symmetry means \(f\) is even, so its polynomial expression contains only even powers of \(x\). 2. Differentiating produces only odd powers of \(x\). Every term of \(f'\) therefore contains a positive power of \(x\), so \(f'(0)=0\). 3. The derivative of an even function is odd, so \(f'(-x)=-f'(x)\). 4. Because \(f\) is nonconstant, the first nonzero term of \(f'\) near \(0\) has odd degree. Thus \(f'\) has opposite signs on the two sides of \(0\): it changes either from negative to positive or from positive to negative. By the first derivative test, \(f\) therefore has a local extremum at \(0\), not a stationary inflection point.

a) An even polynomial contains only even powers, so its derivative contains only odd powers and \(f'(0)=0\). b) The derivative is odd and changes sign at \(0\). Therefore, \(f\) has a local extremum there, not a stationary inflection point.

- Test statements with lines, parabolas, and cubic functions. - Distinguish necessary conditions from sufficient conditions. - An open interval may still contain an interior point where a maximum is attained. - Fermat's theorem applies to differentiable functions at interior local extrema.

1. Statement (1) is true. Strict increase prevents any interior point from being locally greatest or least, and on all of \(\mathbb{R}\) there is no greatest or least value. 2. Statement (2) is false. For \(f(x)=x^3\), \(f'(0)=0\), but \(x=0\) is not a local extremum. 3. Statement (3) is false. The function \(f(x)=-x^2\) has an absolute maximum at \(x=0\) on \((-1,1)\). 4. Statement (4) is true by Fermat's theorem.

(1) True (2) False; for example, \(f(x)=x^3\) (3) False; for example, \(f(x)=-x^2\) on \((-1,1)\) (4) True

- Use continuity on the interval between the two minima. - Consider the end behavior of an even-degree polynomial. - An absolute extremum in the interior is also a local extremum.

1. Statement a) is true. On the closed interval between two local minima, continuity guarantees a maximum. It occurs between the minima; if the maximum is attained on a flat interval, interior points of that interval are still local maxima. 2. Statement b) is true. A quartic has the same end behavior on both sides, approaching either \(+\infty\) at both ends or \(-\infty\) at both ends. By continuity, it attains an absolute minimum or maximum, which is also a local extremum.

a) True b) True

- Collect the terms containing \(a\) to find parameter-independent points. - Use the first derivative condition at \(x = -1\). - Analyze the discriminant of the quadratic derivative. - Show that the discriminant is positive for every real \(a\).

1. Group the parameter terms: \(f_a(x) = x^3 - x + a(x^2 - x)\). A point is shared by every graph when \(x^2 - x = 0\), so \(x = 0\) or \(x = 1\). In both cases, \(f_a(x) = 0\). The common points are \((0, 0)\) and \((1, 0)\). 2. Differentiate: \(f_a'(x) = 3x^2 + 2ax - a - 1\). Require \(f_a'(-1) = 0\): \(3 - 2a - a - 1 = 2 - 3a = 0\), so \(a = \frac{2}{3}\). 3. The second derivative is \(f_a''(x) = 6x + 2a\). At \(a = \frac{2}{3}\) and \(x = -1\), it is \(-\frac{14}{3} \ne 0\), so the critical point is a local extremum. 4. For any \(a\), the discriminant of the quadratic derivative is \(D = (2a)^2 - 4 \cdot 3(-a - 1) = 4(a^2 + 3a + 3)\). 5. Complete the square: \(a^2 + 3a + 3 = \left(a + \frac{3}{2}\right)^2 + \frac{3}{4} > 0\). Thus, \(D > 0\) for every real \(a\). The derivative always has two distinct real zeros and changes sign at each, so every graph has exactly two local extrema.

a) \((0, 0)\) and \((1, 0)\) b) \(a = \frac{2}{3}\) c) The derivative has two distinct real zeros for every \(a\), so every graph has exactly two local extrema.

- Identify the two factors before applying the product rule. - Differentiate the exponential factor with the chain rule. - A local extremum occurs where the first derivative is zero. - A point on the y-axis has x-coordinate zero.

1. By the product rule, \(f_t'(x)=2e^{0.5x}+(2x-t)(0.5)e^{0.5x}=(x+2-0.5t)e^{0.5x}\). 2. Since \(e^{0.5x}>0\), the equation \(f_t'(x)=0\) reduces to \(x+2-0.5t=0\). Thus, the critical number is \(x=0.5t-2\). 3. The second derivative is \(f_t''(x)=(0.5x+2-0.25t)e^{0.5x}\). At \(x=0.5t-2\), it equals \(e^{0.25t-1}>0\), so the extremum is a local minimum. 4. A point on the y-axis has x-coordinate \(0\). Set \(0.5t-2=0\) to obtain \(t=4\).

1. \(f_t'(x)=(x+2-0.5t)e^{0.5x}\) 2. The x-coordinate is \(0.5t-2\). 3. \(t=4\)

- Solve the first-derivative equation using the range of \(e^x\). - For \(u^2=v^2\), consider both \(u=v\) and \(u=-v\). - A horizontal tangent has slope zero.

1. The derivative is \(f_a'(x)=-2e^x(a-e^x)\). Since \(e^x>0\), a critical number requires \(e^x=a\), which is possible exactly when \(a>0\). Then \(x=\ln(a)\). 2. The second derivative is \(f_a''(x)=4e^{2x}-2ae^x\). At \(x=\ln(a)\), it equals \(2a^2>0\), so the point is a local minimum. Its y-coordinate is \(0\), giving \((\ln(a), 0)\). 3. For an intersection, \((a_1-e^x)^2=(a_2-e^x)^2\). Because \(a_1\ne a_2\), the equal-base case is impossible, so \(a_1-e^x=-(a_2-e^x)\). Thus \(e^x=\frac{a_1+a_2}{2}\). An intersection exists exactly when \(a_1+a_2>0\), and then \(x=\ln\left(\frac{a_1+a_2}{2}\right)\). 4. A horizontal tangent at \(x=0\) requires \(f_a'(0)=-2(a-1)=0\), so \(a=1\).

1. For \(a>0\), the local minimum is \((\ln(a), 0)\). 2. An intersection exists when \(a_1+a_2>0\), at \(x=\ln\left(\frac{a_1+a_2}{2}\right)\). 3. \(a=1\)

- Apply the product rule and replace \(\cos^2x\) using the Pythagorean identity. - Horizontal tangents occur where the first derivative is zero. - In part c, substitute \(u=\sin x\). - Reject any solution outside the range \([-1, 1]\) of sine. - A positive sine value occurs twice in one full cycle.

1. Apply the product rule: \(f_k'(x)=-\sin x(k+\sin x)+\cos^2x\). Using \(\cos^2x=1-\sin^2x\), \(f_k'(x)=1-k\sin x-2\sin^2x\). 2. For \(k=0\), horizontal tangents satisfy \(1-2\sin^2x=0\). Since \(1-2\sin^2x=\cos(2x)\), solve \(\cos(2x)=0\). In \([0, 2\pi)\), this gives \(x\in\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}\). 3. For \(k=\frac{7}{3}\), set the derivative equal to zero: \(1-\frac{7}{3}\sin x-2\sin^2x=0\). Let \(u=\sin x\). Then \(6u^2+7u-3=0\), so \(u=\frac{1}{3}\) or \(u=-\frac{3}{2}\). The second value is outside the range of sine. Thus, \(x=\arcsin\left(\frac{1}{3}\right)\approx0.34\) or \(x=\pi-\arcsin\left(\frac{1}{3}\right)\approx2.80\). The second derivative is nonzero at both inputs, so both are local extrema.

a) \(f_k'(x)=1-k\sin x-2\sin^2x\) b) \(x\in\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}\) c) \(x\approx0.34\) and \(x\approx2.80\)

- Since \(a\ne0\), determine when the trigonometric factor equals zero. - Rewrite each trigonometric equation using tangent. - Local extrema can occur where the first derivative equals zero. - Examine what happens to the factor \(a\) after setting the derivative equal to zero.

1. Since \(a\ne0\), the zeros satisfy \(\sin x-\sqrt{3}\cos x=0\). Thus, \(\tan x=\sqrt{3}\), which gives \(x=\frac{\pi}{3}\) and \(x=\frac{4\pi}{3}\) in \([0, 2\pi]\). 2. Differentiate: \(f_a'(x)=a(\cos x+\sqrt{3}\sin x)\). Critical numbers satisfy \(\cos x+\sqrt{3}\sin x=0\), so \(\tan x=-\frac{1}{\sqrt{3}}\). In \([0, 2\pi]\), the solutions are \(x=\frac{5\pi}{6}\) and \(x=\frac{11\pi}{6}\). Both are local extrema. Their classifications reverse when the sign of \(a\) changes. 3. The factor \(a\) can be divided out of the equation \(f_a'(x)=0\) because \(a\ne0\). The remaining equation contains no \(a\), so the critical numbers do not depend on the parameter.

a) \(x=\frac{\pi}{3}\) and \(x=\frac{4\pi}{3}\) b) \(x=\frac{5\pi}{6}\) and \(x=\frac{11\pi}{6}\) c) The nonzero factor \(a\) cancels from the equation \(f_a'(x)=0\), leaving an equation independent of \(a\).

- Factor the first derivative to find the critical numbers. - Use a sign chart to classify the critical points for the different parameter cases. - Eliminate the parameter from the coordinates of the off-axis extremum.

1. Differentiate and factor: \(f_a'(x)=x(x-a+2)e^x\). The critical numbers are \(x=0\) and \(x=a-2\). 2. When \(a=2\), these critical numbers coincide and the derivative does not change sign. The point \((0,2)\) is a stationary inflection point, so there are no local extrema. 3. When \(a>2\), the local maximum is \(E_1=(0,a)\), and the local minimum is \(E_2=\left(a-2,(4-a)e^{a-2}\right)\). When \(a<2\), the classifications are reversed. 4. For the extremum not on the y-axis, let \(x=a-2\), so \(a=x+2\). Then \(y=(4-a)e^{a-2}=(2-x)e^x\). Because \(a\ne2\), the locus is \(y=(2-x)e^x\), with \(x\ne0\).

a) If \(a=2\), there are no local extrema; \((0,2)\) is a stationary inflection point. If \(a>2\), \((0,a)\) is a local maximum and \(\left(a-2,(4-a)e^{a-2}\right)\) is a local minimum. If \(a<2\), the classifications are reversed. b) \(y=(2-x)e^x\), with \(x\ne0\)

- Compare nearby function values on both sides of each point. - A corner can still be a local extremum. - Check whether the graph changes from increasing to decreasing or vice versa. - A horizontal tangent alone does not guarantee an extremum.

1. At \(A\), the graph has a corner at \(x=-2\), so \(f\) is not differentiable there. However, nearby values on both sides are greater than \(f(-2)\), so \(A\) is a local minimum. Differentiability is not required for a local extremum. 2. At \(B\), the graph is smooth and changes from increasing to decreasing at \(x=0\). Thus \(f'(0)=0\) and \(B\) is a local maximum. 3. At \(C\), the tangent is horizontal at \(x=2\), but the graph is decreasing on both sides. Therefore, \(C\) is not an extremum; it is a stationary inflection point. A zero derivative is not sufficient for an extremum.

\(A\): local minimum \(B\): local maximum \(C\): stationary inflection point, not a local extremum

- Read the \(x\)-coordinates of the local maximum and local minimum from the graph. - The first derivative equals zero at a local extremum. - Substitute one critical number into the derivative to solve for \(a\). - Use the discriminant to determine when the quadratic derivative does not have two distinct real zeros.

1. Differentiate: \(f_a'(x) = 0.75x^2 - 1.5ax + 2a\). 2. From the graph, the local maximum occurs at \(x = 2\) and the local minimum occurs at \(x = 4\). Use either critical number. Setting \(f_a'(2) = 0\) gives \(0.75(2)^2 - 1.5a(2) + 2a = 3 - a = 0\), so \(a = 3\). 3. This agrees with the other critical number: \(f_a'(4) = 12 - 4a = 0\), which also gives \(a = 3\). 4. The function has no local extrema when the quadratic derivative has at most one distinct real zero. Using the standard discriminant, \(D = (-1.5a)^2 - 4(0.75)(2a) = 2.25a^2 - 6a = 2.25a\left(a - \frac{8}{3}\right)\). 5. Solve \(D \le 0\): \(0 \le a \le \frac{8}{3}\). At the endpoint values, the derivative has a double zero and does not change sign.

a) \(a = 3\) b) \(0 \le a \le \frac{8}{3}\)

- Use the sign of the leading coefficient to determine the end behavior. - Find the derivative and evaluate the slope at \(x = 0\). - Set the derivative equal to zero at \(x = 2\) and solve for \(k\).

1. Since \(k > 0\), the leading coefficient is positive. Therefore, \(f_k(x) \to \infty\) as \(x \to \infty\), and \(f_k(x) \to -\infty\) as \(x \to -\infty\). Only graph 1 has this end behavior. 2. Also, \(f_k'(x) = 3kx^2 - 3\), so \(f_k'(0) = -3\). The graph must be decreasing as it passes through the origin. Again, only graph 1 has this behavior. 3. For a local extremum at \(x = 2\), require \(f_k'(2) = 0\): \(3k(2)^2 - 3 = 12k - 3 = 0\). Thus, \(k = \frac{1}{4}\). 4. For this value, \(f_k''(2) = 6k(2) = 3 > 0\), so the extremum at \(x = 2\) is a local minimum.

a) Graph 1. It has the required end behavior for a positive leading coefficient and is decreasing at the origin, where the slope is \(-3\). b) \(k = \frac{1}{4}\)

- Use the y-intercept to determine one parameter first. - Differentiate the product and then impose the critical-point condition at the given maximum input. - Check that the derivative changes from positive to negative for the resulting parameters.

1. The y-intercept condition gives \(f(0)=b=2\). 2. By the product and chain rules, \(f'(x)=(a-0.5ax-0.5b)e^{-0.5x}\). 3. A local maximum at \(x=1\) requires \(f'(1)=0\). Substituting \(b=2\) gives \((0.5a-1)e^{-0.5}=0\), so \(a=2\). 4. With \(a=b=2\), \(f'(x)=(1-x)e^{-0.5x}\), which changes from positive to negative at \(x=1\). Therefore, the stated point is a local maximum.

\(a=2\), \(b=2\)

- Apply the definition using function values in a small neighborhood. - Differentiability is not required for a local extremum. - Compare the values on both sides of \(x_1\). - On a constant interval, nearby values equal the value at the marked point.

1. A function has a local maximum at \(x_0\) when \(f(x)\le f(x_0)\) for all \(x\) sufficiently close to \(x_0\). It has a local minimum when \(f(x)\ge f(x_0)\) nearby. 2. At \(x_1=-2\), \(f(-2)=4\), and every nearby function value is less than \(4\). Therefore, \(x=-2\) is a local-maximum location. The corner does not prevent a local extremum. 3. At \(x_2=1.5\), the function is constant with value \(1\) throughout a neighborhood of \(1.5\). Thus, every nearby value is both less than or equal to and greater than or equal to \(f(1.5)\). Under the stated non-strict definition, \(x=1.5\) is both a local-maximum and a local-minimum location.

\(x_1=-2\): local maximum \(x_2=1.5\): both a local maximum and a local minimum under the non-strict definition

- Must a function have derivative \(0\) in order to have a local extremum? - Compare the function value at \(x_1=-1\) with nearby values. - Use the shape and opening direction of the parabola near \(x_2=3\).

1. Near \(x_1=-1\), the graph has a V-shaped point and \(f(x)\ge f(-1)=0\). Therefore, \(f\) has a local minimum at \(x_1=-1\). The function is not differentiable there, but a local extremum is defined by nearby function values and does not require differentiability. 2. At \(x_2=3\), the graph is the vertex of a downward-opening parabola. The function changes from increasing to decreasing, so \(f\) has a local maximum there.

a) Yes. The function has a local minimum at \(x_1=-1\), even though it is not differentiable there. b) The function has a local maximum at \(x_2=3\).

- The exponential factor is never zero. - The zero comes from the linear factor. - Use the product and chain rules. - At a local minimum, the first derivative is zero and the second derivative is positive.

1. Since \(e^{0.5x}>0\), the only zero of \(g_a\) is \(x=a\). Graph 1 has its zero at \(x=1\), so it represents \(a=1\). Graph 2 has its zero at \(x=-1\), so it represents \(a=-1\). 2. Differentiate: \(g_a'(x)=e^{0.5x}(1+0.5x-0.5a)\). Requiring a critical point at \(x=4\) gives \(1+2-0.5a=0\), so \(a=6\). 3. For \(a=6\), \(g_a''(4)=0.5e^2>0\), confirming a local minimum.

a) Graph 1: \(a=1\); Graph 2: \(a=-1\) b) \(a=6\)

- Start with the graph whose y-intercept is zero. - Use a visible extremum to create one function-value equation and one derivative equation. - Add a constant for a vertical translation.

1. Use the graph through the origin as the unshifted base graph. It has a local minimum at \((1, -1)\). 2. Since \(g(1)=-1\), \(a+b=-1\). Since the derivative \(g'(x)=3ax^2+b\) is zero at \(x=1\), \(3a+b=0\). 3. Solving gives \(a=0.5\) and \(b=-1.5\). Thus \(g(x)=0.5x^3-1.5x\). 4. A vertical shift by \(c\) gives \(f_c(x)=0.5x^3-1.5x+c\).

\(f_c(x)=0.5x^3-1.5x+c\)

- A quotient is zero only when its numerator is zero. - Track the sign of the denominator on each side of \(x=-2\). - Compare exponential and linear growth for large positive \(x\). - Use the quotient rule to locate the turning point after the end behavior is established.

1. A quotient is zero only when its numerator is zero. Since \(e^x>0\) for every real \(x\), \(h\) has no zeros. 2. Near \(x=-2\), the numerator approaches \(e^{-2}>0\). The denominator approaches \(0^+\) from the right and \(0^-\) from the left, so \(\lim_{x\to-2^+}h(x)=\infty\) and \(\lim_{x\to-2^-}h(x)=-\infty\). As \(x\to\infty\), exponential growth dominates the linear denominator, so \(h(x)\to\infty\). As \(x\to-\infty\), \(e^x\to0\), so \(h(x)\to0\). 3. By the quotient rule, \(h'(x)=\frac{e^x(x+1)}{(x+2)^2}\). The only critical number is \(x=-1\). The derivative changes from negative to positive, so the graph has a local minimum at \(\left(-1,\frac{1}{e}\right)\).

a) No zeros b) \(\lim_{x\to-2^+}h(x)=\infty\), \(\lim_{x\to-2^-}h(x)=-\infty\), \(\lim_{x\to\infty}h(x)=\infty\), and \(\lim_{x\to-\infty}h(x)=0\) c) Local minimum at \(\left(-1,\frac{1}{e}\right)\)

- Use the initial value at \(x=0\) to find \(b\). - At an interior maximum, the first derivative is zero. - Differentiate with the product and chain rules.

1. The initial condition gives \(c(0)=b=10\,\text{mg/L}\). 2. Differentiate: \(c'(x)=e^{-0.1x}(a-0.1ax-1)\). 3. A maximum at \(x=5\) requires \(c'(5)=0\). Since the exponential factor is nonzero, \(a-0.5a-1=0\), so \(a=2\,\text{mg}/(\text{L}\cdot\text{h})\). 4. With \(a=2\), \(c'(x)=e^{-0.1x}(1-0.2x)\), which changes from positive to negative at \(x=5\). Thus, the point is a maximum.

\(a=2\,\text{mg}/(\text{L}\cdot\text{h})\) and \(b=10\,\text{mg/L}\)

- Use the necessary condition for an interior maximum in part a). - For part b), identify where the graph lies on or above the horizontal threshold line. - For part c), examine what the formula predicts after a very long time. - Also consider what the model predicts at \(t=0\).

1. Differentiate: \(c'(t)=\frac{12(t^2+4)-24t^2}{(t^2+4)^2}=\frac{48-12t^2}{(t^2+4)^2}\). 2. Set the derivative equal to zero: \(48-12t^2=0\), so \(t=2\) for \(t\ge 0\). The derivative changes from positive to negative, so the concentration is maximized at \(t=2\). 3. The maximum concentration is \(c(2)=\frac{24}{8}=3\,\text{mg/L}\). 4. The effectiveness endpoints satisfy \(\frac{12t}{t^2+4}=1.5\). This simplifies to \(t^2-8t+4=0\), giving \(t=4\pm 2\sqrt{3}\). 5. Therefore, the graph is at or above the threshold on \([4-2\sqrt{3},4+2\sqrt{3}]\), approximately \([0.54,7.46]\). The duration is \(4\sqrt{3}\approx 6.93\) hours. 6. As \(t\to\infty\), \(c(t)\to 0\), but the model remains positive for every finite \(t>0\). It is an idealized curve that does not account for individual metabolism, body size, detection limits, or more complex absorption and elimination processes.

a) The maximum occurs at \(t=2\,\text{h}\), and the maximum concentration is \(3\,\text{mg/L}\). b) The medication is effective for \(t\in[4-2\sqrt{3},4+2\sqrt{3}]\), approximately from \(0.54\) to \(7.46\) hours, for a duration of about \(6.93\) hours. c) The model approaches zero but never reaches it at a finite positive time and does not capture individual or physiological complexity.

- A closed, bounded interval is only one part of the theorem's hypotheses. - Compare the right-hand behavior near \(0\) with the assigned value \(f(0)\). - Decide separately whether a least value and a greatest value are actually attained.

1. The function is not continuous at \(x=0\): values \(f(x)=1/x\) become arbitrarily large as \(x\to0^+\), rather than approaching \(f(0)=0\). 2. Therefore, the continuity hypothesis of the Extreme Value Theorem fails. 3. Since \(f(0)=0\) and \(f(x)\ge1\) for \(0<x\le1\), the absolute minimum is \(0\), attained at \(x=0\). 4. As \(x\to0^+\), \(f(x)\to\infty\), so the function has no absolute maximum.

a) The function is discontinuous at \(x=0\), so the Extreme Value Theorem does not apply. b) Absolute minimum: \(0\) at \(x=0\); no absolute maximum

- The absolute-value function is continuous even though it can have a corner. - Build the candidate set from endpoints and every kind of interior critical number. - Rewrite the function on the two sides of the corner to understand its derivative.

1. The function is continuous on the closed, bounded interval \([-1,3]\), so the Extreme Value Theorem guarantees both absolute extrema. 2. For \(x<1\), \(f(x)=1-\frac{x}{2}\), so \(f'(x)=-\frac12\). For \(x>1\), \(f(x)=\frac32x-1\), so \(f'(x)=\frac32\). 3. At \(x=1\), the derivative does not exist, so \(x=1\) is an interior critical number. There are no derivative-zero critical numbers. 4. Evaluate the candidates: \(f(-1)=\frac32\), \(f(1)=\frac12\), and \(f(3)=\frac72\). 5. The absolute minimum is \(\frac12\) at \(x=1\), and the absolute maximum is \(\frac72\) at \(x=3\).

Absolute minimum: \(\frac12\) at \(x=1\); absolute maximum: \(\frac72\) at \(x=3\)

- 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) \(c \ge 3\)

- What happens to the degree of the function when the leading coefficient is zero? - Differentiate and determine how many real zeros the derivative has. - Use the discriminant for the quadratic derivative when \(a \ne 0\). - Check whether the derivative changes sign at each critical number.

1. First consider \(a = 0\). Then \(g_0(x) = 3x^2 + 3x\), a quadratic function. Its derivative is \(g_0'(x) = 6x + 3\), which changes sign at \(x = -\frac{1}{2}\). Thus, \(g_0\) has exactly one local extremum. 2. For \(a \ne 0\), differentiate: \(g_a'(x) = 3ax^2 + 6x + 3\). 3. The discriminant of the derivative equation is \(D = 6^2 - 4 \cdot 3a \cdot 3 = 36(1 - a)\). 4. If \(a < 1\) and \(a \ne 0\), then \(D > 0\). The derivative has two distinct real zeros and changes sign at both, so the graph has two local extrema. 5. If \(a = 1\), then \(g_1'(x) = 3(x + 1)^2\). The derivative has a double zero without a sign change, so the graph has no local extremum. 6. If \(a > 1\), then \(D < 0\), so the derivative has no real zeros and the graph has no local extrema. 7. Combining the cases, there are two local extrema for \(a < 1\) with \(a \ne 0\), one local extremum for \(a = 0\), and no local extrema for \(a \ge 1\).

Two local extrema: \(a < 1\) with \(a \ne 0\) One local extremum: \(a = 0\) No local extrema: \(a \ge 1\)

- Factor the first derivative to find the critical numbers. - Determine when the two critical numbers are equal, then check whether the derivative changes sign. - Group the function into a parameter-free part plus \(a\) times another expression. - Common points occur where the coefficient of \(a\) is zero.

1. Differentiate: \(f_a'(x) = 3x^2 - 6ax + 12a - 12 = 3(x - 2)(x - 2a + 2)\). 2. The critical numbers are \(x = 2\) and \(x = 2a - 2\). They are distinct when \(a \ne 2\), and each then produces a local extremum. 3. When \(a = 2\), the critical numbers coincide at \(x = 2\), and \(f_2'(x) = 3(x - 2)^2\). The derivative does not change sign, so there is no local extremum. Since \(f_2''(x) = 6(x - 2)\) changes sign at \(x = 2\), the graph has a stationary inflection point there. Also, \(f_2(2) = 18\), so the point is \((2, 18)\). 4. Separate the parameter term: \(f_a(x) = x^3 - 12x + 10 + a(-3x^2 + 12x)\). A point is common to every graph when the coefficient of \(a\) is zero. 5. Solve \(-3x^2 + 12x = -3x(x - 4) = 0\): \(x = 0\) or \(x = 4\). The corresponding values are \(f_a(0) = 10\) and \(f_a(4) = 26\). Thus, the common points are \((0, 10)\) and \((4, 26)\).

a) \(x = 2\) and \(x = 2a - 2\); for \(a = 2\), these coincide. b) \(a = 2\); the graph has a stationary inflection point at \((2, 18)\). c) \((0, 10)\) and \((4, 26)\)

- Apply both the product rule and the chain rule. - Factor the intersection equation so that one factor is \(x\). - Analyze the maximum value of \(w(x)=\frac{h(x)}{x}\) for \(x>0\). - Count distinct intersection points, not root multiplicity.

1. Differentiate using the product and chain rules: \(h'(x)=e^{-0.5x}(2x-0.5x^2)=xe^{-0.5x}(2-0.5x)\). 2. The interior critical number is \(x=4\). The derivative changes from positive to negative there, so the point is a local maximum. Its coordinates are \(\left(4,16e^{-2}\right)\). 3. Intersections satisfy \(x^2e^{-0.5x}=kx\), or \(x\left(xe^{-0.5x}-k\right)=0\). Thus \(x=0\) is always one intersection. 4. For positive intersections, analyze \(w(x)=xe^{-0.5x}\). Its derivative is \(w'(x)=(1-0.5x)e^{-0.5x}\), so \(w\) has a maximum at \(x=2\), where \(w(2)=\frac{2}{e}\). 5. If \(0<k<\frac{2}{e}\), the equation \(w(x)=k\) has two positive solutions, for a total of three intersections. If \(k=\frac{2}{e}\), it has one positive solution, for a total of two intersections. If \(k\le0\) or \(k>\frac{2}{e}\), it has no positive solution, so the origin is the only intersection.

a) \(\left(4,16e^{-2}\right)\). b) If \(0<k<\frac{2}{e}\), there are \(3\) intersections. If \(k=\frac{2}{e}\), there are \(2\) intersections. If \(k\le0\) or \(k>\frac{2}{e}\), there is \(1\) intersection.

- Compare the signs of both sides when \(x\leq0\). - For \(x>0\), take logarithms and isolate a constant. - Find the maximum of \(\frac{\ln(x)}{x}\). - Compare \(\frac{1}{n}\) with that maximum.

1. If \(x\leq0\), then \(x^n\leq0\) because \(n\) is odd, while \(e^x>0\). Thus, there are no nonpositive solutions. 2. For \(x>0\), take natural logarithms: \(x=n\ln(x)\), or \(\frac{\ln(x)}{x}=\frac{1}{n}\). 3. Let \(q(x)=\frac{\ln(x)}{x}\). Then \(q'(x)=\frac{1-\ln(x)}{x^2}\), so \(q\) increases on \((0, e)\), decreases on \((e, \infty)\), and has maximum \(q(e)=\frac{1}{e}\). 4. Since \(n\geq3\), \(0<\frac{1}{n}\leq\frac{1}{3}<\frac{1}{e}\). Also, \(q(x)\to-\infty\) as \(x\to0^+\) and \(q(x)\to0^+\) as \(x\to\infty\). 5. Therefore, the horizontal line \(y=\frac{1}{n}\) intersects the graph of \(q\) once in \((1, e)\) and once in \((e, \infty)\). Hence, the original equation has exactly two real solutions.

There are no solutions for \(x\leq0\), and exactly two solutions for \(x>0\). Therefore, \(e^x=x^n\) has exactly two real solutions.

- Substitute the given point into the function. - Compare two functions in the family to analyze common points. - A horizontal stretch by \(c\) replaces \(x\) with \(x/c\). - Use the critical-point condition to simplify the y-coordinate of each extremum.

1. Substitute \((2, 2)\): \(2=2e^{1-4k}\). Thus, \(e^{1-4k}=1\), so \(k=\frac14\). The exponential equation has only this solution. 2. For \(k=0\), \(f_0(x)=ex\), so the slope is \(e\). 3. Every graph passes through the origin. The second statement shows that no two distinct graphs share any other point. Therefore, the origin is the only common point of the family. 4. A horizontal stretch by \(c\) followed by a vertical stretch by \(c\) gives \(y=c f_k\left(\frac{x}{c}\right)\) \(=xe^{1-(k/c^2)x^2}\). This is \(f_{k^*}(x)\) with \(k^*=\frac{k}{c^2}\), so the transformed graph remains in the family. 5. Differentiate: \(f_k'(x)=(1-2kx^2)e^{1-kx^2}\). For \(k>0\), the two critical points satisfy \(kx^2=\frac12\). The derivative changes sign at both points, so they are local extrema. At either critical point, \(f_k(x)=xe^{1-1/2}=\sqrt e\, x\). Therefore, every local extremum lies on \(y=\sqrt e\, x\).

a) \(k=\frac14\) b) The slope is \(e\). c) All graphs intersect only at the origin. d) The stretched graph is \(f_{k/c^2}\). e) Every local extremum satisfies \(y=\sqrt e\, x\).

- Analyze the rational function's domain, zeros, and end behavior first. - Use the derivative sign around the critical point to classify it. - To compare two family members, set their function values equal at the same \(x\). - On the closed interval in part d), list the endpoints and every interior critical number before comparing values.

1. The denominator cannot be zero, so \(D_k=\mathbb R\setminus\{0\}\). The x-intercept is \(x=\frac{k}{2}\), the vertical asymptote is \(x=0\), and \(f_k(x)\to0\) as \(x\to\pm\infty\), so the horizontal asymptote is \(y=0\). 2. \(f_k'(x)=\frac{2x-2k}{x^3}\). The only critical point is \(x=k\), where \(f_k(k)=-\frac1k\). The derivative sign shows a local minimum for \(k>0\) and a local maximum for \(k<0\). 3. If \(f_{k_1}(x)=f_{k_2}(x)\) at a domain point, then \(k_1=k_2\). Thus distinct graphs do not intersect. 4. For \(k>0\), \(f_k\) is continuous on \([k/2,k]\), so the extreme value theorem applies. The candidates are the endpoints and any interior critical numbers. Here \(x=k\) is an endpoint, so compare \(f_k(k/2)=0\) and \(f_k(k)=-\frac1k\). The absolute maximum is \(0\) at \(x=k/2\), and the absolute minimum is \(-\frac1k\) at \(x=k\).

a) \(D_k=\mathbb R\setminus\{0\}\); x-intercept \(x=\frac{k}{2}\); vertical asymptote \(x=0\); horizontal asymptote \(y=0\). b) Extremum: \(\left(k,-\frac1k\right)\); local minimum for \(k>0\), local maximum for \(k<0\). c) Distinct graphs do not intersect. d) For \(k>0\) on \([k/2,k]\): absolute maximum \(0\) at \(x=k/2\); absolute minimum \(-\frac1k\) at \(x=k\).

- Determine when the first derivative can equal zero. - Eliminate the parameter from the coordinates of the extrema. - Write the tangent equation at a general point \(x_0\). - Require the origin to lie on that tangent.

1. Differentiate: \(f_a'(x)=3-ae^x\) and \(f_a''(x)=-ae^x\). The critical-point equation is \(e^x=\frac3a\). This equation has a solution only when \(a>0\), in which case \(x=\ln\left(\frac3a\right)\). At this point, the second derivative is \(-3<0\), so the extremum is a local maximum. Its coordinates are \(H=\left(\ln\left(\frac3a\right), 3\ln\left(\frac3a\right)-3\right)\). For \(a<0\), there are no extrema. 2. Writing the maximum point as \((x, y)\), its coordinates satisfy \(y=3x-3\). Thus, the line containing all extrema is \(y=3x-3\). 3. Let \(x_0\) be a point of tangency. The tangent at \(x_0\) passes through the origin exactly when \(f_a(x_0)-x_0f_a'(x_0)=0\). Substitution gives \(ae^{x_0}(x_0-1)=0\). Since \(a\ne0\) and \(e^{x_0}>0\), the unique solution is \(x_0=1\). Therefore, the unique point of tangency is \(P_a=(1, 3-ae)\).

a) For \(a>0\), the graph has a local maximum at \(\left(\ln\left(\frac3a\right), 3\ln\left(\frac3a\right)-3\right)\). For \(a<0\), it has no extrema. b) \(y=3x-3\) c) \(P_a=(1, 3-ae)\)

- A tangency point on the x-axis is a zero that is also a local minimum. - Factor the first derivative to find its zeros. - Compare the x- and y-coordinates of \(H_a\). - Substitute the specified x-value into the derivative.

1. Because \(x>0\), \(f_a(x)=0\) exactly when \(\ln x-a=0\). Thus, \(x=e^a\). Since \(f_a(x)\ge0\) for all \(x>0\), the graph is tangent to the x-axis at \(B_a=(e^a, 0)\). 2. Differentiate: \(f_a'(x)=(\ln x-a)(\ln x-a+2)\). Besides \(x=e^a\), the other critical point satisfies \(\ln x=a-2\), so \(x=e^{a-2}\). The derivative changes from positive to negative there, so this point is a local maximum. Its y-coordinate is \(f_a(e^{a-2})=4e^{a-2}\). Therefore, \(H_a=(e^{a-2}, 4e^{a-2})\). 3. The coordinates of \(H_a\) satisfy \(y=4x\), so the locus is the line \(y=4x\). 4. Substitute \(x=e^{a-1}\) into the derivative: \(f_a'(e^{a-1})=(-1)(1)=-1\). The tangent slope is always \(-1\).

a) \(B_a=(e^a, 0)\) b) \(H_a=(e^{a-2}, 4e^{a-2})\) c) \(y=4x\) d) The tangent slope is \(-1\).

- Use \(a>0\) to determine the domain and the sign of the derivative denominator. - Classify the critical points from the derivative sign changes. - Eliminate the parameter from the maximum coordinates. - For part d), compare endpoint values with the function values at both interior critical numbers.

1. Since \(a>0\), \(x^2+a>0\) for all real \(x\), so \(D_a=\mathbb R\). The only zero is \(x=0\), and \(f_a(-x)=-f_a(x)\), so the graph has origin symmetry. 2. \(f_a'(x)=\frac{a^2(a-x^2)}{(x^2+a)^2}\). The derivative changes from negative to positive at \(-\sqrt a\) and from positive to negative at \(\sqrt a\). Thus \(T=\left(-\sqrt a,-\frac12a\sqrt a\right)\) is a local minimum and \(H=\left(\sqrt a,\frac12a\sqrt a\right)\) is a local maximum. Eliminating \(a\) from the maximum coordinates gives \(y=0.5x^3\). 3. \(f_a(x)\to0\) as \(x\to\pm\infty\), so the horizontal asymptote is \(y=0\). 4. On the closed interval, the candidates are \(x=-2\sqrt a,-\sqrt a,\sqrt a,2\sqrt a\). The endpoint values are \(f_a(\pm2\sqrt a)=\pm\frac25a^{3/2}\), while the function values at the interior critical numbers are \(\pm\frac12a^{3/2}\). Therefore, the absolute minimum is \(-\frac12a^{3/2}\) at \(x=-\sqrt a\), and the absolute maximum is \(\frac12a^{3/2}\) at \(x=\sqrt a\).

a) \(D_a=\mathbb R\); zero \(x=0\); origin symmetry. b) Local minimum \(\left(-\sqrt a,-\frac12a\sqrt a\right)\); local maximum \(\left(\sqrt a,\frac12a\sqrt a\right)\); the maxima lie on \(y=0.5x^3\). c) Horizontal asymptote: \(y=0\). d) Absolute minimum \(-\frac12a^{3/2}\) at \(x=-\sqrt a\); absolute maximum \(\frac12a^{3/2}\) at \(x=\sqrt a\).

- Test odd or even symmetry by replacing \(x\) with \(-x\). - Use the derivative sign to classify the two critical numbers. - Compare two parameter values algebraically to find common points. - For the closed interval, compare both endpoint values with the function values at both critical numbers.

1. Since \(g_k(-x)=-g_k(x)\), the graph has origin symmetry. The only zero is \(x=0\). 2. \(g_k'(x)=\frac{2k-6x^2}{(x^2+k)^3}\), so the critical numbers are \(x=\pm\sqrt{\frac{k}{3}}\). The derivative changes from negative to positive at the negative critical number and from positive to negative at the positive one. Thus the extrema are \(T=\left(-\sqrt{\frac{k}{3}},-\frac{3\sqrt3}{8k^{3/2}}\right)\) and \(H=\left(\sqrt{\frac{k}{3}},\frac{3\sqrt3}{8k^{3/2}}\right)\). 3. Distinct family members can meet only at \(x=0\), so their only common point is \((0,0)\). Also \(g_k(x)\to0\) as \(x\to\pm\infty\), so the horizontal asymptote is \(y=0\). 4. On \([-\sqrt k,\sqrt k]\), the endpoint values are \(\pm\frac{1}{2k^{3/2}}\), while the function values at the interior critical numbers are \(\pm\frac{3\sqrt3}{8k^{3/2}}\). Since \(\frac{3\sqrt3}{8}>\frac12\), the interior critical points give the absolute extrema. The absolute minimum is the negative critical point and the absolute maximum is the positive critical point.

a) Origin symmetry; zero \(x=0\). b) Local minimum \(\left(-\sqrt{\frac{k}{3}},-\frac{3\sqrt3}{8k^{3/2}}\right)\); local maximum \(\left(\sqrt{\frac{k}{3}},\frac{3\sqrt3}{8k^{3/2}}\right)\). c) Common point \((0,0)\); horizontal asymptote \(y=0\). d) These same two interior critical points are the absolute minimum and maximum on \([-\sqrt k,\sqrt k]\).

- Use the first derivative for the maximum and the second derivative for the inflection point. - Substitute the critical x-value before deciding whether the maximum height depends on the parameter. - For the closed interval, list endpoints and interior critical numbers before comparing values.

1. \(f_k'(x)=k(1-kx)e^{-kx}\) and \(f_k''(x)=k^2(kx-2)e^{-kx}\). The derivative changes from positive to negative at \(x=1/k\), so \(H_k=\left(\frac1k,\frac1e\right)\). The concavity changes at \(x=2/k\), giving \(W_k=\left(\frac2k,\frac2{e^2}\right)\). 2. \(f_k(x)\to0\) as \(x\to\infty\), while \(f_k(x)\to-\infty\) as \(x\to-\infty\). 3. At the critical point \(x=1/k\), substitution gives \(f_k(1/k)=k(1/k)e^{-1}=1/e\), so the maximum height does not depend on \(k\). 4. The function is continuous on \([0,3/k]\), so absolute extrema exist. The candidates are \(x=0,1/k,3/k\). Their values are \(0\), \(1/e\), and \(3/e^3\). Since \(0<3/e^3<1/e\), the absolute minimum is \(0\) at \(x=0\), and the absolute maximum is \(1/e\) at \(x=1/k\).

a) \(H_k=\left(\frac1k,\frac1e\right)\), \(W_k=\left(\frac2k,\frac2{e^2}\right)\). b) \(f_k(x)\to0\) as \(x\to\infty\) and \(f_k(x)\to-\infty\) as \(x\to-\infty\). c) The maximum y-coordinate is always \(1/e\). d) On \([0,3/k]\): absolute minimum \(0\) at \(x=0\); absolute maximum \(1/e\) at \(x=1/k\).

- Analyze the derivative separately for positive and nonpositive values of \(a\). - Determine the end behavior of the function. - For \(a>0\), find the minimum value. - Compare the minimum value with zero.

1. Differentiate: \(f_a'(x)=ae^x-1\). 2. If \(a<0\), then \(f_a'(x)<0\) for every \(x\), so the function is strictly decreasing. Also, \(\lim_{x\to-\infty}f_a(x)=\infty\) and \(\lim_{x\to\infty}f_a(x)=-\infty\). Therefore, there is exactly one zero. If \(a=0\), then \(f_0(x)=-x\), which also has exactly one zero. 3. If \(a>0\), the only critical point satisfies \(ae^x=1\), so \(x=-\ln a\). Since \(f_a(x)\to\infty\) as \(x\to\pm\infty\), this critical point is the absolute minimum. Its value is \(f_a(-\ln a)=1+\ln a\). 4. The minimum is positive when \(a>1/e\), zero when \(a=1/e\), and negative when \(0<a<1/e\). Therefore: - there are no zeros when \(a>1/e\); - there is exactly one zero when \(a=1/e\) or \(a\le0\); - there are exactly two zeros when \(0<a<1/e\).

There are no real zeros for \(a>\frac1e\). There is exactly one real zero for \(a=\frac1e\) or \(a\le0\). There are exactly two real zeros for \(0<a<\frac1e\).

- Keep the domain \(x>0\) in mind. - Separate the cases \(k<0\), \(k=0\), and \(k>0\). - Use end behavior and monotonicity when \(k<0\). - For \(k>0\), find the minimum and compare it with zero.

1. Differentiate: \(h_k'(x)=2x-\frac{k}{x}\). 2. If \(k<0\), then \(h_k'(x)=2x+\frac{|k|}{x}>0\) for all \(x>0\). The function is strictly increasing, with \(h_k(x)\to-\infty\) as \(x\to0^+\) and \(h_k(x)\to\infty\) as \(x\to\infty\). Therefore, it has exactly one zero. 3. If \(k=0\), then \(h_0(x)=x^2\), which has no zero in the domain \(x>0\). 4. If \(k>0\), the only critical point is \(x=\sqrt{\frac{k}{2}}\). The function approaches \(\infty\) at both ends of its domain, so this point is the absolute minimum. Its value is \(\frac{k}{2}\left(1-\ln\left(\frac{k}{2}\right)\right)\). This value is zero when \(k=2e\), positive when \(0<k<2e\), and negative when \(k>2e\). 5. Therefore: - there are no zeros when \(0\le k<2e\); - there is exactly one zero when \(k<0\) or \(k=2e\); - there are exactly two zeros when \(k>2e\).

There are no real zeros for \(0\le k<2e\). There is exactly one real zero for \(k<0\) or \(k=2e\). There are exactly two real zeros for \(k>2e\).

- Rewrite the square root as a power before differentiating. - A fraction with a nonzero denominator equals zero exactly when its numerator equals zero. - Maximize the quadratic exponent before considering the exponential and square root. - Use the definition of a strictly increasing function to compare nearby function values.

1. a) Write \(h(x)=(f(x))^{\frac{1}{2}}\). By the chain rule, \(h'(x)=\frac{1}{2}(f(x))^{-\frac{1}{2}}f'(x)=\frac{f'(x)}{2\sqrt{f(x)}}\). 2. b) Since \(f(x)>0\), the denominator is always positive. Therefore, \(h'(x)=0\) if and only if \(f'(x)=0\). 3. c) The exponential function is strictly increasing, so \(f\) is maximized where the exponent \(-x^2+4x\) is maximized. Its vertex is at \(x=2\). The square-root function is also strictly increasing, so \(h\) has its maximum at \(x=2\). 4. d) A strictly increasing function preserves inequalities. If \(f(x)\le f(x_0)\) near \(x_0\), then \(\varphi(f(x))\le\varphi(f(x_0))\), so a local maximum remains a local maximum. The same argument applies to local minima.

a) \(h'(x)=\frac{f'(x)}{2\sqrt{f(x)}}\) b) \(h'(x)=0\iff f'(x)=0\) c) The maximum occurs at \(x=2\). d) Strictly increasing composition preserves local order, so local maxima and minima occur at the same inputs and keep their types.

- Each local extremum requires a sign-changing zero of the first derivative. - Compare the degree of a polynomial with the degree of its derivative. - To show the lower bound is attainable, test the derivative of an explicit fifth-degree polynomial. - Relate the parity of the degree to derivative end signs.

1. Four local extrema require at least four distinct zeros of \(g'\) at which the derivative changes sign. Therefore, \(g'\) must have degree at least \(4\), so \(g\) must have degree at least \(5\). 2. Degree \(5\) is attainable. For example, \(g(x)=\frac15x^5-\frac53x^3+4x\) has derivative \(g'(x)=x^4-5x^2+4=(x+2)(x+1)(x-1)(x-2)\). The derivative changes sign at all four simple zeros, so this fifth-degree polynomial has exactly four local extrema. 3. If \(g\) had even degree, then \(g'\) would have odd degree and opposite signs at the two ends. A continuous derivative with four sign-changing zeros would have the same sign after four changes, which is incompatible with opposite end signs. 4. If \(g\) has odd degree, then \(g'\) has even degree and can have the same sign at both ends, allowing four sign changes. Thus the degree must be odd.

a) The least possible degree is \(5\). b) The degree must be odd.

- Substitute the given x-coordinates to check the common points. - Factor the first derivative to find all horizontal tangents. - Use the second derivative or a sign chart to classify critical points. - Evaluate \(f_k(k)\), then eliminate the parameter.

1. Direct substitution gives \(f_k(0)=0\) and \(f_k(4)=\frac{64}{6}-4(k+2)+4k=\frac83\). Both values are independent of \(k\). 2. Differentiate and factor: \(f_k'(x)=\frac12x^2-\frac12(k+2)x+k=\frac12(x-2)(x-k)\). Therefore, \(f_k'(2)=0\) for every \(k\), so every graph has a horizontal tangent at \(x=2\). The second derivative is \(f_k''(x)=x-\frac12(k+2)\), so \(f_k''(2)=1-\frac k2\). Thus, \(x=2\) is a local maximum when \(k>2\), a local minimum when \(k<2\), and a stationary inflection point when \(k=2\). 3. The factorization of the first derivative shows that its other zero is \(x=k\). When \(k\ne2\), this zero is simple, so the derivative changes sign and the point is a local extremum. 4. Evaluate the function at \(x=k\): \(f_k(k)=-\frac1{12}k^3+\frac12k^2\). Since the x-coordinate is \(k\), replace \(k\) by \(x\): \(h(x)=-\frac1{12}x^3+\frac12x^2\), with \(x\ne2\).

a) The common points are \((0, 0)\) and \(\left(4, \frac83\right)\). b) \(f_k'(2)=0\). The point is a local extremum for \(k\ne2\); for \(k>2\), it is a local maximum. c) For \(k\ne2\), the other local extremum occurs at \(x=k\). d) \(h(x)=-\frac1{12}x^3+\frac12x^2\), for \(x\ne2\).

- Translate each point and extremum condition into an equation. - A local extremum at \(x = a\) requires \(f'(a) = 0\). - Solve enough independent conditions to determine the polynomial, then test the remaining condition. - A contradiction shows that the full set of requirements is impossible.

1. Let \(f(x) = ax^3 + bx^2 + cx + d\). Then \(f'(x) = 3ax^2 + 2bx + c\). 2. The conditions at \((0, 4)\) give \(f(0) = 4\), so \(d = 4\), and \(f'(0) = 0\), so \(c = 0\). 3. The conditions at \((2, 0)\) give \(f(2) = 0\), so \(8a + 4b = -4\), and \(f'(2) = 0\), so \(12a + 4b = 0\). 4. Solving gives \(a = 1\) and \(b = -3\), so the only possible polynomial is \(f(x) = x^3 - 3x^2 + 4\). 5. Its second derivative is \(f''(x) = 6x - 6\). Since \(f''(0) < 0\) and \(f''(2) > 0\), the required maximum and minimum classifications are satisfied. 6. However, \(f(1) = 1 - 3 + 4 = 2\), not \(3\). Therefore, no cubic polynomial satisfies all three conditions.

No such cubic polynomial exists. The extremum conditions force \(f(x) = x^3 - 3x^2 + 4\), but this polynomial passes through \((1, 2)\), not \((1, 3)\).

- For part a, think about a greatest limiting value located at a missing endpoint. - For part b, keep the interval closed but make the function misbehave near one point. - Your examples must fail to attain a maximum, not merely fail a theorem hypothesis. - Verify each requested property explicitly after choosing your examples.

1. For part a, one example is \(f(x)=x^2\) on \((-1,1)\). The function is continuous and the interval is bounded, but it is not closed. Values approach \(1\) near either missing endpoint, yet \(1\) is never attained, so there is no absolute maximum. 2. For part b, define \(g(x)=x\) for \(0\le x<1\) and \(g(1)=0\), on \([0,1]\). The interval is closed and bounded, but \(g\) is discontinuous at \(x=1\). 3. Values of \(g(x)\) approach \(1\) as \(x\to1^-\), but \(g(1)=0\), so no absolute maximum is attained. 4. These examples show why neither the closed-interval condition nor the continuity condition can simply be dropped from the theorem's guarantee.

One valid pair is: a) \(f(x)=x^2\) on \((-1,1)\); it has no absolute maximum. b) \(g(x)=x\) for \(0\le x<1\) and \(g(1)=0\), on \([0,1]\); it has no absolute maximum. Other correctly justified examples are also valid.