Free Math Worksheets

- What does the derivative represent at a point on a graph? - First find the derivative function. - Substitute the given \(x\)-value into the derivative.

1. Differentiate the function: \(f'(x)=10x^4-6x\). 2. The slope at \(x=1\) is \(f'(1)\). 3. Evaluate: \(f'(1)=10(1)^4-6(1)=4\).

\(4\)

- Which quantity gives the slope of a function at a specific input? - Differentiate the function term by term. - Substitute the point's \(x\)-coordinate into the derivative.

1. Differentiate: \(f'(x)=x^3-4x+3\). 2. Evaluate the derivative at the point's \(x\)-coordinate: \(f'(2)=2^3-4(2)+3=3\).

The slope at \(P\) is \(3\).

- The slope of the tangent line at \(x=a\) is the derivative value at \(a\). - Differentiate the function term by term. - Use parentheses when substituting a negative value.

1. Differentiate: \(g'(x)=\frac{1}{2}x^2+\frac{1}{2}x-1\). 2. Evaluate at \(x=-2\): \(g'(-2)=\frac{1}{2}(-2)^2+\frac{1}{2}(-2)-1=0\).

The slope at \(x=-2\) is \(0\).

- What does the first derivative represent at a point? - Differentiate the polynomial term by term. - Substitute the given input into the derivative.

1. Differentiate: \(f'(x)=\frac{1}{2}x^2-6x+10\). 2. Evaluate at \(x=6\): \(f'(6)=\frac{1}{2}(6)^2-6(6)+10=-8\).

The slope at \(x=6\) is \(-8\).

- The slope of a tangent line is a derivative value. - Use parentheses when substituting a negative input. - A constant term has derivative \(0\).

1. Differentiate: \(h'(x)=2x^3-2x^2+1\). 2. Evaluate at \(x=-1\): \(h'(-1)=2(-1)^3-2(-1)^2+1=-3\).

The tangent line has slope \(-3\) at \(x=-1\).

- What does the sign of the derivative tell you about the slope of a graph? - Use parentheses when substituting a negative value. - Find the derivative function before evaluating it at the given input.

1. Differentiate: \(g'(x)=-x^5+8x\). 2. Evaluate at \(x=-1\): \(g'(-1)=-(-1)^5+8(-1)=-7\). 3. Since \(-7<0\), the slope is negative at \(x=-1\).

\(g'(x)=-x^5+8x\); the slope is negative at \(x=-1\).

- How is a specified slope represented using the derivative? - Differentiate the quadratic function. - Set the derivative equal to the given value. - Solve the resulting equation for \(x\).

1. Differentiate: \(f'(x)=3x-7\). 2. Set the derivative equal to the required value: \(3x-7=5\). 3. Solve: \(3x=12\), so \(x=4\).

\(x=4\)

- Evaluate the first derivative at each given value. - A positive derivative means the graph rises; a negative derivative means it falls.

1. Differentiate: \(f'(x)=3x^2-6x-9\). 2. At \(x=-2\), \(f'(-2)=3\cdot4-6(-2)-9=15>0\), so the graph has positive slope. 3. At \(x=2\), \(f'(2)=3\cdot4-6\cdot2-9=-9<0\), so the graph has negative slope.

At \(x=-2\), the slope is positive: \(f'(-2)=15\). At \(x=2\), the slope is negative: \(f'(2)=-9\).

- Use the power rule to find each derivative function. - Rewrite a variable in the denominator using a negative exponent. - A derivative value gives the graph's slope at that input. - Use parentheses when substituting negative values.

1. For part a), \(f'(x)=5x^4\). Therefore, \(f'(1)=5\), \(f'(-2)=80\), and \(f'(0)=0\). 2. For part b), rewrite \(g(x)=2x^{-1}\), so \(g'(x)=-2x^{-2}=-\frac{2}{x^2}\). Thus \(g'(2)=-0.5\) and \(g'(-4)=-0.125\).

a) \(f'(1)=5\), \(f'(-2)=80\), and \(f'(0)=0\) b) \(g'(2)=-0.5\) and \(g'(-4)=-0.125\)

- Differentiate using the sum and constant multiple rules. - Recall the derivatives of sine and cosine. - Use exact unit-circle values at the two inputs. - The derivative value is the graph’s slope.

1. Differentiate: \(f'(x)=3\cos x+2\sin x\). 2. At \(x=0\), \(f'(0)=3\). 3. At \(x=\frac{\pi}{2}\), \(f'\left(\frac{\pi}{2}\right)=2\).

At \(x=0\), the slope is \(3\). At \(x=\frac{\pi}{2}\), the slope is \(2\).

- Differentiate first. - Substitute each candidate input into the derivative. - Use exact unit-circle values. - Select the inputs where the derivative equals \(1\).

1. Differentiate: \(h'(x)=\cos x-\sin x\). 2. Evaluate at the four inputs: \(h'(0)=1\), \(h'\left(\frac{\pi}{2}\right)=-1\), \(h'(\pi)=-1\), and \(h'\left(\frac{3\pi}{2}\right)=1\). Therefore, the tangent slope is \(1\) at \(x=0\) and \(x=\frac{3\pi}{2}\).

\(x=0\) and \(x=\frac{3\pi}{2}\)

- A tangent slope is a value of the derivative. - Find the derivative function first. - Solve one linear equation for each requested slope.

1. Differentiate: \(f'(x)=2x-3\). 2. Set \(2x-3\) equal to each requested slope and solve. 3. For slope \(0\), \(2x-3=0\), so \(x=1.5\). 4. For slope \(1\), \(2x-3=1\), so \(x=2\). 5. For slope \(-1\), \(2x-3=-1\), so \(x=1\). 6. For slope \(\frac{1}{2}\), \(2x-3=\frac{1}{2}\), so \(x=1.75\). 7. For slope \(-\frac{1}{2}\), \(2x-3=-\frac{1}{2}\), so \(x=1.25\).

Slope \(0\): \(x=1.5\) Slope \(1\): \(x=2\) Slope \(-1\): \(x=1\) Slope \(\frac{1}{2}\): \(x=1.75\) Slope \(-\frac{1}{2}\): \(x=1.25\)

- The derivative gives the graph's slope at each input. - Substitute a specified input into the derivative to find the slope there. - Rewrite the reciprocal power using a negative exponent before differentiating. - Use parentheses when substituting negative values.

1. Rewrite the function as \(f(x)=x^4-x^{-2}\). Then \(f'(x)=4x^3+2x^{-3}=4x^3+\frac{2}{x^3}\). 2. Evaluate: \(f'(1)=4+2=6\). 3. At \(x=-1\), \(f'(-1)=4(-1)^3+\frac{2}{(-1)^3}=-6\). Since \(-6<0\), the slope is negative.

a) \(f'(x)=4x^3+\frac{2}{x^3}\) b) \(6\) c) \(f'(-1)=-6<0\), so the slope is negative.

- The derivative gives the slope at each input. - Use the power rule for a function with a negative exponent. - Set each derivative equal to the specified slope. - A quadratic equation can have two real solutions.

1. Differentiate: \(f'(x)=x-3\) and \(g'(x)=-\frac{2}{x^2}\). 2. For part a), solve \(x-3=2\), giving \(x=5\). 3. For part b), solve \(-\frac{2}{x^2}=-0.5\). This gives \(x^2=4\), so \(x=-2\) or \(x=2\).

a) \(x=5\) b) \(x=-2\) and \(x=2\)

- The derivative value at a point is the graph's slope there. - Rewrite the radical and reciprocal using exponents before differentiating. - After finding the derivative function, substitute the given input.

1. Rewrite the function as \(f(x)=4x^{1/2}-9x^{-1}+2\). 2. Differentiate: \(f'(x)=2x^{-1/2}+9x^{-2}=\frac{2}{\sqrt{x}}+\frac{9}{x^2}\). 3. Evaluate at \(x=1\): \(f'(1)=2+9=11\).

a) \(f'(x)=\frac{2}{\sqrt{x}}+\frac{9}{x^2}\), where \(x>0\) b) \(11\)

- Rewrite the expression so that the term containing \(k\) is isolated. - Find the x-value that makes the coefficient of \(k\) equal to zero. - Evaluate the derivative at that x-value.

1. Rewrite the function as \(f_k(x)=k(x-2)^2+2x-3\). The parameter term is zero when \(x=2\). 2. Then \(f_k(2)=1\), so every graph passes through \(P(2, 1)\). 3. Differentiate: \(f_k'(x)=2kx+2-4k\). At \(x=2\), \(f_k'(2)=4k+2-4k=2\), independent of \(k\). Therefore, every graph has slope \(2\) at \(P\).

a) \(P(2, 1)\) b) Every graph has slope \(2\) at \(P\).

- Apply the power rule term by term. - Substitute \(x=0\) into the derivative. - Use two points on the tangent line to compute rise over run.

1. Differentiate: \(f'(x)=0.5\cdot2x-2=x-2\). 2. Evaluate at \(x=0\): \(f'(0)=0-2=-2\). 3. The tangent line passes through \((0,0)\). Moving \(1\) unit to the right along the line corresponds to moving \(2\) units down, so its slope is \(\frac{-2}{1}=-2\). This agrees with the derivative.

a) \(f'(x)=x-2\) b) \(f'(0)=-2\) c) The tangent line has slope \(-2\), confirming the calculated value.

- The derivative gives the slope of the tangent line. - A horizontal line has slope \(0\). - Differentiate first, and then evaluate at \(x=1\).

1. Differentiate: \(f'(x)=3x^2-6x+3\). 2. Evaluate the derivative at \(x=1\): \(f'(1)=3-6+3=0\). 3. Because the tangent slope is \(0\), the tangent line is horizontal at \(x=1\).

\(f'(x)=3x^2-6x+3\) and \(f'(1)=0\), so the tangent line is horizontal at \(x=1\).

- Apply the power rule to each term. - Keep constant factors when differentiating. - Evaluate the derivative at \(x=1\).

1. Differentiate using the power rule: \(f'(x)=4x-3\). 2. Evaluate the derivative: \(f'(1)=4-3=1\). 3. The instantaneous rate of change at \(x=1\) is \(1\).

\(f'(1)=1\)

- Differentiate each term separately. - The derivative of a constant is \(0\). - Substitute \(x=2\) after differentiating.

1. Differentiate: \(g'(x)=3x^2-8x\). 2. Evaluate at \(x=2\): \(g'(2)=3\cdot2^2-8\cdot2=12-16=-4\). 3. The graph has slope \(-4\) at \(x=2\).

\(g'(2)=-4\)

- Find the y-intercept by setting \(x=0\). - Find the x-intercepts by solving \(f(x)=0\). - The slope at an intercept is the derivative evaluated at that intercept's \(x\)-coordinate. - Be sure to evaluate the derivative at every distinct intercept input.

1. Differentiate: \(f'(x)=4x^3-15x^2+8x\). 2. The graph intersects the y-axis at \((0,0)\). 3. Factor to find the x-intercepts: \(f(x)=x^2(x^2-5x+4)=x^2(x-1)(x-4)\). Thus the distinct intercept inputs are \(x=0\), \(x=1\), and \(x=4\). 4. Evaluate the derivative: \(f'(0)=0\), \(f'(1)=-3\), and \(f'(4)=48\).

At \((0,0)\), the slope is \(0\). At \((1,0)\), the slope is \(-3\). At \((4,0)\), the slope is \(48\).

- First find the coordinates of all intercepts. - Differentiate the function to obtain its slope function. - Check the signs carefully when solving the quadratic equation. - Substitute each intercept's \(x\)-coordinate into the derivative.

1. Differentiate: \(g'(x)=x-2\). 2. The y-intercept is \((0,-6)\), and \(g'(0)=-2\). 3. To find the x-intercepts, solve \(0.5x^2-2x-6=0\). Multiplying by \(2\) gives \(x^2-4x-12=0=(x-6)(x+2)\), so \(x=6\) or \(x=-2\). 4. Evaluate the derivative: \(g'(6)=4\) and \(g'(-2)=-4\).

At the y-intercept \((0,-6)\), the slope is \(-2\). At the x-intercepts \((6,0)\) and \((-2,0)\), the slopes are \(4\) and \(-4\), respectively.

- Parallel lines have equal slopes. - Which coefficient in slope-intercept form gives the slope? - The derivative gives the tangent slope at each input. - After setting the derivative equal to the line's slope, solve the resulting quadratic equation.

1. The line \(y=8x-1\) has slope \(8\). 2. Differentiate: \(f'(x)=x^2+x-4\). 3. Parallel tangent lines must have slope \(8\), so solve \(x^2+x-4=8\). 4. This gives \(x^2+x-12=0=(x+4)(x-3)\). 5. Therefore, \(x=-4\) or \(x=3\).

\(x=-4\) and \(x=3\)

- The derivative gives the slope at a specific input. - Set a derivative equal to a required slope to form an equation. - Equal slopes at the same input mean the derivative values are equal. - Set the two derivative functions equal and solve for \(x\).

1. Differentiate: \(f'(x)=2x-4\) and \(g'(x)=x^2-4x+1\). 2. For part a), solve \(2x-4=6\), giving \(x=5\). 3. For part b), \(g'(5)=25-20+1=6\), so both graphs have slope \(6\) at \(x=5\). 4. For part c), set the derivatives equal: \(2x-4=x^2-4x+1\). 5. Rearranging gives \(x^2-6x+5=0=(x-1)(x-5)\). The solutions are \(x=1\) and \(x=5\), so the additional value is \(x=1\).

a) \(x=5\) b) Yes. Both graphs have slope \(6\) at \(x=5\). c) The other value is \(x=1\).

- What is the slope of a horizontal line? - After finding an \(x\)-coordinate, substitute it into the original function to find the point. - If two graphs have the same slope at the same input, their derivatives are equal there. - Solve the resulting quadratic equations by factoring or another valid method.

1. Differentiate: \(h'(x)=3x^2-12x+9\) and \(k'(x)=-3x+3\). 2. For part a), horizontal tangent lines have slope \(0\), so solve \(3x^2-12x+9=0\). 3. Dividing by \(3\) gives \(x^2-4x+3=0=(x-1)(x-3)\), so \(x=1\) or \(x=3\). 4. Evaluate the function: \(h(1)=6\) and \(h(3)=2\). The points are \((1,6)\) and \((3,2)\). 5. For part b), set the derivatives equal: \(3x^2-12x+9=-3x+3\). 6. Simplifying gives \(x^2-3x+2=0=(x-1)(x-2)\), so \(x=1\) or \(x=2\).

a) \((1,6)\) and \((3,2)\) b) \(x=1\) and \(x=2\)

- Use the point-slope form of a tangent line. - Find both the function value and derivative value at \(x=0\). - A point on the tangent line must satisfy its equation. - Express the tangent slope and y-intercept in terms of \(k\).

1. Differentiate: \(f_k'(x)=ke^x-1\). 2. At \(x=0\), \(f_k(0)=k\) and \(f_k'(0)=k-1\). 3. The tangent line is \(y=(k-1)x+k\). 4. Substitute \((2, 5)\): \(5=2(k-1)+k\). 5. Solve \(5=3k-2\) to obtain \(k=\frac{7}{3}\).

a) \(f_k'(x)=ke^x-1\) b) \(k=\frac{7}{3}\)

- Find both the function value and slope at \(x=1\). - Simplify the tangent equation before applying the conditions. - Positive slope determines the sign of \(a\). - Set the tangent equation equal to \(0\) to find its x-intercept. - Translate the x-intercept condition into an inequality for \(a\).

1. At \(x=1\), \(g_a(1)=a+2\). 2. Differentiate: \(g_a'(x)=ae^{x-1}\), so the tangent slope at \(x=1\) is \(a\). 3. The tangent line is \(y-(a+2)=a(x-1)\), which simplifies to \(y=ax+2\). 4. Positive slope requires \(a>0\). 5. The x-intercept is \(x=-\frac{2}{a}\). Require \(-\frac{2}{a}>-1\). 6. Since \(a>0\), multiplication by \(a\) preserves the inequality: \(-2>-a\), so \(a>2\).

a) \(y=ax+2\) b) \(a>2\)

- A horizontal tangent has slope \(0\). - Treat the parameter \(k\) as a constant when differentiating with respect to \(x\). - Substitute the specified input into the derivative before solving for \(k\).

1. Treat \(k\) as a constant and differentiate: \(f_k'(x)=3kx^2-6x\). 2. A horizontal tangent has slope \(0\), so require \(f_k'(1)=0\). 3. Substitute \(x=1\): \(3k-6=0\). 4. Solving gives \(k=2\).

a) \(f_k'(x)=3kx^2-6x\) b) \(k=2\)

- The derivative gives the slope at each input. - What value must the derivative equal at the requested points? - Solve the resulting quadratic equation. - Check that you found every solution.

1. Differentiate: \(f'(x)=x^2-2x-5\). 2. Set the derivative equal to the required slope: \(x^2-2x-5=3\). 3. Rearrange: \(x^2-2x-8=0\). 4. Factor: \((x-4)(x+2)=0\). 5. Therefore, \(x=4\) or \(x=-2\).

\(x=-2\) and \(x=4\)

- Rewrite the function before differentiating. - Write the tangent line at a general x-coordinate \(x_0\). - Use the fact that the origin lies on the tangent line. - Substitute the resulting x-coordinate into the original function.

1. Rewrite the function as \(f_k(x)=1+\frac{k}{x}\), so \(f_k'(x)=-\frac{k}{x^2}\). 2. Let the point of tangency have x-coordinate \(x_0\). The tangent line is \(y=f_k'(x_0)(x-x_0)+f_k(x_0)\). 3. Since the line passes through \((0, 0)\), \(0=-\frac{k}{x_0^2}(0-x_0)+1+\frac{k}{x_0}\). 4. Simplify: \(0=1+\frac{2k}{x_0}\), so \(x_0=-2k\). Since \(k>0\), this is a valid domain value. 5. The y-coordinate is \(f_k(-2k)=\frac{-2k+k}{-2k}=\frac{1}{2}\). Therefore, \(B=\left(-2k, \frac{1}{2}\right)\).

\(B=\left(-2k, \frac{1}{2}\right)\)

- Tangent slopes are the values of the first derivative. - Find the minimum value of \(3x^2+3\). - Interpret a derivative that is always positive.

1. Differentiate: \(f'(x)=3x^2+3\). 2. Since \(x^2\ge0\), \(f'(x)\ge3\) for every real \(x\). Therefore, the possible slopes are \([3,\infty)\), and every slope \(m<3\) is impossible. 3. Geometrically, the graph is strictly increasing everywhere. It has no horizontal or negative-slope tangent, and its smallest slope is \(3\) at \(x=0\).

1) \(f'(x)=3x^2+3\) 2) Every slope \(m<3\) is impossible. 3) The graph rises everywhere, and its minimum tangent slope is \(3\) at \(x=0\).

- Rewrite the function with a negative exponent. - Determine the sign of \(x^3\) on each side of zero. - A fraction with a nonzero constant numerator cannot equal zero.

1. Rewrite and differentiate: \(g(x)=-x^{-2}\), so \(g'(x)=2x^{-3}=\frac{2}{x^3}\). 2. For \(x<0\), \(x^3<0\), so \(g'(x)<0\). The left branch is strictly decreasing. 3. For \(x>0\), \(x^3>0\), so \(g'(x)>0\). The right branch is strictly increasing. 4. The derivative can never equal zero because its numerator is the nonzero constant \(2\). Every nonzero real slope occurs, but \(m=0\) is excluded. 5. Geometrically, neither branch has a horizontal tangent.

For \(x<0\), slopes are negative. For \(x>0\), slopes are positive. The only excluded slope is \(m=0\).

- First find the derivative function. - Set the derivative equal to each required value. - Solve each resulting equation for \(x\). - An equation of the form \(x^2=a\) can have two real solutions.

1. Differentiate: \(f'(x)=x^2-\frac{1}{4}\). 2. For \(f'(x)=0\), solve \(x^2=\frac{1}{4}\), giving \(x=\pm\frac{1}{2}\). 3. For \(f'(x)=\frac{3}{4}\), solve \(x^2=1\), giving \(x=\pm1\). 4. For \(f'(x)=2\), solve \(x^2=\frac{9}{4}\), giving \(x=\pm\frac{3}{2}\).

For \(f'(x)=0\): \(x\in\left\{-\frac{1}{2},\frac{1}{2}\right\}\) For \(f'(x)=\frac{3}{4}\): \(x\in\{-1,1\}\) For \(f'(x)=2\): \(x\in\left\{-\frac{3}{2},\frac{3}{2}\right\}\)

- Which function gives the tangent slope at each input? - Set the derivative equal to the proposed slope. - What values can the square of a real number take?

1. Differentiate: \(f'(x)=3x^2+3\). 2. A tangent line with slope \(2\) would require \(3x^2+3=2\). 3. This equation simplifies to \(x^2=-\frac{1}{3}\), which has no real solution. 4. Equivalently, since \(x^2\geq0\), \(f'(x)=3x^2+3\geq3\) for every real \(x\).

\(f'(x)=3x^2+3\geq3\) for every \(x\in\mathbb{R}\), so the derivative never equals \(2\).

- Find the derivative. - Use the range of sine. - Account for the constant factor and negative sign. - Report the values outside the derivative’s range.

1. Differentiate: \(g'(x)=-4\sin x\). 2. Since \(-1\le\sin x\le 1\), the derivative satisfies \(-4\le g'(x)\le 4\). Therefore, tangent slopes outside this interval cannot occur.

The impossible slopes are \(m<-4\) or \(m>4\).

- The tangent slope at \(x=-1\) is \(g'(-1)\). - Use the given slope to form an equation containing only \(a\). - Find the point of tangency after determining \(a\). - Use the point-slope form of a line.

1. Differentiate: \(g'(x)=2ax+4\). 2. Use the given slope: \(g'(-1)=2\). Therefore, \(-2a+4=2\), so \(a=1\). 3. With \(a=1\), the point of tangency is \((-1,g(-1))=(-1,-5)\). 4. A line with slope \(2\) through \((-1,-5)\) satisfies \(y+5=2(x+1)\), so \(y=2x-3\).

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

- Use only odd powers for origin symmetry. - Translate the slope condition into an equation involving the derivative. - Leave one coefficient as a parameter when finding all possible functions. - Use the additional point to determine the parameter.

1. A cubic polynomial symmetric about the origin has the form \(f(x)=ax^3+cx\), where \(a\ne0\). 2. Since \(f'(x)=3ax^2+c\), the slope condition \(f'(1)=6\) gives \(3a+c=6\), so \(c=6-3a\). 3. Therefore, all such cubic polynomials are \(f_a(x)=ax^3+(6-3a)x\), where \(a\in\mathbb{R}\setminus\{0\}\). 4. To pass through \(Q(2, 20)\), the polynomial must satisfy \(8a+2(6-3a)=20\). Thus, \(2a+12=20\), so \(a=4\) and \(c=-6\). 5. The required polynomial is \(f(x)=4x^3-6x\).

All such polynomials: \(f_a(x)=ax^3+(6-3a)x\), where \(a\in\mathbb{R}\setminus\{0\}\) Polynomial through \(Q(2, 20)\): \(f(x)=4x^3-6x\)

- Tangency to the x-axis gives both a zero and a zero slope. - Use the origin to determine the constant term. - Translate each remaining condition into an equation. - Solve the resulting linear system.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Passing through the origin gives \(d=0\). 2. The point \(P(2, 4)\) gives \(8a+4b+2c=4\). Tangency to the x-axis at \(x=4\) gives \(f(4)=0\) and \(f'(4)=0\), so \(64a+16b+4c=0\) and \(48a+8b+c=0\). 3. Solving the system gives \(a=\frac{1}{2}\), \(b=-4\), and \(c=8\). 4. Therefore, \(f(x)=\frac{1}{2}x^3-4x^2+8x\).

\(f(x)=\frac{1}{2}x^3-4x^2+8x\)

- Recall the derivative of \(\log_bx\). - Substitute the given input and slope. - Solve the resulting equation for \(\ln b\). - Exponentiate to solve for \(b\).

1. Differentiate: \(h'(x)=\frac{1}{x\ln b}\). 2. Use the given slope: \(\frac{1}{2\ln b}=0.5=\frac{1}{2}\). Therefore, \(\ln b=1\), so \(b=e\).

\(b=e\)

- Use the given point to form one parameter equation. - A horizontal tangent means the first derivative is zero. - Use polynomial division after finding the parameters. - Check the signs of the numerator and denominator on each side of the vertical asymptote.

1. The point condition gives \(f(4)=8\): \(\frac{16a+4b}{2}=8\), so \(4a+b=4\). 2. Differentiate: \(f'(x)=\frac{(2ax+b)(x-2)-(ax^2+bx)}{(x-2)^2}=\frac{ax^2-4ax-2b}{(x-2)^2}\). 3. A horizontal tangent at \(x=0\) requires \(f'(0)=0\), so \(b=0\). Then \(4a=4\), giving \(a=1\). 4. The function is \(f(x)=\frac{x^2}{x-2}=x+2+\frac{4}{x-2}\). Therefore, the slant asymptote is \(y=x+2\). 5. The vertical asymptote is \(x=2\). Since the numerator is positive near \(2\), \(\lim_{x\to 2^-}f(x)=-\infty\) and \(\lim_{x\to 2^+}f(x)=\infty\).

a) \(a=1\), \(b=0\) b) Slant asymptote: \(y=x+2\); \(\lim_{x\to 2^-}f(x)=-\infty\) and \(\lim_{x\to 2^+}f(x)=\infty\)

- Convert each graph condition into an equation involving \(f\) or \(f'\). - Differentiate the general function before using the slope condition. - Use exact sine and cosine values at \(0\), \(\frac{\pi}{2}\), and \(\pi\). - Solve the resulting linear system for the three parameters.

1. Translate the conditions into equations: \(f(0)=4\), \(f\left(\frac{\pi}{2}\right)=1\), and \(f'(0)=2\). 2. Differentiate: \(f'(x)=a\cos x-2b\sin(2x)\). From \(f'(0)=2\), \(a=2\). 3. From \(f(0)=4\), \(b+c=4\). 4. From \(f\left(\frac{\pi}{2}\right)=1\), \(2-b+c=1\), so \(-b+c=-1\). 5. Solve the system \(b+c=4\) and \(-b+c=-1\). This gives \(b=\frac{5}{2}\) and \(c=\frac{3}{2}\).

\(a=2\), \(b=\frac{5}{2}\), and \(c=\frac{3}{2}\) Therefore, \(f(x)=2\sin x+\frac{5}{2}\cos(2x)+\frac{3}{2}\).

- A logarithm requires a positive argument. - Apply the chain rule to \(\ln(u(x))\). - Use the monotonicity and zero of \(f\) to determine where \(f(x)>0\). - Express \(g'\) in terms of \(f\) and \(f'\), then simplify the equation. - Read the required function value from the graph.

1. For part a, require \(3x-5>0\), so \(D_h=\left(\frac{5}{3}, \infty\right)\). The zero satisfies \(3x-5=1\), giving \(x=2\). By the chain rule, \(h'(x)=\frac{3}{3x-5}\). 2. Since \(f\) is strictly increasing and has its only zero at \(-2\), \(f(x)>0\) exactly when \(x>-2\). Therefore, \(D_g=(-2, \infty)\). Also, \(g'(x)=\frac{f'(x)}{f(x)}\). The equation \(\frac{f'(x)}{f(x)}=\frac{1}{2}f'(x)\) can be divided by \(f'(x)>0\), giving \(f(x)=2\). From the graph, \(f(x)=2\) at \(x=2\).

a) \(D_h=\left(\frac{5}{3}, \infty\right)\), zero at \(x=2\), and \(h'(x)=\frac{3}{3x-5}\) b) 1. \(D_g=(-2, \infty)\) 2. \(x=2\)

- Substitute each point into both functions. - Differentiate to find the tangent slopes. - Use the acute-angle formula for two lines with known slopes. - At the origin, one tangent is horizontal.

1. Set the functions equal: \(\frac{1}{4}x^3=x\). Then \(x(x^2-4)=0\), so \(x=-2\), \(x=0\), or \(x=2\). The intersection points are \(Q(-2,-2)\), \(O(0,0)\), and \(P(2,2)\). 2. The tangent slopes are given by \(f'(x)=\frac{3}{4}x^2\) and \(g'(x)=1\). 3. At \(O\), the slopes are \(0\) and \(1\), so the acute angle is \(45^\circ\). 4. At both \(Q\) and \(P\), the slopes are \(3\) and \(1\). Thus, \(\tan\theta=\left|\frac{3-1}{1+3\cdot1}\right|=\frac{1}{2}\), so \(\theta=\arctan(\frac{1}{2})\approx26.6^\circ\).

a) \(Q(-2,-2)\), \(O(0,0)\), and \(P(2,2)\) b) At \(O\), the acute angle is \(45^\circ\). At \(Q\) and \(P\), it is \(\arctan(\frac{1}{2})\approx26.6^\circ\).

- Use the power rule to find the derivative. - The derivative value is the tangent slope. - Parallel lines have equal slopes. - Substitute each x-value into \(f\) to find the point's y-coordinate.

1. Differentiate: \(f'(x)=2x-2\). 2. For slope \(2\), solve \(2x-2=2\), giving \(x=2\). Then \(f(2)=-1\), so \(P=(2,-1)\). 3. A line parallel to \(y=-4x+5\) has slope \(-4\). Solve \(2x-2=-4\), giving \(x=-1\). Then \(f(-1)=2\), so \(Q=(-1,2)\).

a) \(f'(x)=2x-2\) b) \(P=(2,-1)\) c) \(Q=(-1,2)\)

- Differentiate each term. - A horizontal tangent has slope \(0\). - Parallel lines have the same slope; \(y=x\) has slope \(1\). - Substitute each x-value into \(f\) to find the corresponding point.

1. Differentiate: \(f'(x)=-x+3\). 2. For a horizontal tangent, solve \(-x+3=0\), giving \(x=3\). Since \(f(3)=4.5\), \(S=(3,4.5)\). 3. The line \(y=x\) has slope \(1\). Solve \(-x+3=1\), giving \(x=2\). Since \(f(2)=4\), \(R=(2,4)\).

a) \(f'(x)=-x+3\) b) \(S=(3,4.5)\) c) \(x=2\), and \(R=(2,4)\)

- Add the y-values of the two component functions. - Parallel lines have equal slopes. - Look for where the cosine term in the derivative is zero. - Use \(-1\le\sin x\le1\) to bound \(f(x)\).

1. Since \(f(x)=g(x)+h(x)\), the graph of \(f\) is obtained by adding the y-values of the line \(y=x\) and the sine curve \(y=1.5\sin x\). Thus, it oscillates around the line \(y=x\). 2. The line \(g(x)=x\) has slope \(1\). Differentiate: \(f'(x)=1+1.5\cos x\). Parallel tangents satisfy \(1+1.5\cos x=1\), so \(\cos x=0\). On \([0, 2\pi]\), this occurs at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). 3. For \(0<x\le\pi\), \(\sin x\ge0\), so \(f(x)>0\). For \(x>\pi\), \(f(x)=x+1.5\sin x\ge x-1.5>\pi-1.5>0\). Therefore, \(f(x)>0\) for all \(x>0\).

a) The graph of \(f\) is the pointwise sum of the graphs of \(g\) and \(h\), so it oscillates around \(y=x\). b) \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) c) For \(0<x\le\pi\), both terms are nonnegative. For \(x>\pi\), \(f(x)\ge x-1.5>0\).

- Compare the shapes and formulas of the three functions. - Set the derivative equal to the desired slope in part b. - A horizontal tangent requires the derivative to equal zero. - Use the unit circle to solve the resulting sine equations.

1. The line through the origin is \(g(x)=0.5x\), the reflected cosine curve is \(h(x)=-\cos x\), and their pointwise sum is \(f(x)=0.5x-\cos x\). 2. Differentiate: \(f'(x)=0.5+\sin x\). For slope \(0.5\), solve \(0.5+\sin x=0.5\), so \(\sin x=0\). On \([0, 2\pi]\), the inputs are \(0\), \(\pi\), and \(2\pi\). The corresponding points are \((0, -1)\), \(\left(\pi, \frac{\pi}{2}+1\right)\), and \((2\pi, \pi-1)\). 3. Horizontal tangents satisfy \(0.5+\sin x=0\), so \(\sin x=-\frac{1}{2}\). Therefore, \(x=\frac{7\pi}{6}+2\pi n\) or \(x=\frac{11\pi}{6}+2\pi n\), where \(n\in\mathbb{Z}\).

a) \(g\) is the line through the origin, \(h\) is the reflected cosine curve, and \(f\) is their sum. b) \((0, -1)\), \(\left(\pi, \frac{\pi}{2}+1\right)\), and \((2\pi, \pi-1)\) c) \(x=\frac{7\pi}{6}+2\pi n\) or \(x=\frac{11\pi}{6}+2\pi n\), where \(n\in\mathbb{Z}\)

- Use the y-intercept to determine \(a\). - A \(45^\circ\) angle with the y-axis corresponds to slopes \(1\) or \(-1\). - The derivative at the y-intercept gives the tangent slope. - Apply the restriction \(k\neq0\).

1. The y-intercept gives \(f_{a,k}(0)=a=3\). 2. Differentiate: \(f_{a,k}'(x)=(1+kx+ak)e^{kx}\). 3. At \(x=0\), the tangent slope is \(1+3k\). 4. A line making a \(45^\circ\) angle with the y-axis has slope \(1\) or \(-1\). 5. The equation \(1+3k=1\) gives \(k=0\), which is excluded. The equation \(1+3k=-1\) gives \(k=-\frac{2}{3}\). 6. Therefore, \(a=3\) and \(k=-\frac{2}{3}\).

\(a=3\) and \(k=-\frac{2}{3}\)

- Differentiate the general function \(f_n(x)=x^n\). - How many real solutions can \(x^k=c\), with \(c>0\), have when \(k\) is even or odd? - Connect the parity of the exponent in the derivative to the number of solutions. - Check \(n=1\) separately.

1. Differentiate: \(f_n'(x)=nx^{n-1}\). The slope condition is \(nx^{n-1}=10\), or \(x^{n-1}=\frac{10}{n}\). 2. Since \(\frac{10}{n}>0\), the equation has exactly two real solutions when the positive exponent \(n-1\) is even. 3. The case \(n=1\) must be checked separately: \(f_1'(x)=1\), so there are no solutions. 4. For \(n=3\), \(x^2=\frac{10}{3}\) has two solutions. For \(n=5\), \(x^4=2\) has two solutions. 5. For \(n=2,4,6\), the exponent \(n-1\) is odd, so each equation has exactly one real solution.

\(n=3\) and \(n=5\)