Free math worksheets

- Identify the exponent of the power. - The power rule multiplies by the exponent and then reduces the exponent by one.

1. Apply the power rule \(\frac{d}{dx}x^n=nx^{n-1}\). 2. With \(n=7\), \(f'(x)=7x^6\).

\(f'(x)=7x^6\)

- Rewrite the reciprocal power using a negative exponent. - The power rule changes both the coefficient and the exponent.

1. Rewrite \(g(x)=x^{-2}\). 2. The power rule gives \(g'(x)=-2x^{-3}=-\frac{2}{x^3}\). 3. Therefore, choice B is correct.

B) \(g'(x)=-\frac{2}{x^3}\)

- What does the derivative represent at a point? - Differentiate each term of the function. - Substitute \(x=5\) into the derivative.

1. Differentiate the function: \(y'=0.4x+5\). 2. Evaluate the derivative at \(x=5\): \(y'(5)=0.4\cdot 5+5\). 3. Simplify: \(y'(5)=2+5=7\).

\(7\)

- Differentiate each term separately using the power, constant-multiple, and sum rules. - What is the derivative of a term that does not contain the variable? - Treat \(k\) and \(n\) as parameters and \(\pi\) as a constant.

1. Differentiate each term in part a): \(f'(x)=\frac{1}{4}(4x^3)-2(3x^2)+5=x^3-6x^2+5\). 2. Differentiate each term in part b). The constant term has derivative \(0\): \(f'(x)=\frac{2}{3}(6x^5)+2\sqrt{5}x=4x^5+2\sqrt{5}x\). 3. Treat \(k\) and \(n\) as constants in part c): \(f'(x)=knx^{n-1}+2\).

a) \(f'(x)=x^3-6x^2+5\) b) \(f'(x)=4x^5+2\sqrt{5}x\) c) \(f'(x)=knx^{n-1}+2\)

- Rewrite each expression as a sum of simple power functions before differentiating. - Expand the squared binomial in part b). - Identify the variable of differentiation and treat every parameter as a constant.

1. Rewrite part a) as \(f(x)=\frac{1}{3}x^3-2x\). Then \(f'(x)=x^2-2\). 2. Expand part b): \(f(x)=x^2+6x+9\). Therefore, \(f'(x)=2x+6\). 3. In part c), treat \(a\) as a constant. Then \(f'(x)=2a^2x+a\).

a) \(f'(x)=x^2-2\) b) \(f'(x)=2x+6\) c) \(f'(x)=2a^2x+a\)

- Use the constant-multiple rule when a coefficient multiplies a power of \(x\). - In the power rule, how does the exponent change? - What is the derivative of a constant? - Differentiate a sum term by term.

1. Apply the power rule to the first term: \(\frac{d}{dx}\left(\frac{1}{2}x^4\right)=2x^3\). 2. Differentiate the second term: \(\frac{d}{dx}(-4x^3)=-12x^2\). 3. The derivative of \(12x\) is \(12\), and the derivative of \(-5\) is \(0\). 4. Combine the results: \(f'(x)=2x^3-12x^2+12\).

\(f'(x)=2x^3-12x^2+12\)

- 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: \(f'(1)=10(1)^4-6(1)=4\).

\(4\)

- Differentiate only with respect to the variable shown in the function notation. - A constant term has derivative \(0\). - For the power rule, multiply by the exponent and then decrease the exponent by \(1\). - Treat parameters as constants.

1. For part a), apply the power rule term by term: \(f'(x)=x^4-6x\). 2. For part b), differentiate with respect to \(t\): \(f'(t)=t^3-2t^2+7\). 3. For part c), treat \(m\), \(n\), and \(p\) as constants: \(f'(x)=2mx+n\). 4. For part d), differentiate with respect to \(k\) and treat \(x^3\) as a constant coefficient: \(f'(k)=2x^3k-5\).

a) \(f'(x)=x^4-6x\) b) \(f'(t)=t^3-2t^2+7\) c) \(f'(x)=2mx+n\) d) \(f'(k)=2x^3k-5\)

- For each term, decide whether it contains the variable of differentiation. - A term that does not contain the variable has derivative \(0\). - Simplify numerical coefficients after applying the power rule. - In part d), \(x\) is the variable and \(k\) is a parameter.

1. For part a), differentiate term by term: \(f'(z)=\frac{1}{2}z^2-2z+\frac{1}{2}\). 2. In part b), treat \(a\) as a constant: \(f'(x)=28x^6-6ax^2\). 3. In part c), \(v\) and \(s_0\) are constants, so \(s'(t)=v\). 4. In part d), treat \(k\) as a constant. Since \(-5k\) does not contain \(x\), its derivative is \(0\), and \(f'(x)=3k^2x^2\).

a) \(f'(z)=\frac{1}{2}z^2-2z+\frac{1}{2}\) b) \(f'(x)=28x^6-6ax^2\) c) \(s'(t)=v\) d) \(f'(x)=3k^2x^2\)

- 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 \(0\).

- Differentiate a sum one term at a time. - Apply the power rule to both the exponent and coefficient. - A constant has derivative \(0\). - Treat a decimal coefficient the same way as any other constant coefficient.

1. Differentiate each power term: \(\frac{d}{dx}\left(\frac{5}{6}x^6\right)=5x^5\), \(\frac{d}{dx}(-2x^3)=-6x^2\), and \(\frac{d}{dx}(0.4x^2)=0.8x\). 2. The derivative of \(-9\) is \(0\). 3. Combine the terms: \(f'(x)=5x^5-6x^2+0.8x\).

\(f'(x)=5x^5-6x^2+0.8x\)

- What role does \(a\) play when differentiating with respect to \(t\)? - The power rule still applies when a coefficient contains a parameter. - What is the derivative of the linear term \(5t\)? - Since \(a\) is part of a constant coefficient, it remains in the denominator.

1. Differentiate with respect to \(t\), treating \(a\) as a constant. 2. The first term gives \(at^2\), the second gives \(-\frac{2}{a}t\), and the third gives \(5\). 3. Therefore, \(g_a'(t)=at^2-\frac{2}{a}t+5\).

\(g_a'(t)=at^2-\frac{2}{a}t+5\)

- 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 \(x=-1\).

- What is the derivative of a term of the form \(cx\)? - How does the exponent change under the power rule? - Check every term of the original function separately.

1. Neha left the variable in the derivative of the linear term. Since \(\frac{d}{dx}(x)=1\), the derivative of \(7x\) is \(7\), not \(7x\). 2. Differentiate all terms: \(\frac{d}{dx}\left(\frac{1}{3}x^3\right)=x^2\), \(\frac{d}{dx}(-4x^2)=-8x\), \(\frac{d}{dx}(7x)=7\), and \(\frac{d}{dx}(-5)=0\). 3. Therefore, \(f'(x)=x^2-8x+7\).

The derivative of \(7x\) is \(7\), not \(7x\). The correct derivative is \(f'(x)=x^2-8x+7\).

- What is the derivative of a term that does not contain \(x\)? - Rewrite a power of \(x\) in the denominator using a negative exponent. - When applying the power rule to a negative exponent, check the signs in the coefficient multiplication.

1. In (1), the constant rule was applied incorrectly. The derivative of \(12\) is \(0\), so \(g'(x)=2x^3-4x\). 2. In (2), rewrite \(-\frac{3}{x^2}\) as \(-3x^{-2}\). Its derivative is \((-3)(-2)x^{-3}=6x^{-3}\), so \(h'(x)=10x^4+6x^{-3}=10x^4+\frac{6}{x^3}\).

(1) The constant \(12\) should have derivative \(0\). The correct derivative is \(g'(x)=2x^3-4x\). (2) The sign is wrong because \((-3)(-2)=6\). The correct derivative is \(h'(x)=10x^4+6x^{-3}\), or \(h'(x)=10x^4+\frac{6}{x^3}\).

- 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 \(x=4\).

\(x=4\)

- Substitute the given x-value into the function to find the point of tangency. - The derivative gives the slope of the tangent line. - Use point-slope form after finding the point and slope. - A tangent line matches the curve's instantaneous slope at the point of tangency.

1. Find the point of tangency: \(f(1)=1^3-3(1)^2+2=0\), so the point is \((1,0)\). 2. Differentiate: \(f'(x)=3x^2-6x\). 3. Find the tangent slope: \(f'(1)=3(1)^2-6(1)=-3\). 4. Use point-slope form: \(y-0=-3(x-1)\). 5. Simplify: \(y=-3x+3\).

\(y=-3x+3\)

- First find the point through which the normal line passes. - Tangent and normal slopes are negative reciprocals. - Use the derivative to find the tangent slope. - Substitute the slope and point into a line equation.

1. Find the point: \(f(4)=0.5\cdot4^2-2\cdot4+4=4\), so the point is \((4,4)\). 2. Differentiate: \(f'(x)=x-2\). The tangent slope is \(f'(4)=2\). 3. The normal slope is the negative reciprocal: \(m_n=-\frac{1}{2}\). 4. Use point-slope form: \(y-4=-\frac{1}{2}(x-4)\), so \(y=-0.5x+6\).

\(n: y=-0.5x+6\)

- A line is determined by its slope and one point. - Use the derivative to find the curve's slope at \(x=2\). - Substitute \(x=2\) into the original function to find the point of tangency. - Use point-slope form to write the line equation.

1. Find the point of tangency: \(f(2)=0.5\cdot 2^4-3\cdot 2^2+4=0\), so the point is \((2,0)\). 2. Differentiate: \(f'(x)=2x^3-6x\). 3. Find the tangent slope: \(f'(2)=2\cdot 2^3-6\cdot 2=4\). 4. Use point-slope form: \(y-0=4(x-2)\). 5. Simplify: \(y=4x-8\).

\(y=4x-8\)

- The derivative at a point gives the tangent line's slope. - How are a line's slope \(m\) and its angle \(\alpha\) related? - Use degree mode when evaluating the inverse tangent.

1. Differentiate: \(f'(x)=-1.5x^2+4x\). 2. Find the tangent slope at \(x=1\): \(m=f'(1)=-1.5(1)^2+4(1)=2.5\). 3. The angle \(\alpha\) satisfies \(\tan(\alpha)=m\), so \(\alpha=\arctan(2.5)\). 4. In degree mode, \(\alpha\approx 68.2^\circ\).

\(68.2^\circ\)

- The derivative at a point gives the tangent line's slope. - Use point-slope form to write the tangent line. - Relate a line's slope to its angle with the positive x-axis.

1. Find the point of tangency: \(f(3)=\frac{1}{3}\cdot 3^3-3^2-2\cdot 3=-6\), so the point is \((3,-6)\). 2. Differentiate: \(f'(x)=x^2-2x-2\). 3. Find the slope: \(f'(3)=3^2-2\cdot 3-2=1\). 4. Use point-slope form: \(y+6=x-3\), so \(y=x-9\). 5. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

The tangent line is \(y=x-9\), and its angle with the positive x-axis is \(45^\circ\).

- A line is determined by a point and a slope. - Use the derivative to find the tangent slope. - First calculate the complete point of tangency. - Use point-slope form to write the line.

1. Find the point of tangency: \(f(1)=1^3-4(1)^2+5=2\), so \(P=(1,2)\). 2. Differentiate: \(f'(x)=3x^2-8x\). 3. Find the tangent slope: \(f'(1)=3(1)^2-8(1)=-5\). 4. Use point-slope form: \(y-2=-5(x-1)\). 5. Simplify: \(y=-5x+7\).

\(y=-5x+7\)

- Evaluate the function to find the point on the graph. - Tangent and normal slopes are negative reciprocals. - Use point-slope form after finding the normal slope.

1. Find the point: \(f(3)=3^2-4\cdot3+1=-2\), so the point is \((3,-2)\). 2. Differentiate: \(f'(x)=2x-4\). The tangent slope is \(f'(3)=2\). 3. The normal slope is \(-\frac{1}{2}\). 4. Use point-slope form: \(y+2=-\frac{1}{2}(x-3)\), so \(y=-0.5x-0.5\).

\(n: y=-0.5x-0.5\)

- Use the derivative to find the tangent slope at the given point. - The normal line is perpendicular to the tangent line. - Perpendicular nonvertical lines have slopes whose product is \(-1\).

1. Find the point: \(f(2)=\frac{1}{4}\cdot2^3+2=4\), so \(P=(2,4)\). 2. Differentiate: \(f'(x)=\frac{3}{4}x^2+1\). The tangent slope is \(f'(2)=4\). 3. The normal slope is \(-\frac{1}{4}\). 4. Use point-slope form: \(y-4=-\frac{1}{4}(x-2)\), so \(y=-0.25x+4.5\).

\(n: y=-0.25x+4.5\)

- Evaluate the function to find the point on the graph. - Use the derivative for the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Write the line using its slope and the point.

1. Find the point: \(f(2)=\frac{1}{2}\cdot2^3-2\cdot2=0\), so the point is \((2,0)\). 2. Differentiate: \(f'(x)=\frac{3}{2}x^2-2\). The tangent slope is \(f'(2)=4\). 3. The normal slope is \(-\frac{1}{4}\). 4. Use point-slope form: \(y=-\frac{1}{4}(x-2)\), so \(y=-0.25x+0.5\).

\(n: y=-0.25x+0.5\)

- Identify the variable and treat all parameters as constants. - A constant term has derivative \(0\). - Apply the constant-multiple rule to each power term. - Differentiate a sum term by term.

1. For part a), differentiate term by term: \(f'(x)=2x^4-6x^2\). 2. For part b), \(g'(t)=\frac{1}{2}t^2-\frac{3}{2}t+1\). 3. For part c), treat \(k\) as a constant: \(h'(x)=2k^2x-2k\).

a) \(f'(x)=2x^4-6x^2\) b) \(g'(t)=\frac{1}{2}t^2-\frac{3}{2}t+1\) c) \(h'(x)=2k^2x-2k\)

- Can you rewrite the expression as a sum of simple powers before differentiating? - A term with no variable has derivative \(0\). - Fractions and symbols such as \(\pi\) can be constant coefficients or constant terms.

1. Expand part a): \(f(x)=x^3+2x^2-5x-10\). Then \(f'(x)=3x^2+4x-5\). 2. Differentiate part b) term by term. Since \(\pi\) is constant, \(g'(x)=x^3-2x^2\). 3. In part c), differentiate with respect to \(z\), treating \(c\) and \(\frac{1}{c}\) as constants: \(h'(z)=3cz^2-\frac{1}{c}\).

a) \(f'(x)=3x^2+4x-5\) b) \(g'(x)=x^3-2x^2\) c) \(h'(z)=3cz^2-\frac{1}{c}\)

- Parallel lines have equal slopes. - Set the derivative equal to the slope of the given line. - Find the point of tangency after finding its x-coordinate. - Use point-slope form to write the line.

1. Differentiate: \(f'(x)=0.5x+1\). 2. A parallel tangent must have slope \(2\), so solve \(0.5x+1=2\). This gives \(x=2\). 3. Find the point of tangency: \(f(2)=0.25(2)^2+2-3=0\), so the point is \((2,0)\). 4. Write the tangent line: \(y-0=2(x-2)\), so \(y=2x-4\).

\(y=2x-4\)

- Substitute a point into the function to require that it lies on the graph. - Differentiate while treating the parameter as a constant. - A specified tangent slope gives an equation involving the derivative. - Notice whether the derivative at a given input depends on the parameter.

1. For the graph to pass through \((2,5)\), require \(5=4a-8+1\). Thus, \(4a=12\), so \(a=3\). 2. Differentiate: \(f_a'(x)=2ax-4\). At \(x=0\), \(f_a'(0)=-4\), independent of \(a\). 3. For slope \(10\) at \(x=1\), solve \(f_a'(1)=10\): \(2a-4=10\). Thus, \(a=7\).

a) \(a=3\) b) \(-4\) c) \(a=7\)

- Apply the power rule to each term separately. - What happens to a constant term that does not contain \(x\)? - Treat decimal and fractional coefficients as constant factors.

1. Differentiate each term using the power, constant-multiple, and sum rules. 2. The constant term has derivative \(0\), so \(f'(x)=x^4-3x^3+7\).

\(f'(x)=x^4-3x^3+7\)

- Identify the variable of differentiation and treat the other letters as constants. - A parameter multiplying a power remains as a constant coefficient. - A term without the variable has derivative \(0\). - Expanding the squared binomial may simplify part b).

1. In part a), differentiate with respect to \(x\) and treat \(a\) as a constant: \(h'(x)=\frac{3}{a}x^2+2ax-a^3\). 2. For part b), expand first: \(p(t)=9t^2-6kt+k^2\). 3. Differentiate with respect to \(t\), treating \(k\) as a constant: \(p'(t)=18t-6k\).

a) \(h'(x)=\frac{3}{a}x^2+2ax-a^3\) b) \(p'(t)=18t-6k\)

- Decide which symbol is the variable and which is treated as a constant. - Differentiate the sum term by term. - What is the derivative of a term that contains no \(x\)?

1. Treat \(k\) as a constant and differentiate each term. 2. By the power rule, \(\frac{d}{dx}\left(\frac{1}{2k}x^4\right)=\frac{2}{k}x^3\). 3. Also, \(\frac{d}{dx}(-k^2x^2)=-2k^2x\), and \(\frac{d}{dx}(7k)=0\). 4. Therefore, \(f_k'(x)=\frac{2}{k}x^3-2k^2x\).

\(f_k'(x)=\frac{2}{k}x^3-2k^2x\)

- The derivative gives the tangent slope. - Use the tangent function to relate slope and angle. - Find the point of tangency from the original function. - Use point-slope form to write the tangent line.

1. Differentiate: \(f'(x)=-x\). Therefore, the slope at \(x=1\) is \(f'(1)=-1\). 2. Since \(\tan(\alpha)=-1\) and \(\alpha\in[0^\circ,180^\circ)\), \(\alpha=135^\circ\). 3. The point of tangency is \((1,f(1))=(1,2.5)\). Using slope \(-1\), \(y-2.5=-(x-1)\), so \(y=-x+3.5\).

1. \(-1\) 2. \(135^\circ\) 3. \(y=-x+3.5\)

- Expanding a product can turn it into a sum of powers. - A constant term has derivative \(0\). - The value of the derivative at a given input is the graph's slope there.

1. For part a), differentiate term by term: \(f'(x)=3x^5-12x^2+2\). 2. For part b), the constant \(\pi\) has derivative \(0\), so \(f'(x)=2x^2-x\). 3. For part c), expand: \(f(x)=2x^3-x^2\). Then \(f'(x)=6x^2-2x\). 4. Evaluate at \(x=2\): \(f'(2)=6(2)^2-2(2)=20\).

a) \(f'(x)=3x^5-12x^2+2\) b) \(f'(x)=2x^2-x\) c) \(f'(x)=6x^2-2x\); the slope at \(x=2\) is \(20\).

- The derivative gives the slope at a point. - A tangent line requires both a point and a slope. - The tangent and the curve have the same slope at the point of tangency.

1. Differentiate: \(f'(x)=4x-8\). 2. Evaluate the derivative: \(f'(3)=4(3)-8=4\). 3. Find the point of tangency: \(f(3)=2(3)^2-8(3)+5=-1\), so the point is \((3,-1)\). 4. Use point-slope form: \(y+1=4(x-3)\), so \(y=4x-13\).

a) \(4\) b) \(y=4x-13\)

- Use the power rule and constant-multiple rule. - What is the derivative of a term with no variable? - You can distribute division across the terms in a numerator. - Differentiate with respect to the variable used in each function.

1. Differentiate the first function term by term: \(y'=x^3-4x\). 2. The derivative of \(15\) is \(0\), so \(y'=-6\). 3. Differentiate with respect to \(t\): \(z'=12t^2-10t\). 4. Rewrite \(f(x)=\frac{2}{3}x^3-3x\). Then \(f'(x)=2x^2-3\). 5. Treat \(a\), \(b\), and \(c\) as constants: \(y'=2ax+b\).

1) \(y'=x^3-4x\) 2) \(y'=-6\) 3) \(z'=12t^2-10t\) 4) \(f'(x)=2x^2-3\) 5) \(y'=2ax+b\)

- 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: \(f'(2)=-9\).

- Rewrite the fifth root as a power with a rational exponent. - Apply the power rule to each term. - It is often easier to put every term in the form \(ax^r\) before differentiating.

1. Rewrite \(\sqrt[5]{x^3}=x^{\frac{3}{5}}\), so \(f(x)=4x^{\frac{3}{5}}-1.2x^{-\frac{2}{5}}\). 2. Apply the power rule: \(f'(x)=4\left(\frac{3}{5}\right)x^{-\frac{2}{5}}-1.2\left(-\frac{2}{5}\right)x^{-\frac{7}{5}}=2.4x^{-\frac{2}{5}}+0.48x^{-\frac{7}{5}}\).

\(f'(x)=2.4x^{-\frac{2}{5}}+0.48x^{-\frac{7}{5}}\)

- Use the quotient of powers property to simplify the function first. - Rewrite \(\frac{x^a}{x^b}\) as \(x^{a-b}\). - Apply the power rule after simplifying.

1. Simplify with exponent rules: \(f(x)=\frac{6x}{x^{\frac{2}{3}}}=6x^{\frac{1}{3}}\). 2. Apply the power rule: \(f'(x)=6\left(\frac{1}{3}\right)x^{-\frac{2}{3}}=2x^{-\frac{2}{3}}\).

\(f'(x)=2x^{-\frac{2}{3}}=\frac{2}{\sqrt[3]{x^2}}\)

- Substitute the given point into the function to find the parameter. - Differentiate the power function after finding the parameter. - A point and slope determine a line. - An x-intercept has y-coordinate \(0\). - Solve the line equation after setting \(y=0\).

1. Substitute \((2,2)\) into \(f(x)=ax^3\): \(2=8a\), so \(a=\frac{1}{4}\). 2. Differentiate: \(f'(x)=3ax^2\). With \(a=\frac{1}{4}\), \(f'(2)=3\), so the tangent slope is \(3\). 3. A line with slope \(3\) through \((2,2)\) satisfies \(y-2=3(x-2)\). Thus, \(y=3x-4\). 4. At the x-intercept, \(y=0\). Solve \(0=3x-4\), giving \(x=\frac{4}{3}\). The intercept is \((\frac{4}{3},0)\).

a) \(a=\frac{1}{4}\) b) \(f'(2)=3\), and \(y=3x-4\) c) \((\frac{4}{3},0)\)

- Use the derivative to find each tangent slope. - Find the corresponding point on the graph. - Use point-slope form for each tangent line. - The tangent slope equals the derivative at the input.

1. Differentiate: \(f'(x)=3x^2-4x\). 2. At \(x=2\), \(f(2)=0\) and \(f'(2)=4\). The tangent line is \(y=4(x-2)=4x-8\). 3. At \(x=1\), \(f(1)=-1\) and \(f'(1)=-1\). The tangent line is \(y+1=-(x-1)\), so \(y=-x\).

a) \(y=4x-8\) b) \(y=-x\)

- Write the general tangent line at \(x=a\). - The derivative gives the slope in terms of \(a\). - Evaluate the original function at \(a\). - Expand and combine like terms.

1. Differentiate: \(f'(x)=4x-5\). 2. At \(x=a\), the tangent slope is \(f'(a)=4a-5\), and the point is \((a,2a^2-5a)\). 3. Use point-slope form: \(y=(4a-5)(x-a)+2a^2-5a\). 4. Expand and simplify: \(y=(4a-5)x-4a^2+5a+2a^2-5a=(4a-5)x-2a^2\).

\(t_a: y=(4a-5)x-2a^2\)

- 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 \(g'(-4)=-0.125\)

- The derivative at \(x=4\) gives the tangent slope. - Use point-slope form for the line. - Relate slope and angle using tangent.

1. Find the point of tangency: \(f(4)=2\), so the point is \((4,2)\). 2. Differentiate: \(f'(x)=\frac{1}{4}x\). Thus, \(f'(4)=1\). 3. Write the tangent line: \(y-2=x-4\), so \(y=x-2\). 4. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

1. \(y=x-2\) 2. \(45^\circ\)

- 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 \(-\frac{1}{2}\): \(x=1.25\)

- Compare how each exponent changes and what coefficient appears. - Look for a rule connecting the original exponent to the new coefficient and exponent. - To reverse the derivative in part c), which original exponent would produce \(x^{11}\)?

1. Apply the power rule: \(f_1'(x)=3x^2\), \(f_2'(x)=5x^4\), and \(f_3'(x)=7x^6\). 2. The pattern is \(\frac{d}{dx}(x^n)=nx^{n-1}\). 3. To reverse the power rule in part c), increase the exponent from \(11\) to \(12\) and divide the coefficient by \(12\). One possible function is \(h(x)=x^{12}\). More generally, \(h(x)=x^{12}+C\).

a) \(f_1'(x)=3x^2\), \(f_2'(x)=5x^4\), and \(f_3'(x)=7x^6\) b) \(\frac{d}{dx}(x^n)=nx^{n-1}\) c) One possible function is \(h(x)=x^{12}\).

- Rewrite variables in denominators using negative exponents. - Use exponent rules to simplify products with the same base. - Apply the power rule by multiplying by the exponent and subtracting \(1\) from it.

1. For part a), the power rule gives \(f'(x)=13x^{12}\). 2. Rewrite part b) as \(g(x)=x^{-5}\). Then \(g'(x)=-5x^{-6}=-\frac{5}{x^6}\). 3. For part c), \(h'(x)=-9x^{-10}=-\frac{9}{x^{10}}\). 4. Simplify part d): \(s(t)=t^{-4}\). Then \(s'(t)=-4t^{-5}=-\frac{4}{t^5}\).

a) \(f'(x)=13x^{12}\) b) \(g'(x)=-\frac{5}{x^6}\), where \(x\neq0\) c) \(h'(x)=-\frac{9}{x^{10}}\), where \(x\neq0\) d) \(s'(t)=-\frac{4}{t^5}\), where \(t\neq0\)

- 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.

- First find the complete point of tangency. - Use the power rule to find the derivative. - Evaluate the derivative at \(x=-2\). - Use point-slope form for the tangent line.

1. Find the point of tangency: \(f(-2)=\frac{1}{4}(-2)^4-(-2)^2=0\), so \(P=(-2,0)\). 2. Differentiate: \(f'(x)=x^3-2x\). 3. Find the slope: \(f'(-2)=-8+4=-4\). 4. Use point-slope form: \(y=-4(x+2)\), so \(y=-4x-8\).

\(y=-4x-8\)

- 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 \(x=2\)

- Rewrite reciprocals and radicals using negative or rational exponents. - A constant factor remains when differentiating. - Apply \(\frac{d}{dx}(x^n)=nx^{n-1}\).

1. For part a), \(f'(x)=2.5(4x^3)=10x^3\). 2. For part b), \(f'(x)=\frac{5}{12}(6x^5)=\frac{5}{2}x^5\). 3. Rewrite part c) as \(f(x)=-3x^{-2}\). Then \(f'(x)=6x^{-3}=\frac{6}{x^3}\). 4. Rewrite part d) as \(f(x)=6x^{1/2}\). Then \(f'(x)=3x^{-1/2}=\frac{3}{\sqrt{x}}\).

a) \(f'(x)=10x^3\) b) \(f'(x)=\frac{5}{2}x^5\) c) \(f'(x)=\frac{6}{x^3}\), where \(x\neq0\) d) \(f'(x)=\frac{3}{\sqrt{x}}\), where \(x>0\)

- Differentiate a sum one term at a time. - Keep constant coefficients when applying the power rule. - What is the derivative of a constant? - Recall that \(x=x^1\).

1. Differentiate \(4x^6\) to get \(24x^5\). 2. Differentiate \(-\frac{1}{2}x^4\) to get \(-2x^3\). 3. The derivative of \(3x\) is \(3\), and the derivative of \(-5\) is \(0\). 4. Therefore, \(f'(x)=24x^5-2x^3+3\).

\(f'(x)=24x^5-2x^3+3\)

- Track how each exponent and coefficient changes under the power rule. - A term with no \(x\) has derivative \(0\). - Differentiate the function one term at a time. - Keep constant coefficients, including radicals, when differentiating.

1. Differentiate term by term using the power and constant-multiple rules. 2. The derivative is \(f'(x)=5x^5-6x^4+2\sqrt{7}x\).

\(f'(x)=5x^5-6x^4+2\sqrt{7}x\)

- Radicals and \(\pi\) are constants with respect to \(x\). - What is the derivative of a term of the form \(cx\)? - Differentiate a sum term by term. - Decrease each power exponent by exactly \(1\).

1. Differentiate each term using the power, constant-multiple, and sum rules. 2. The polynomial terms give \(2x^7-4x^5+3x^3\). 3. The derivative of \(-\sqrt[5]{10}x\) is \(-\sqrt[5]{10}\), and the derivative of \(\pi\) is \(0\). 4. Therefore, \(f'(x)=2x^7-4x^5+3x^3-\sqrt[5]{10}\).

\(f'(x)=2x^7-4x^5+3x^3-\sqrt[5]{10}\)

- 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) \(11\)

- Convert the angle to a slope using tangent. - Set the derivative equal to that slope. - Find the y-coordinate after solving for the x-coordinate. - A point requires both coordinates.

1. An angle of \(135^\circ\) corresponds to slope \(\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=2x-5\). 3. Set the derivative equal to the required slope: \(2x-5=-1\). Thus, \(x=2\). 4. Evaluate the function: \(f(2)=4-10+2=-4\). Therefore, the point is \((2,-4)\).

\((2,-4)\)

- Expand products into sums of powers before differentiating. - Distribute a negative sign carefully across parentheses. - Use binomial identities when helpful. - After simplifying, differentiate each term with the power rule.

1. Expand part a): \(f(x)=4x^3-3x^2+20x-15\). Then \(f'(x)=12x^2-6x+20\). 2. Simplify part b): \(h(t)=2t^2+8t-(t^2-2t+1)=t^2+10t-1\). Then \(h'(t)=2t+10\). 3. Simplify part c): \(g(z)=\frac{1}{5}z^3-2z+z^2-4\). Then \(g'(z)=\frac{3}{5}z^2+2z-2\).

a) \(f'(x)=12x^2-6x+20\) b) \(h'(t)=2t+10\) c) \(g'(z)=\frac{3}{5}z^2+2z-2\)

- Treat parameters as constants. - Use a difference-of-squares identity in part b). - Expand the squared binomial in part c) before multiplying. - Distinguish the variable from constant factors in each term.

1. Expand part a): \(f(x)=kx^2+2akx+ka^2-bx\). Therefore, \(f'(x)=2kx+2ak-b\). 2. Use the difference of squares in part b): \((t^2+4)(t^2-4)=t^4-16\). Thus \(s(t)=t^4+\frac{2}{3}t^3-16\), and \(s'(t)=4t^3+2t^2\). 3. Expand part c): \(q(r)=3r^3-12r^2+12r\). Therefore, \(q'(r)=9r^2-24r+12\).

a) \(f'(x)=2kx+2ak-b\) b) \(s'(t)=4t^3+2t^2\) c) \(q'(r)=9r^2-24r+12\)

- Rewrite reciprocals and radicals as powers before differentiating. - A constant term has derivative \(0\). - Differentiate a sum term by term. - Treat parameters as constants with respect to \(x\).

1. Differentiate each term in part a): \(f'(x)=0.2\cdot5x^4-2\cdot3x^2+4=x^4-6x^2+4\). 2. For part b), rewrite \(f(x)=4x^{-1}+6x^{1/2}\). Then \(f'(x)=-4x^{-2}+3x^{-1/2}=-\frac{4}{x^2}+\frac{3}{\sqrt{x}}\), for \(x>0\). 3. For part c), treat \(a\), \(c\), and \(n\) as constants. The power rule gives \(f'(x)=ncx^{n-1}\).

a) \(f'(x)=x^4-6x^2+4\) b) \(f'(x)=-\frac{4}{x^2}+\frac{3}{\sqrt{x}}\), for \(x>0\) c) \(f'(x)=ncx^{n-1}\)

- A tangent line requires a point and a slope. - Use the derivative to find the slope at \(x=1\). - Evaluate the original function to find the point. - Use point-slope form.

1. Find the point of tangency: \(f(1)=0.25-2+5=3.25\), so the point is \((1,3.25)\). 2. Differentiate: \(f'(x)=x^3-4x\). 3. Find the slope: \(f'(1)=1-4=-3\). 4. Use point-slope form: \(y-3.25=-3(x-1)\), so \(y=-3x+6.25\).

\(y=-3x+6.25\)

- 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.

- Use the drawn tangent to make a rough angle estimate before calculating. - The derivative gives the exact tangent slope at the specified input. - Relate slope to inclination with \(m=\tan(\alpha)\).

1. From the graph, the tangent rises about one unit for each unit it moves to the right, so an estimate near \(45^\circ\) is reasonable. 2. Differentiate: \(f'(x)=x\). 3. At \(x=1\), the tangent slope is \(m=f'(1)=1\). 4. Since \(m=\tan(\alpha)\), \(\alpha=\arctan(1)=45^\circ\).

Graphical estimate: about \(45^\circ\); exact angle: \(\alpha=45^\circ\).

- First use the drawn tangent to determine the sign of its slope. - Compute the exact slope from the derivative at \(x=0.5\). - Convert a negative inverse-tangent value to the standard inclination interval \([0^\circ,180^\circ)\).

1. The drawn tangent falls from left to right, so its slope is negative and its angle of inclination is obtuse. 2. Differentiate: \(f'(x)=-2x\). 3. At \(x=0.5\), the tangent slope is \(m=f'(0.5)=-1\). 4. The calculator value \(\arctan(-1)=-45^\circ\) corresponds to an angle of inclination of \(135^\circ\) in \([0^\circ,180^\circ)\).

The angle is obtuse, and its exact measure is \(135^\circ\).

- Differentiate the polynomial term by term. - Evaluate the derivative at the specified input. - Compare the result with the slope of a horizontal line.

1. Differentiate term by term: \(f'(x)=3x^2-6x+3\). 2. Evaluate at \(x=1\): \(f'(1)=0\). 3. Since a horizontal tangent has slope \(0\), the graph has a horizontal tangent at \(x=1\).

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

- Rewrite the reciprocal using a negative exponent. - Differentiate both terms and evaluate at \(x=1\). - Use the original function, not the derivative, for the y-coordinate.

1. Rewrite \(f(x)=x^{-1}+x\). Then \(f'(x)=-x^{-2}+1=-\frac{1}{x^2}+1\). 2. Since \(f'(1)=0\), the tangent is horizontal at \(x=1\). 3. The original function gives \(f(1)=2\), so the point of tangency is \((1,2)\).

\(f'(x)=-\frac{1}{x^2}+1\), \(f'(1)=0\), and the point of tangency is \((1,2)\).

- Evaluate the original function to locate the point of tangency. - Differentiate the polynomial term by term for the tangent slope. - Use point-slope form to write the line.

1. The point of tangency is \((3,-0.5)\). 2. Differentiate term by term: \(f'(x)=x-2\), so the tangent slope is \(f'(3)=1\). 3. Using point-slope form, \(y+0.5=x-3\), so \(y=x-3.5\).

\(t:y=x-3.5\)

- Apply the Power Rule to the quadratic term. - Keep the constant factor attached to each differentiated term. - Evaluate only after finding the derivative function.

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

\(f'(1)=1\)

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

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

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

- Rewrite the reciprocal as a negative power before differentiating. - Parallel lines have equal slopes. - Use the original function to find the tangency point, then write the tangent line.

1. Rewrite \(f(x)=8x^{-1}\) and differentiate: \(f'(x)=-\frac{8}{x^2}\). 2. At \(x=2\), the tangent slope is \(f'(2)=-2\), the same as the slope of \(g\), so the lines are parallel. 3. Since \(f(2)=4\), the tangency point is \((2,4)\). 4. Point-slope form gives \(y-4=-2(x-2)\), so \(t:y=-2x+8\).

a) Since \(f'(2)=-2\), the tangent line has the same slope as \(g\), so the lines are parallel. b) \(t:y=-2x+8\)

- Differentiate the polynomial term by term. - A horizontal tangent corresponds to a derivative value of \(0\). - For a prescribed slope, set the derivative equal to that slope. - Substitute each resulting x-coordinate into the original function to find the point.

1. Differentiate using the power, constant, and sum rules: \(f'(x)=-2x+4\). 2. A horizontal tangent has slope \(0\), so solve \(-2x_0+4=0\). This gives \(x_0=2\). 3. Evaluate the function: \(f(2)=2\). Therefore, \(P=(2,2)\). 4. For slope \(6\), solve \(-2q+4=6\). This gives \(q=-1\). 5. Evaluate the function: \(f(-1)=-7\). Therefore, \(Q=(-1,-7)\).

a) \(f'(x)=-2x+4\) b) \(P=(2,2)\) c) \(Q=(-1,-7)\)

- What must be true about the slopes of parallel lines? - Read the slope of the given line from its equation. - Use the slope and point of tangency to write the tangent line equation.

1. Differentiate: \(g'(x)=x-3\). 2. Parallel lines have equal slopes. The given line has slope \(2\), so solve \(g'(b)=2\): \(b-3=2\), giving \(b=5\). 3. Evaluate the function: \(g(5)=\frac{1}{2}\cdot 5^2-3\cdot 5=-2.5\). Thus, \(B=(5,-2.5)\). 4. Use slope \(2\) and point \((5,-2.5)\): \(y+2.5=2(x-5)\). 5. Simplify: \(y=2x-12.5\).

a) \(g'(x)=x-3\) b) \(B=(5,-2.5)\) c) \(y=2x-12.5\)

- Begin by differentiating the original polynomial rather than using a supplied derivative. - Set the derivative equal to each required slope. - Parallel lines have equal slopes. - After finding an input, use the original function to recover the corresponding point.

1. Differentiate term by term: \(f'(x)=3x^2-12\). 2. Evaluate the derivative: \(f'(-1)=-9\) and \(f'(3)=15\). 3. For slope \(15\), solve \(3x^2-12=15\). Then \(x^2=9\), so \(x=-3\) or \(x=3\). 4. A line parallel to \(y=-9x+2\) must have slope \(-9\). Solving \(3x^2-12=-9\) gives \(x=-1\) or \(x=1\). Since \(f(-1)=16\) and \(f(1)=-6\), the points are \((-1,16)\) and \((1,-6)\). 5. At \(x=0\), \(f(0)=5\) and \(f'(0)=-12\). The tangent line is \(y=-12x+5\).

a) \(f'(-1)=-9\) and \(f'(3)=15\) b) \(x=-3\) and \(x=3\) c) \((-1,16)\) and \((1,-6)\) d) \(y=-12x+5\)

- Differentiate the original polynomial term by term first. - Set the derivative equal to the requested slope and solve. - Read the comparison line's slope from its equation. - Substitute each relevant input into the original function to find the point.

1. Differentiate term by term: \(g'(x)=x^2+2x-3\). 2. Evaluate: \(g'(-2)=-3\) and \(g'(1)=0\). 3. For slope \(5\), solve \(x^2+2x-3=5\). Factoring \(x^2+2x-8=(x+4)(x-2)\) gives \(x=-4\) or \(x=2\). 4. A tangent parallel to \(y=-3x+10\) must have slope \(-3\). Solve \(x^2+2x-3=-3\), giving \(x=0\) or \(x=-2\). 5. Since \(g(0)=0\) and \(g(-2)=\frac{22}{3}\), the points are \((0,0)\) and \((-2,\frac{22}{3})\).

a) \(g'(-2)=-3\) and \(g'(1)=0\) b) \(x=-4\) and \(x=2\) c) \((0,0)\) and \((-2,\frac{22}{3})\)

- Recall the algebraic tests for symmetry using \(f(-x)\). - Consider what happens to the exponent of \(x^n\) when you differentiate it. - Substitute \(-x\) into the derivative and compare the result with \(f'(x)\) and \(-f'(x)\).

1. Evaluate \(f(-x)\): \(f(-x)=-(-x)^4+4(-x)^2-2=-x^4+4x^2-2=f(x)\). Therefore, \(f\) is even, and its graph is symmetric about the \(y\)-axis. 2. Apply the power rule and sum rule: \(f'(x)=-4x^3+8x\). 3. Evaluate the derivative at \(-x\): \(f'(-x)=-4(-x)^3+8(-x)=4x^3-8x=-f'(x)\). Therefore, \(f'\) is odd, and its graph is symmetric about the origin. 4. Differentiating an even polynomial changes each positive even exponent to an odd exponent, while the constant term disappears. Thus, the derivative of an even polynomial is odd.

a) The graph of \(f\) is symmetric about the \(y\)-axis because \(f(-x)=f(x)\). b) \(f'(x)=-4x^3+8x\). Its graph is symmetric about the origin because \(f'(-x)=-f'(x)\). c) The derivative of an even polynomial is an odd polynomial.

- Recall which powers can appear in an odd polynomial. - Choose simple coefficients for your cubic example. - Compare the parity of an odd integer \(k\) with the parity of \(k-1\).

1. A polynomial symmetric about the origin is an odd function, so every nonzero term has an odd exponent. One possible example is \(g(x)=x^3+2x\). 2. For this example, \(g'(x)=3x^2+2\). Since \(g'(-x)=3(-x)^2+2=3x^2+2=g'(x)\), the derivative is even and its graph is symmetric about the \(y\)-axis. 3. In general, every nonzero term of an odd polynomial has the form \(a_kx^k\) with odd \(k\). Its derivative is \(ka_kx^{k-1}\), and \(k-1\) is even. Therefore, the derivative contains only even powers and is symmetric about the \(y\)-axis.

a) Every nonzero term must have an odd exponent. One possible example is \(g(x)=x^3+2x\). b) For this example, \(g'(x)=3x^2+2\), which is symmetric about the \(y\)-axis. c) Differentiating each odd power produces an even power, so the derivative is an even polynomial.

- 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 \(x=2\)

- Apply the power rule to the leading term \(a_nx^n\). - Can a function be strictly increasing while having a horizontal tangent at one point? - Which powers occur in an even polynomial, and what happens to them after differentiation? - What derivative value corresponds to a horizontal tangent?

1. Statement a) is true. If the leading term is \(a_nx^n\), where \(a_n\ne 0\), then its derivative is \(na_nx^{n-1}\). Since \(n\ge 1\), the new leading coefficient is nonzero, so the derivative has degree \(n-1\). 2. Statement b) is false. The function \(f(x)=x^3\) is strictly increasing on all real numbers, but \(f'(x)=3x^2\), and \(f'(0)=0\). 3. Statement c) is true. Such a polynomial is even and contains only even powers. Its derivative contains only odd powers, so \(f'(0)=0\). Therefore, the tangent line at \(x=0\) is horizontal.

a) True b) False c) True

- Find the point of tangency and the derivative at \(x=3\). - A line is determined by a point and its slope. - The derivative gives the tangent line's slope. - What is the y-coordinate of every point on the x-axis?

1. Find the point of tangency: \(f(3)=\frac{1}{2}\cdot 3^2-2\cdot 3+1=-0.5\), so \(P=(3,-0.5)\). 2. Differentiate: \(f'(x)=x-2\). 3. Find the tangent slope: \(f'(3)=1\). 4. Write the tangent line: \(y+0.5=x-3\), so \(y=x-3.5\). 5. At an x-intercept, \(y=0\). Solve \(0=x-3.5\) to get \(x=3.5\).

\((3.5,0)\)

- First find the point where the tangent touches the graph. - Use the derivative to find the tangent slope. - Write the tangent line equation. - A point on the x-axis has y-coordinate \(0\).

1. Find the point of tangency: \(g(1)=1^3-4(1)=-3\), so the point is \((1,-3)\). 2. Differentiate: \(g'(x)=3x^2-4\). 3. Find the tangent slope: \(g'(1)=3(1)^2-4=-1\). 4. Write the tangent line: \(y+3=-(x-1)\), so \(y=-x-2\). 5. Set \(y=0\) to find the x-intercept: \(0=-x-2\), giving \(x=-2\).

\((-2,0)\)

- Find the slope of a line whose angle of inclination is \(135^\circ\). - Express the tangent slope at an unknown input using the derivative. - Set the derivative equal to the required slope and solve.

1. Convert the angle to a slope: \(m=\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=\frac{1}{2}x+1\). 3. Set the derivative equal to the required slope: \(\frac{1}{2}x_0+1=-1\). 4. Solve: \(\frac{1}{2}x_0=-2\), so \(x_0=-4\).

\(x_0=-4\)

- Convert the given angle to a slope. - Set the derivative equal to that slope to find the x-coordinate. - Substitute the x-coordinate into the original function, then write the tangent line.

1. Convert the angle to a slope: \(m=\tan(135^\circ)=-1\). 2. Differentiate: \(f'(x)=2x-4\). 3. Set the derivative equal to the tangent slope: \(2x-4=-1\). Solving gives \(x=1.5\). 4. Find the y-coordinate: \(f(1.5)=(1.5)^2-4(1.5)+2=-1.75\). Thus, \(B=(1.5,-1.75)\). 5. Use point-slope form: \(y+1.75=-(x-1.5)\), so \(y=-x-0.25\).

\(B=(1.5,-1.75)\) and \(y=-x-0.25\)

- Evaluate the function to find the point of tangency. - Use the derivative for the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Use inverse tangent to find each angle of inclination.

1. Find the point: \(f(4)=0.5\cdot4^2-2\cdot4=0\), so the point is \((4,0)\). 2. Differentiate: \(f'(x)=x-2\). The tangent slope is \(m_t=f'(4)=2\). 3. The tangent line is \(y=2(x-4)=2x-8\). 4. The normal slope is \(m_n=-\frac{1}{2}\), so the normal line is \(y=-0.5(x-4)=-0.5x+2\). 5. The tangent angle is \(\alpha_t=\arctan(2)\approx63.43^\circ\). The normal angle is \(\alpha_n=\arctan(-0.5)+180^\circ\approx153.43^\circ\).

\(t: y=2x-8\) \(n: y=-0.5x+2\) \(\alpha_t\approx63.43^\circ\), \(\alpha_n\approx153.43^\circ\)

- What relationship must the slopes of two perpendicular nonvertical lines satisfy? - Use the derivative to express the tangent slopes at the two intercepts. - Substitute those slopes into the perpendicularity condition. - Use the condition \(k>0\) when choosing the solution.

1. Differentiate: \(f_k'(x)=2kx\). 2. The tangent slopes are \(m_P=f_k'(-2)=-4k\) and \(m_Q=f_k'(2)=4k\). 3. A right angle at \(S\) means the two tangent lines are perpendicular, so their slopes satisfy \(m_Pm_Q=-1\). 4. Substitute: \((-4k)(4k)=-1\), so \(-16k^2=-1\) and \(k^2=\frac{1}{16}\). 5. Since \(k>0\), \(k=\frac{1}{4}\).

\(k=\frac{1}{4}\)

- The tangent slope is the derivative at the point. - The slopes of a tangent and its normal are negative reciprocals when both are defined. - Convert an angle of inclination to a slope using tangent. - To verify a tangent line, check both the point and the slope.

1. Differentiate: \(f'(x)=x-3\). 2. At \(x=4\), \(f(4)=-3\) and \(f'(4)=1\). The tangent line is \(y+3=x-4\), or \(y=x-7\). The normal slope is \(-1\), so the normal line is \(y+3=-(x-4)\), or \(y=-x+1\). 3. An angle of \(135^\circ\) corresponds to slope \(\tan(135^\circ)=-1\). Solve \(x-3=-1\) to get \(x=2\). 4. At \(x=2\), \(f(2)=-3\) and \(f'(2)=-1\). The tangent line is \(y+3=-(x-2)\), which simplifies to \(y=-x-1\). Therefore, the given line is tangent at \(x=2\).

a) Tangent: \(y=x-7\); normal: \(y=-x+1\) b) \(x=2\) c) Yes. At \(x=2\), the point is \((2,-3)\) and the tangent slope is \(-1\), giving \(y=-x-1\).

- A line parallel to the x-axis has slope \(0\). - Substitute the given input into the original function to find the point. - A normal line has slope equal to the negative reciprocal of the tangent slope. - Use point-slope form to write each line.

1. Differentiate: \(f'(x)=x^2-1\). 2. At \(x=2\), \(f(2)=\frac{2}{3}\) and \(f'(2)=3\). The tangent line is \(y-\frac{2}{3}=3(x-2)\), or \(y=3x-\frac{16}{3}\). 3. A tangent parallel to the x-axis has slope \(0\). Solve \(x^2-1=0\), giving \(x=-1\) and \(x=1\). 4. At \(x=0\), \(f(0)=0\) and the tangent slope is \(-1\). The normal slope is \(1\), so the normal line is \(y=x\).

a) \(y=3x-\frac{16}{3}\) b) \(x=-1\) and \(x=1\) c) \(y=x\)

- First verify that the two functions have the same value at \(x=1\). - Then compare their derivative values at that input. - A common tangent must share both the point and the slope.

1. The function values agree: \(f(1)=3\) and \(g(1)=3\), so both graphs pass through \((1,3)\). 2. Differentiate term by term: \(f'(x)=2x+1\) and \(g'(x)=-2x+5\). 3. The slopes also agree at \(x=1\): \(f'(1)=g'(1)=3\). Therefore, the graphs are tangent at \((1,3)\). 4. The common tangent line is \(y-3=3(x-1)\), so \(y=3x\).

1. Both graphs pass through \((1,3)\), and \(f'(1)=g'(1)=3\), so they are tangent there. 2. \(y=3x\)

- The derivative gives the graph's slope at a point. - A tangent line with slope \(0\) is horizontal. - The normal slope is the negative reciprocal of the tangent slope. - Use point-slope form for each line.

1. Differentiate: \(f'(x)=x^3-4x\). 2. At \(x=1\), \(f'(1)=1-4=-3\). 3. At \(x=2\), \(f(2)=1\) and \(f'(2)=0\). The tangent line is horizontal, so its equation is \(y=1\). 4. At \(x=-1\), \(f(-1)=\frac{13}{4}\) and the tangent slope is \(f'(-1)=3\). The normal slope is \(-\frac{1}{3}\). 5. The normal line is \(y-\frac{13}{4}=-\frac{1}{3}(x+1)\), which simplifies to \(y=-\frac{1}{3}x+\frac{35}{12}\).

a) \(-3\) b) \(y=1\) c) \(y=-\frac{1}{3}x+\frac{35}{12}\)

- Convert the known normal slope to the tangent slope. - The derivative gives the tangent slope at each input. - Set the derivative equal to the required tangent slope. - Evaluate the function before writing the line equation.

1. The tangent slope is the negative reciprocal of the normal slope: \(m_t=-\frac{1}{0.5}=-2\). 2. Differentiate: \(f'(x)=2x-4\). Set \(f'(x_0)=-2\): \(2x_0-4=-2\), so \(x_0=1\). 3. Evaluate the function: \(f(1)=2\), so the point is \((1,2)\). 4. The normal line is \(y-2=0.5(x-1)\), or \(y=0.5x+1.5\).

a) \(x_0=1\) b) \(n: y=0.5x+1.5\)

- Find the tangent slope at \(A\) using the derivative. - Convert the tangent slope to the normal slope. - Write the normal line through \(A\). - Solve the two line equations as a system.

1. Differentiate: \(p'(x)=2x-4\). The tangent slope at \(x=3\) is \(p'(3)=2\). 2. The normal slope is \(-\frac{1}{2}\). 3. Through \(A(3,3)\), the normal line is \(y-3=-0.5(x-3)\), or \(y=-0.5x+4.5\). 4. Set the normal equation equal to \(y=x-3\): \(-0.5x+4.5=x-3\). Thus, \(7.5=1.5x\), so \(x=5\). 5. Then \(y=5-3=2\), so \(S=(5,2)\).

\(S=(5,2)\)

- Set the derivative equal to the given slope. - Substitute each solution into the original function to find the points. - Use a coordinate-area formula for a triangle whose one vertex is the origin. - Take the absolute value so the area is positive.

1. Differentiate: \(f'(x)=x^2-2x-2\). 2. Set the derivative equal to \(1\): \(x^2-2x-2=1\), or \(x^2-2x-3=0\). Thus, \(x=-1\) or \(x=3\). 3. Find the corresponding points: \(f(-1)=\frac{2}{3}\) and \(f(3)=-6\). Thus, the points are \((-1,\frac{2}{3})\) and \((3,-6)\). 4. With one vertex at the origin, the area is \(\frac{1}{2}|x_1y_2-x_2y_1|\). 5. Therefore, \(A=\frac{1}{2}|(-1)(-6)-3(\frac{2}{3})|=\frac{1}{2}|6-2|=2\).

\(2\) square units

- The y-intercept occurs where \(x=0\). - A horizontal tangent has derivative \(0\). - The derivative at a point gives the tangent slope. - Use point-slope form to write each tangent line.

1. The y-intercept occurs at \(x=0\), and \(f(0)=11\). Differentiate: \(f'(x)=3x^2-6x-9\), so \(f'(0)=-9\). The tangent line is \(y-11=-9(x-0)\), or \(y=-9x+11\). 2. A horizontal tangent has slope \(0\). Solve \(3x^2-6x-9=0\). Dividing by \(3\) gives \(x^2-2x-3=0\), so \(x=-1\) or \(x=3\). 3. At \(x=4\), \(f(4)=-9\) and \(f'(4)=15\). Thus, \(P=(4,-9)\), and the tangent line is \(y+9=15(x-4)\), or \(y=15x-69\).

1. At \(x=0\), \(f(0)=11\) and \(f'(0)=-9\), so the tangent is \(y=-9x+11\). 2. \(x=-1\) and \(x=3\) 3. \(P=(4,-9)\) and \(y=15x-69\)

- Differentiate the polynomial term by term to obtain the tangent slope. - Use the original function to locate the point of tangency. - Write the line with point-slope form. - Relate slope to inclination with the tangent function and use the requested precision.

1. Find the point of tangency: \(f(2)=1\), so \(P=(2,1)\). 2. Differentiate term by term: \(f'(x)=3x^2-4x\). 3. The tangent slope is \(f'(2)=4\). 4. The tangent line is \(y-1=4(x-2)\), or \(y=4x-7\). 5. The angle \(\alpha\) satisfies \(\tan(\alpha)=4\), so \(\alpha=\arctan(4)\approx75.96^\circ\).

Tangent line: \(y=4x-7\) Angle: \(\alpha\approx75.96^\circ\)

- Use the derivative to find the tangent slope. - Substitute \(x=1\) into the original function to find the point. - Relate the slope to the angle using tangent. - Write the line in slope-intercept or point-slope form.

1. Find the point of tangency: \(f(1)=-0.25(1)^4+1^2+1=1.75\), so \(P=(1,1.75)\). 2. Differentiate: \(f'(x)=-x^3+2x\). 3. Find the slope: \(f'(1)=1\). 4. Write the tangent line: \(y-1.75=x-1\), so \(y=x+0.75\). 5. Since \(\tan(\alpha)=1\), the angle is \(\alpha=45^\circ\).

Tangent line: \(y=x+0.75\) Angle: \(45^\circ\)

- Parallel lines have equal slopes. - Set the derivative equal to the specified slope. - Find the corresponding points on the graph. - Use each point and the common slope to write a tangent line.

1. Differentiate: \(f'(x)=x^2-4\). 2. The given line has slope \(5\). Set \(f'(x)=5\): \(x^2-4=5\), so \(x=-3\) or \(x=3\). 3. Evaluate the function: \(f(-3)=5\) and \(f(3)=-1\). 4. At \((-3,5)\), the tangent line is \(y-5=5(x+3)\), or \(y=5x+20\). 5. At \((3,-1)\), the tangent line is \(y+1=5(x-3)\), or \(y=5x-16\).

\(y=5x+20\) and \(y=5x-16\)

- Treat the parameter as a constant when differentiating. - Set the derivative at \(x=2\) equal to the required slope. - Find the point of tangency using the corresponding function. - Use point-slope form for the tangent line.

1. Differentiate while treating \(k\) as a constant: \(g_k'(x)=3x^2-2kx\). 2. Require slope \(4\) at \(x=2\): \(3(2)^2-2k(2)=4\). Thus, \(12-4k=4\), so \(k=2\). 3. For \(k=2\), \(g_2(2)=2^3-2(2)^2+4=4\). The point of tangency is \((2,4)\). 4. Using slope \(4\), the tangent line is \(y-4=4(x-2)\), or \(y=4x-4\).

a) \(g_k'(x)=3x^2-2kx\) b) \(k=2\) c) \(y=4x-4\)

- Use the derivative to find the tangent slope. - Find the complete point of tangency. - Use point-slope form to write the line. - Set \(x=0\) to find the y-intercept.

1. Differentiate: \(f'(x)=x-2\). 2. At \(x=-1\), \(f(-1)=1.5\) and \(f'(-1)=-3\). 3. Write the tangent line: \(y-1.5=-3(x+1)\), so \(y=-3x-1.5\). 4. At the y-intercept, \(x=0\). Thus, \(y=-1.5\), and the intercept is \((0,-1.5)\).

Tangent line: \(y=-3x-1.5\) Y-intercept: \((0,-1.5)\)

- Evaluate the function to find the full point. - Use the derivative for the tangent slope. - Perpendicular line slopes are negative reciprocals. - Use point-slope form for both lines.

1. Differentiate: \(g'(x)=x^2-2x\). 2. Evaluate the function and derivative: \(g(3)=2\) and \(g'(3)=3\). 3. The tangent line is \(y-2=3(x-3)\), or \(y=3x-7\). 4. The normal slope is \(-\frac{1}{3}\). Thus, the normal line is \(y-2=-\frac{1}{3}(x-3)\), or \(y=-\frac{1}{3}x+3\).

a) \(t: y=3x-7\) b) \(n: y=-\frac{1}{3}x+3\)

- What value must the derivative have at a horizontal tangent? - Treat \(a\) as a constant while differentiating with respect to \(x\). - Substitute the given x-coordinate into the derivative equation.

1. Treat \(a\) as a constant and apply the power rule: \(g_a'(x)=3ax^2-8x+a\). 2. A horizontal tangent at \(x=1\) requires \(g_a'(1)=0\). 3. Substitute \(x=1\): \(3a-8+a=0\). 4. Solve \(4a-8=0\) to obtain \(a=2\).

a) \(g_a'(x)=3ax^2-8x+a\) b) \(a=2\)

- Substitute a point into the original function to find a parameter. - The derivative gives the slope at a specified x-value. - Apply the power rule term by term. - The origin has x-coordinate \(0\).

1. Substitute \((2, 8)\): \(8-4k+4=8\). Solving gives \(k=1\). 2. Differentiate: \(f_k'(x)=3x^2-2kx+2\). 3. At the origin, \(f_k'(0)=2\), independent of \(k\). 4. For slope \(11\) at \(x=3\), solve \(f_k'(3)=11\): \(27-6k+2=11\). 5. This gives \(6k=18\), so \(k=3\).

a) \(k=1\) b) The slope is \(2\). c) \(k=3\)

- Substitute the given point into the function. - Differentiate to find a slope formula. - Evaluate the derivative at the specified x-value. - Decide whether the requested slope is constant or depends on the parameter.

1. Substitute \((1, 2)\): \(1-3+t=2\), so \(t=4\). 2. Differentiate: \(g_t'(x)=3x^2-6x+t\). 3. At the origin, \(g_t'(0)=t\). 4. For slope \(5\) at \(x=2\), solve \(g_t'(2)=5\): \(12-12+t=5\). 5. Therefore, \(t=5\).

a) \(t=4\) b) The slope is \(t\). c) \(t=5\)

- Convert each angle to a slope using tangent. - Set the derivative equal to the required slope. - Check whether the resulting squared equation has real solutions. - Consider the range of \(g'(x)=x^2-1\).

1. An angle of \(45^\circ\) corresponds to slope \(\tan(45^\circ)=1\). Differentiate: \(g'(x)=x^2-1\). 2. Solve \(x^2-1=1\). Then \(x^2=2\), so \(x=\pm\sqrt{2}\). 3. Evaluate the function: \(g(\sqrt{2})=-\frac{\sqrt{2}}{3}\) and \(g(-\sqrt{2})=\frac{\sqrt{2}}{3}\). The points are \((\sqrt{2},-\frac{\sqrt{2}}{3})\) and \((-\sqrt{2},\frac{\sqrt{2}}{3})\). 4. An angle of \(120^\circ\) corresponds to slope \(\tan(120^\circ)=-\sqrt{3}\). The equation \(x^2-1=-\sqrt{3}\) becomes \(x^2=1-\sqrt{3}\). 5. Since \(1-\sqrt{3}<0\), there is no real solution. Equivalently, \(g'(x)=x^2-1\ge -1\), so it can never equal \(-\sqrt{3}\).

1. \((\sqrt{2},-\frac{\sqrt{2}}{3})\) and \((-\sqrt{2},\frac{\sqrt{2}}{3})\) 2. No such point exists because \(g'(x)\ge -1\), while \(\tan(120^\circ)=-\sqrt{3}<-1\).

- 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) \(k=2\)

- A horizontal tangent has slope \(0\). - Evaluate the derivative at both specified inputs. - Compare absolute values when comparing steepness.

1. Differentiate: \(g'(x)=x^2-4x+3\). 2. Horizontal tangent lines occur where \(g'(x)=0\). Factor \(x^2-4x+3=(x-1)(x-3)\), so \(x=1\) and \(x=3\). 3. Evaluate the slopes: \(g'(0)=3\) and \(g'(4)=3\). 4. Steepness is determined by the absolute value of slope. Since both slopes have magnitude \(3\), the graph is equally steep at the two points.

a) \(x=1\) and \(x=3\) b) The graph is equally steep at \(x=0\) and \(x=4\); the slope is \(3\) at each point.

- 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 \(x=4\)

- Find the point and the derivative at \(x=1\) in terms of \(k\). - Write the tangent line in point-slope form. - Substitute the coordinates of \(Q\) into the tangent-line equation. - Solve the resulting equation for \(k\).

1. At \(x=1\), the point on the graph is \(B(1, f_k(1))=(1, 5-k)\). 2. Differentiate: \(f_k'(x)=4x^3-2kx\), so the tangent slope at \(x=1\) is \(4-2k\). 3. The tangent line is \(y-(5-k)=(4-2k)(x-1)\). 4. Because the line passes through \(Q(0, 1)\), \(1-(5-k)=(4-2k)(-1)\). 5. Simplifying gives \(k-4=-4+2k\), so \(k=0\).

\(k=0\)

- Simplify the rational expression before differentiating. - Use the derivative value at \(x=1\) as the tangent slope. - For the asymptote, identify where the derivative denominator is zero while the numerator is nonzero. - For the sign question, analyze the numerator and denominator only near \(x=0\).

1. Simplify before differentiating: \(f(x)=4.5x-6+\frac{2}{x}\). 2. Differentiate term by term: \(f'(x)=4.5-\frac{2}{x^2}=\frac{9x^2-4}{2x^2}\). 3. At \(x=1\), the slope is \(f'(1)=2.5\). Through \((1,0.5)\), the tangent is \(y-0.5=2.5(x-1)\), so \(y=2.5x-2\). 4. The derivative is undefined at \(x=0\), and its numerator approaches \(-4\ne0\), so \(x=0\) is a vertical asymptote of \(f'\). For \(x\) sufficiently close to \(0\), \(9x^2-4<0\) while \(2x^2>0\), so \(f'(x)<0\) on both sides. Thus, there is no sign change across the asymptote sufficiently close to \(0\).

a) \(y=2.5x-2\) b) The vertical asymptote of \(f'\) is \(x=0\). The derivative is negative on both sides sufficiently close to the asymptote, so it does not change sign there.

- Rewrite each radical or reciprocal using a rational exponent first. - In the Power Rule, compare the original exponent with the coefficient of the derivative. - Check that the exponent in the derivative is exactly one less than the original exponent.

1. \(g(x)=x^{5/4}\). Applying the Power Rule gives \(g'(x)=\frac54x^{1/4}\). 2. \(k(x)=x^{-1/4}\). Applying the Power Rule gives \(k'(x)=-\frac14x^{-5/4}\). 3. In each case, the original exponent becomes the coefficient and the new exponent is one less. Thus, for rational \(r\) and \(x>0\), \(\frac{d}{dx}x^r=rx^{r-1}\).

1. \(g(x)=x^{5/4}\), so \(g'(x)=\frac54x^{1/4}\) 2. \(k(x)=x^{-1/4}\), so \(k'(x)=-\frac14x^{-5/4}\) 3. For rational \(r\) and \(x>0\), \(\frac{d}{dx}x^r=rx^{r-1}\)

- Rewrite each radical and reciprocal as a power with a rational exponent. - The derivative evaluated at a point gives the slope of the tangent line there. - Any power of \(1\) equals \(1\). - Watch the signs when differentiating negative exponents.

1. Rewrite the function using rational exponents: \(f(x)=\frac{3}{4}x^{\frac{3}{2}}-2x^{-\frac{1}{4}}\). 2. Apply the power rule: \(f'(x)=\frac{3}{4}\cdot\frac{3}{2}x^{\frac{1}{2}}-2\left(-\frac{1}{4}\right)x^{-\frac{5}{4}}=\frac{9}{8}x^{\frac{1}{2}}+\frac{1}{2}x^{-\frac{5}{4}}\). 3. Evaluate at \(x=1\): \(f'(1)=\frac{9}{8}+\frac{1}{2}=\frac{13}{8}=1.625\).

The tangent line has slope \(\frac{13}{8}=1.625\).

- First rewrite radicals and reciprocals in the form \(x^r\). - Apply the power rule to each term. - When differentiating a power, multiply by the exponent and subtract \(1\) from the exponent. - Substituting \(x=1\) simplifies every power of \(1\) to \(1\).

1. a) By the power rule, \(f'(x)=1.2\left(\frac{1}{2}\right)x^{-\frac{1}{2}}=0.6x^{-\frac{1}{2}}\). Therefore, \(f'(1)=0.6\). 2. b) Rewrite \(f(x)=x^{\frac{7}{3}}\). Then \(f'(x)=\frac{7}{3}x^{\frac{4}{3}}\), so \(f'(1)=\frac{7}{3}\approx2.333\). 3. c) Rewrite \(f(x)=6x^{-\frac{3}{2}}\). Then \(f'(x)=-9x^{-\frac{5}{2}}\), so \(f'(1)=-9\).

a) \(f'(x)=0.6x^{-\frac{1}{2}}\); \(f'(1)=0.6\) b) \(f'(x)=\frac{7}{3}x^{\frac{4}{3}}\); \(f'(1)=\frac{7}{3}\approx2.333\) c) \(f'(x)=-9x^{-\frac{5}{2}}\); \(f'(1)=-9\)

- Simplify each expression with exponent rules before differentiating. - Recall that \(\sqrt[n]{x^m}=x^{\frac{m}{n}}\). - A power in the denominator can be written with a negative exponent. - When multiplying powers with the same base, add their exponents.

1. a) Combine the powers: \(g(x)=x^{\frac{7}{5}}x^{\frac{4}{5}}=x^{\frac{11}{5}}\). Thus, \(g'(x)=\frac{11}{5}x^{\frac{6}{5}}\), and \(g'(1)=\frac{11}{5}=2.2\). 2. b) Rewrite \(g(x)=\frac{5}{2}x^{-\frac{3}{4}}\). Thus, \(g'(x)=-\frac{15}{8}x^{-\frac{7}{4}}\), and \(g'(1)=-\frac{15}{8}=-1.875\). 3. c) Rewrite \(g(x)=x^{\frac{2}{3}-2}=x^{-\frac{4}{3}}\). Thus, \(g'(x)=-\frac{4}{3}x^{-\frac{7}{3}}\), and \(g'(1)=-\frac{4}{3}\approx-1.333\).

a) \(g'(x)=\frac{11}{5}x^{\frac{6}{5}}\); \(g'(1)=\frac{11}{5}=2.2\) b) \(g'(x)=-\frac{15}{8}x^{-\frac{7}{4}}\); \(g'(1)=-\frac{15}{8}=-1.875\) c) \(g'(x)=-\frac{4}{3}x^{-\frac{7}{3}}\); \(g'(1)=-\frac{4}{3}\approx-1.333\)

- Use the fact that \(P\) lies on the graph. - The point \(P(2, 0)\) is also the tangent line's x-intercept. - Use the triangle's area to find the tangent line's y-intercept. - Find the tangent slope from the intercepts and match it to the derivative at \(x=2\).

1. Since \(P(2, 0)\) lies on the graph, \(f_{a,b}(2)=0\). Therefore, \(4a+2b+5=0\), or \(4a+2b=-5\). 2. The tangent line crosses the x-axis at \((2, 0)\). Let its y-intercept be \((0, y_0)\). The triangle's area is \(\frac{1}{2}\cdot2\cdot y_0=6\), so \(y_0=6\). 3. The tangent line through \((0, 6)\) and \((2, 0)\) has slope \(\frac{0-6}{2-0}=-3\). 4. Differentiate: \(f_{a,b}'(x)=2ax+b\). At \(x=2\), the tangent slope gives \(4a+b=-3\). 5. Solve the system \(4a+2b=-5\) and \(4a+b=-3\). Subtraction gives \(b=-2\), and then \(a=-\frac{1}{4}\).

\(a=-\frac{1}{4}\) and \(b=-2\)

- Rewrite the quadratic in vertex form. - Substitute the point's x-coordinate into the function. - Use the derivative to verify the tangent slope at \(P\). - Use the point and slope to write the tangent line. - Test the origin by substituting \((0,0)\) into the line equation.

1. Complete the square: \(f(x)=(x-2)^2+1\). Therefore, the vertex is \((2,1)\). 2. Evaluate the function: \(f(3)=9-12+5=2\). Thus, \(P=(3,2)\) lies on the graph. 3. Differentiate: \(f'(x)=2x-4\). Then \(f'(3)=2\), so the tangent slope at \(P\) is \(2\). 4. A line with slope \(2\) through \((3,2)\) satisfies \(y-2=2(x-3)\), so \(y=2x-4\). 5. At \(x=0\), the line has value \(-4\), not \(0\). Therefore, it does not pass through the origin.

a) \((2,1)\) b) Yes, because \(f(3)=2\). c) \(f'(3)=2\), so the tangent line is \(y=2x-4\); it does not pass through the origin.

- Evaluate the function to find the full point. - The derivative value gives the tangent slope. - The normal slope is the negative reciprocal of the tangent slope. - Use point-slope form.

1. Find the point: \(f(2)=\frac{1}{2}\cdot2^2+2=4\), so \(P=(2,4)\). 2. Differentiate: \(f'(x)=x\). The tangent slope is \(f'(2)=2\). 3. The tangent line is \(y-4=2(x-2)\), or \(y=2x\). 4. The normal slope is \(-\frac{1}{2}\). Thus, \(y-4=-\frac{1}{2}(x-2)\), or \(y=-\frac{1}{2}x+5\).

\(t: y=2x\) \(n: y=-\frac{1}{2}x+5\)

- The derivative gives the tangent slope. - A horizontal tangent has slope \(0\). - Write the general tangent line at \(x=u\). - Expand and combine like terms carefully.

1. Differentiate: \(f'(x)=3x^2-3\). 2. At \(x=2\), \(f(2)=2\) and \(f'(2)=9\). Thus, \(y-2=9(x-2)\), so \(y=9x-16\). 3. A horizontal tangent has slope \(0\). Solve \(3x^2-3=0\), giving \(x=-1\) and \(x=1\). 4. At an arbitrary input \(u\), the point is \((u,u^3-3u)\) and the slope is \(3u^2-3\). 5. Substitute into point-slope form: \(y=(3u^2-3)(x-u)+u^3-3u\). Expanding gives \(y=(3u^2-3)x-2u^3\).

a) \(y=9x-16\) b) \(x=-1\) and \(x=1\) c) Substituting \(f(u)=u^3-3u\) and \(f'(u)=3u^2-3\) into point-slope form simplifies to \(y=(3u^2-3)x-2u^3\).

- The normal slope is the negative reciprocal of the tangent slope. - Find both axis intercepts of the tangent line. - The axes and the line form a right triangle. - Use \(\frac{1}{2}(\text{base})(\text{height})\).

1. Differentiate: \(f'(x)=x\). 2. At \(x=2\), \(f(2)=0\) and the tangent slope is \(2\). Thus, the tangent line is \(y=2(x-2)=2x-4\). 3. The normal slope is the negative reciprocal, \(-\frac{1}{2}\). Thus, the normal line is \(y=-\frac{1}{2}(x-2)=-\frac{1}{2}x+1\). 4. The tangent line intersects the x-axis at \((2,0)\) and the y-axis at \((0,-4)\). 5. The right triangle has leg lengths \(2\) and \(4\), so its area is \(\frac{1}{2}\cdot 2\cdot 4=4\).

a) Tangent: \(y=2x-4\); normal: \(y=-\frac{1}{2}x+1\) b) \(4\) square units

- Rewrite the rational expression using a negative exponent. - Find \(f'(a)\) and \(f(a)\). - Use point-slope form. - Apply exponent rules carefully when simplifying.

1. Rewrite the function as \(f(x)=4x^{-2}\). 2. Differentiate using the power rule: \(f'(x)=-8x^{-3}=-\frac{8}{x^3}\). 3. At \(x=a\), the tangent slope is \(-\frac{8}{a^3}\), and the point is \((a,\frac{4}{a^2})\). 4. Use point-slope form: \(y=-\frac{8}{a^3}(x-a)+\frac{4}{a^2}\). 5. Simplify: \(y=-\frac{8}{a^3}x+\frac{8}{a^2}+\frac{4}{a^2}=-\frac{8}{a^3}x+\frac{12}{a^2}\).

\(y=-\frac{8}{a^3}x+\frac{12}{a^2}\)

- Tangent slopes are the values taken by the first derivative. - After differentiating, determine the smallest value the derivative can have without substituting trial inputs. - Connect the sign and minimum value of the derivative to the directions of possible tangent lines.

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 \(m=0\).

- Rewrite the reciprocal term using a negative exponent. - The derivative gives the tangent slope. - Use the point and slope to write the tangent line.

1. Find the point of tangency: \(f(1)=3\), so \(P=(1,3)\). 2. Rewrite \(f(x)=2x^{-1}+x\) and differentiate: \(f'(x)=-2x^{-2}+1=-\frac{2}{x^2}+1\). 3. Find the slope: \(f'(1)=-1\). 4. Use point-slope form: \(y-3=-(x-1)\), so \(y=-x+4\).

\(y=-x+4\)

- Find both tangent slopes using the derivative. - What slope relationship characterizes perpendicular lines? - You may also compare the lines' angles of inclination.

1. Differentiate: \(g'(x)=2x-3\). 2. The tangent slopes are \(g'(1)=-1\) and \(g'(2)=1\). 3. Since the product of the slopes is \((-1)(1)=-1\), the tangent lines are perpendicular. 4. Therefore, their angle of intersection is \(90^\circ\).

\(90^\circ\)

- Parallel lines have equal slopes. - The derivative gives the tangent slope at \(a\). - Set the derivative equal to \(2\) and solve. - Check for more than one solution.

1. Differentiate: \(f'(x)=x^2+2x-6\). 2. The given line has slope \(2\), so require \(f'(a)=2\). 3. Solve \(a^2+2a-6=2\), or \(a^2+2a-8=0\). 4. Factor: \((a+4)(a-2)=0\). Thus, \(a=-4\) or \(a=2\).

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

- 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 \(2\).

- Rewrite reciprocals and radicals using negative or rational exponents. - Use \(\frac{1}{x^n}=x^{-n}\). - Use \(\sqrt{x}=x^{1/2}\). - Check the sign when differentiating a negative exponent.

1. Rewrite the function as \(g(x)=2x^{-2}+6x^{1/2}-x^3\). 2. Differentiate: \(g'(x)=-4x^{-3}+3x^{-1/2}-3x^2\). 3. Rewrite without negative or fractional exponents: \(g'(x)=-\frac{4}{x^3}+\frac{3}{\sqrt{x}}-3x^2\).

\(g'(x)=-\frac{4}{x^3}+\frac{3}{\sqrt{x}}-3x^2\), where \(x>0\)

- The derivative represents the tangent slope. - Convert the angle to a slope. - Solve the derivative equation for all possible inputs. - Substitute each input into the original function.

1. An angle of \(45^\circ\) corresponds to slope \(1\). 2. Differentiate: \(f'(x)=x^2-x-1\). 3. Set the derivative equal to \(1\): \(x^2-x-1=1\), so \(x^2-x-2=0\). 4. Factor: \((x-2)(x+1)=0\). Thus, \(x=2\) or \(x=-1\). 5. Evaluate the function: \(f(2)=-\frac{4}{3}\) and \(f(-1)=\frac{1}{6}\). The points are \((2,-\frac{4}{3})\) and \((-1,\frac{1}{6})\).

\((2,-\frac{4}{3})\) and \((-1,\frac{1}{6})\)

- Set the two function values equal to find the intersections. - Differentiate both polynomials term by term to obtain their tangent slopes. - Use the angle-between-lines formula for the two slopes at each intersection. - Round only the final angle to the requested precision.

1. Set the functions equal: \(x^2-2x+2=-x^2+4x-2\). This simplifies to \(x^2-3x+2=0\), so \(x=1\) or \(x=2\). The points are \((1,1)\) and \((2,2)\). 2. Differentiate term by term: \(f'(x)=2x-2\) and \(g'(x)=-2x+4\). 3. At \(x=1\), the slopes are \(0\) and \(2\). Thus, \(\tan(\gamma)=2\), so \(\gamma=\arctan(2)\approx63.43^\circ\). 4. At \(x=2\), the slopes are \(2\) and \(0\), giving the same acute angle, \(63.43^\circ\).

1. \((1,1)\) and \((2,2)\) 2. The acute intersection angle is approximately \(63.43^\circ\) at both points.

- Differentiate both functions term by term and evaluate the slopes at each intersection. - Equal tangent slopes give an intersection angle of \(0^\circ\). - Otherwise use the acute angle-between-lines formula and the requested precision.

1. Differentiate term by term: \(f'(x)=3x^2-1\) and \(g'(x)=2x\). 2. At \(P=(-1,0)\), the slopes are \(2\) and \(-2\). Therefore, \(\tan(\gamma)=\frac{4}{3}\), so \(\gamma=\arctan\!\left(\frac{4}{3}\right)\approx53.13^\circ\). 3. At \(Q=(1,0)\), both slopes equal \(2\). The tangent lines coincide, so the intersection angle is \(0^\circ\); the graphs are tangent at \(Q\).

At \(P=(-1,0)\), the acute intersection angle is approximately \(53.13^\circ\). At \(Q=(1,0)\), the angle is \(0^\circ\); the graphs are tangent there.

- Tangent and normal slopes are negative reciprocals. - A line equation requires a point and a slope. - What does a tangent slope of \(0\) imply about the normal line? - Apply the vertical line test to a vertical line.

1. At \(x=1\), \(f(1)=-2\). Also, \(f'(x)=4x\), so the tangent slope is \(f'(1)=4\). 2. The normal slope is \(-\frac{1}{4}\). Through \((1,-2)\), the normal line is \(y+2=-\frac{1}{4}(x-1)\), or \(y=-\frac{1}{4}x-\frac{7}{4}\). 3. Setting \(x=0\) gives the y-intercept \(\left(0,-\frac{7}{4}\right)\). Setting \(y=0\) gives \(x=-7\), so the x-intercept is \((-7,0)\). 4. At \(x=0\), \(f'(0)=0\), so the tangent is horizontal. The normal is the vertical line through \((0,-4)\), namely \(x=0\). 5. A vertical line has undefined slope and assigns many y-values to the same x-value, so it cannot be written as a function \(y=mx+b\).

a) \(n: y=-0.25x-1.75\) b) The intercepts are \((0,-1.75)\) and \((-7,0)\). c) The normal is the y-axis, \(x=0\). It is vertical, so its slope is undefined and it is not a function of \(x\).

- Parallel lines have equal slopes. - Express each tangent slope using a derivative. - Set the derivative expressions equal. - Solve the resulting quadratic equation.

1. Differentiate: \(f'(x)=x^2-4\) and \(g'(x)=2x-1\). 2. Parallel tangent lines have equal slopes, so solve \(x^2-4=2x-1\). 3. Rearrange: \(x^2-2x-3=0\). 4. Factor: \((x-3)(x+1)=0\). Therefore, \(x=-1\) or \(x=3\).

\(x=-1\) and \(x=3\)

- Equal slopes mean equal derivative values. - Set the derivative functions equal and solve. - For each tangent line, use its own point on the corresponding graph. - Use point-slope form.

1. Differentiate: \(f'(x)=3x^2-6x\) and \(g'(x)=6x-9\). 2. Set the slopes equal: \(3x^2-6x=6x-9\). Dividing by \(3\) gives \(x^2-4x+3=0\), so \(x=1\) or \(x=3\). 3. The smaller value is \(x=1\). The common slope is \(-3\). 4. Since \(f(1)=3\), the tangent to \(f\) is \(y-3=-3(x-1)\), or \(y=-3x+6\). 5. Since \(g(1)=-4\), the tangent to \(g\) is \(y+4=-3(x-1)\), or \(y=-3x-1\).

a) \(x=1\) and \(x=3\) b) At \(x=1\), the tangent lines are \(y=-3x+6\) and \(y=-3x-1\).

- Set the function values equal to locate the common point. - Expand the translated square before differentiating so only the rules from this topic are needed. - Find each tangent line from its point and slope, then set \(y=0\) for the x-intercepts. - Use the x-axis segment as the triangle's base.

1. Set the functions equal: \(x^2=(x-2)^2\). This gives \(x=1\), and the common value is \(1\), so \(P=(1,1)\). 2. Expand \(g(x)=x^2-4x+4\). Then \(f'(x)=2x\) and \(g'(x)=2x-4\) by the rules available in this topic. At \(x=1\), the slopes are \(2\) and \(-2\). 3. The tangent lines are \(y=2x-1\) and \(y=-2x+3\). 4. Their x-intercepts are \(\left(\frac{1}{2},0\right)\) and \(\left(\frac{3}{2},0\right)\). 5. The base on the x-axis has length \(1\), and the height from \(P\) is \(1\), so the area is \(\frac{1}{2}\) square unit.

a) \(P=(1,1)\) b) \(\left(\frac{1}{2},0\right)\) and \(\left(\frac{3}{2},0\right)\) c) \(\frac{1}{2}\) square unit

- Differentiate the square-root and reciprocal functions using power notation. - Use point-slope form at \((1,1)\). - The triangle's base lies on the y-axis. - The x-coordinate of \(P\) gives the height relative to that base.

1. Differentiate: \(f'(x)=\frac{1}{2\sqrt{x}}\) and \(g'(x)=-\frac{1}{x^2}\). 2. At \(x=1\), the slopes are \(\frac{1}{2}\) and \(-1\). 3. The tangent to \(f\) is \(y-1=\frac{1}{2}(x-1)\), or \(y=\frac{1}{2}x+\frac{1}{2}\). 4. The tangent to \(g\) is \(y-1=-(x-1)\), or \(y=-x+2\). 5. Their y-intercepts are \(\frac{1}{2}\) and \(2\), so the vertical base has length \(\frac{3}{2}\). The horizontal distance from \(P\) to the y-axis is \(1\). 6. The area is \(\frac{1}{2}\cdot\frac{3}{2}\cdot 1=\frac{3}{4}\).

a) \(y=\frac{1}{2}x+\frac{1}{2}\) and \(y=-x+2\) b) \(\frac{3}{4}\) square unit

- Use each derivative to express a tangent slope. - What relationship do the slopes of perpendicular lines satisfy? - Write one equation involving \(x_1\) and \(x_2\). - Solve for one input in terms of the other.

1. Differentiate: \(f'(x)=x\) and \(g'(x)=-\frac{1}{x^2}\). 2. Perpendicular nonvertical lines have slopes whose product is \(-1\). Thus, \(f'(x_1)g'(x_2)=-1\). 3. Substitute: \(x_1(-\frac{1}{x_2^2})=-1\). 4. Since \(x_2\ne 0\), this simplifies to \(x_1=x_2^2\).

\(x_1=x_2^2\), with \(x_2\ne 0\)

- Set the function values equal to find intersections. - Equal tangent slopes mean equal derivative values. - The normal slope is the negative reciprocal of the tangent slope. - Use point-slope form for the normal line.

1. Set the functions equal: \(x^2+3x=-x^2+5\), so \(2x^2+3x-5=0\). The solutions are \(x=1\) and \(x=-\frac{5}{2}\). 2. The intersection points are \((1,4)\) and \((-\frac{5}{2},-\frac{5}{4})\). 3. Differentiate: \(f'(x)=2x+3\) and \(g'(x)=-2x\). Equal slopes require \(2x+3=-2x\), so \(x=-\frac{3}{4}\). 4. At \(x=2\), \(g(2)=1\) and the tangent slope is \(-4\). The normal slope is \(\frac{1}{4}\). 5. The normal line is \(y-1=\frac{1}{4}(x-2)\), or \(y=\frac{1}{4}x+\frac{1}{2}\).

a) \((1,4)\) and \((-\frac{5}{2},-\frac{5}{4})\) b) \(x=-\frac{3}{4}\) c) \(y=\frac{1}{4}x+\frac{1}{2}\)

- A tangent parallel to the x-axis has slope \(0\). - Write the tangent line at \(x=a\) using point-slope form. - Treat \(a\) as a fixed parameter while simplifying. - Set \(x=0\) to find the y-intercept.

1. Differentiate: \(h'(x)=\frac{1}{2}x-1\). A horizontal tangent has slope \(0\), so \(\frac{1}{2}x-1=0\), giving \(x=2\). 2. Since \(h(2)=-1\), the point is \(P=(2,-1)\). 3. At \(x=a\), the slope is \(\frac{1}{2}a-1\), and the point is \((a,\frac{1}{4}a^2-a)\). 4. Point-slope form gives \(y=(\frac{1}{2}a-1)(x-a)+\frac{1}{4}a^2-a\). 5. Simplifying gives \(t_a: y=(\frac{1}{2}a-1)x-\frac{1}{4}a^2\). 6. Setting \(x=0\) gives \(y=-\frac{1}{4}a^2\), so the y-intercept is \((0,-\frac{1}{4}a^2)\).

a) \(P=(2,-1)\) b) \(t_a: y=(\frac{1}{2}a-1)x-\frac{1}{4}a^2\) c) Setting \(x=0\) gives \(S=(0,-\frac{1}{4}a^2)\).

- 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) \(y=2x-3\)

- Set the derivative equal to the required slope. - Solve for every possible input. - Substitute each input into the original function. - Use the common slope and each point to write the tangent lines.

1. Differentiate: \(f'(x)=3x^2-6x-6\). 2. Set the derivative equal to \(3\): \(3x^2-6x-6=3\). Dividing by \(3\) gives \(x^2-2x-3=0\). 3. Factor: \((x-3)(x+1)=0\), so \(x=3\) or \(x=-1\). 4. Evaluate the function: \(f(3)=-16\) and \(f(-1)=4\). The points are \((3,-16)\) and \((-1,4)\). 5. At \((3,-16)\), the tangent is \(y+16=3(x-3)\), or \(y=3x-25\). 6. At \((-1,4)\), the tangent is \(y-4=3(x+1)\), or \(y=3x+7\).

The points are \((3,-16)\) and \((-1,4)\). The tangent lines are \(y=3x-25\) and \(y=3x+7\), respectively.

- What must a function value equal where its graph crosses the x-axis? - Recall the power rule, including its use with negative exponents. - What relationship between two nonzero slopes indicates perpendicular tangent lines? - Compare the products of the derivatives at the common point.

1. Solve \(0.5x^2-2=0\): \(x^2=4\), so \(x=\pm2\). The positive zero is \(x_0=2\). 2. Evaluate the other functions at \(x=2\): \(g(2)=-0.125\cdot2^2+0.5=0\) and \(h(2)=\frac{2}{2}-1=0\). Therefore, all three graphs pass through \((2,0)\). 3. Differentiate: \(f'(x)=x\), \(g'(x)=-0.25x\), and \(h'(x)=-\frac{2}{x^2}\). At \(x=2\), the slopes are \(f'(2)=2\) and \(g'(2)=h'(2)=-0.5\). Since \(2\cdot(-0.5)=-1\), each of the other two graphs intersects the graph of \(f\) at a right angle.

1. \(x_0=2\) 2. Yes. All three graphs cross the x-axis at \((2,0)\). 3. Yes. The products of the tangent slopes are \(f'(2)g'(2)=-1\) and \(f'(2)h'(2)=-1\).

- 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 \(Q(2, 20)\): \(f(x)=4x^3-6x\)

- A common tangent requires a common point and a common derivative value. - First find the input that makes the parameter-dependent term vanish. - Expand the shifted cube before differentiating so only previously established rules are needed.

1. For the function value to be independent of \(k\), the parameter term must be zero. Since \((x-2)^3=0\), use \(x=2\). 2. Evaluate the function: \(g_k(2)=5\). Every graph passes through \((2,5)\). 3. Expand the parameter term before differentiating: \(g_k(x)=k(x^3-6x^2+12x-8)+x^2+1\). 4. Apply the power, sum, and constant-multiple rules: \(g_k'(x)=k(3x^2-12x+12)+2x\). 5. At \(x=2\), the coefficient of \(k\) is \(12-24+12=0\), so \(g_k'(2)=4\), independent of \(k\). 6. The line with slope \(4\) through \((2,5)\) is \(y-5=4(x-2)\), or \(y=4x-3\).

Every graph has the same tangent line at \((2,5)\), and the line is \(y=4x-3\).

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

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

- Tangent graphs have the same function value and slope at the point of tangency. - Use the logarithm derivative rule to determine the required slope at \(x=1\). - Start with a general quadratic and translate the two tangency conditions plus the extra point into three equations.

1. Write \(p(x)=ax^2+bx+c\). 2. Tangency at \((1,0)\) requires \(p(1)=0\) and \(p'(1)=f'(1)\). 3. Since \(f'(x)=\frac1x\), \(f'(1)=1\). Therefore, \(a+b+c=0\) and \(2a+b=1\). 4. The point \(A(2,4)\) gives \(4a+2b+c=4\). 5. Solving the three equations gives \(a=3\), \(b=-5\), and \(c=2\). Thus \(p(x)=3x^2-5x+2\).

\(p(x)=3x^2-5x+2\)

- Start with the general form of a cubic polynomial. - Tangency at \(x=0\) gives one condition for the value and one for the slope. - A horizontal tangent means the derivative is zero. - Use all conditions to form and solve a linear system.

1. Write \(p(x)=ax^3+bx^2+cx+d\), so \(p'(x)=3ax^2+2bx+c\). 2. Tangency at \(x=0\) requires equal function values and slopes. Since \(f(0)=4\), \(d=4\). 3. Differentiate \(f\): \(f'(x)=-2e^{-0.5x}\), so \(f'(0)=-2\). Therefore, \(c=-2\). 4. The horizontal tangent at \(x=-1\) gives \(p'(-1)=3a-2b-2=0\), so \(3a-2b=2\). 5. Since \(p(1)=2\), \(a+b-2+4=2\), so \(a+b=0\). 6. Solving the system gives \(a=\frac{2}{5}\) and \(b=-\frac{2}{5}\). 7. Therefore, \(p(x)=\frac{2}{5}x^3-\frac{2}{5}x^2-2x+4\).

\(p(x)=\frac{2}{5}x^3-\frac{2}{5}x^2-2x+4\)

- A tangent line through the origin has equation \(y=mx\). - At the point of tangency, use \(m=f'(x_0)\) and \(f(x_0)=mx_0\). - Perpendicular nonvertical lines have slopes whose product is \(-1\).

1. Differentiate: \(f'(x)=\frac{3}{2}x^2-4x+1\). 2. A tangent line through the origin at \(x=x_0\) must satisfy \(f(x_0)=f'(x_0)x_0\). 3. Substitute: \(\frac{1}{2}x_0^3-2x_0^2+x_0=x_0\left(\frac{3}{2}x_0^2-4x_0+1\right)\). 4. Simplifying gives \(x_0^2(x_0-2)=0\). The solutions are \(x_0=0\) and \(x_0=2\). Since the point of tangency is not the origin, \(x_0=2\). 5. Since \(f(2)=-2\), the point of tangency is \(B(2, -2)\). The tangent slope is \(f'(2)=-1\), so the tangent line is \(t: y=-x\). 6. At the origin, the tangent slope is \(f'(0)=1\). Because \((-1)(1)=-1\), the two tangent lines are perpendicular.

a) \(B(2, -2)\); \(t: y=-x\) b) The slopes are \(-1\) and \(1\), whose product is \(-1\), so the tangent lines are perpendicular.

- Set the two function expressions equal to locate the intersections. - Differentiate both formulas term by term to obtain tangent slopes. - Use the acute-angle formula only after the two slopes at each intersection are known.

1. Set the functions equal: \(x^2-6x+9=-x+5\). Then \(x^2-5x+4=0\), so \(x=1\) or \(x=4\). Thus, \(P_1=(1,4)\) and \(P_2=(4,1)\). 2. Differentiate term by term: \(f'(x)=2x-6\) and \(g'(x)=-1\). At \(P_1\), the slopes are \(-4\) and \(-1\). At \(P_2\), the slopes are \(2\) and \(-1\). 3. Using \(\tan(\alpha)=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\), at \(P_1\), \(\tan(\alpha_1)=0.6\), so \(\alpha_1\approx30.96^\circ\). 4. At \(P_2\), \(\tan(\alpha_2)=3\), so \(\alpha_2\approx71.57^\circ\).

a) \(P_1=(1,4)\) and \(P_2=(4,1)\) b) At \(P_1\), the slopes are \(-4\) for \(f\) and \(-1\) for \(g\). At \(P_2\), the slopes are \(2\) for \(f\) and \(-1\) for \(g\). c) \(\alpha_1\approx30.96^\circ\) and \(\alpha_2\approx71.57^\circ\)

- Write the tangent-line equation at a general input \(u\). - A tangent through \(Q\) must produce the point's y-value when \(x=0\). - Solve the resulting equation for the possible tangency inputs. - Use each input in the original function and derivative to finish the tangent line.

1. The tangent line at \(x=u\) is \(t(x)=f'(u)(x-u)+f(u)\), with \(f'(x)=-0.5x+1\). 2. Requiring \(t(0)=7\) gives \((-0.5u+1)(-u)-0.25u^2+u+3=7\), which simplifies to \(0.25u^2=4\). Thus, \(u=4\) or \(u=-4\). 3. Since \(f(4)=3\) and \(f(-4)=-5\), the tangency points are \(B_1=(4,3)\) and \(B_2=(-4,-5)\). 4. The corresponding slopes are \(-1\) and \(3\), so the tangent lines are \(y=-x+7\) and \(y=3x+7\).

a) \(B_1=(4,3)\) and \(B_2=(-4,-5)\) b) \(t_1:y=-x+7\) and \(t_2:y=3x+7\)

- The derivative gives the slope of the graph at an input. - Translate each requested tangent condition into an equation involving \(f'(x)\). - After solving for an input, use the original function to obtain the corresponding point.

1. Differentiate: \(f'(x)=\frac{1}{2}x^2-2x\). 2. Set the derivative equal to \(-2\): \(\frac{1}{2}a^2-2a=-2\). This gives \(a^2-4a+4=0\), so \(a=2\). 3. Evaluate the function: \(f(2)=-\frac{2}{3}\). Thus, \(A\left(2,-\frac{2}{3}\right)\). 4. A line parallel to \(y=6x-1\) has slope \(6\). Set \(\frac{1}{2}x^2-2x=6\). Then \(x^2-4x-12=0\), so \(x=6\) or \(x=-2\). 5. Evaluate the function: \(f(6)=2\) and \(f(-2)=-\frac{10}{3}\). The points are \((6,2)\) and \(\left(-2,-\frac{10}{3}\right)\).

a) \(f'(x)=\frac{1}{2}x^2-2x\) b) \(A\left(2,-\frac{2}{3}\right)\) c) \((6,2)\) and \(\left(-2,-\frac{10}{3}\right)\)

- Read two exact points from the graph and substitute them into \(f(x)=ax^2+bx\). - Evaluate \(f(3)\) and use the derivative for the tangent slope. - Convert the tangent slope to the normal slope. - Use the normal line's axis intercepts as the triangle's legs.

1. From the graph, use \((2,2)\) and \((4,0)\). Substitution gives \(4a+2b=2\) and \(16a+4b=0\). 2. Simplifying gives \(2a+b=1\) and \(4a+b=0\). Subtracting yields \(a=-0.5\), and then \(b=2\). Thus, \(f(x)=-0.5x^2+2x\). 3. Evaluate the function: \(f(3)=1.5\), so \(P=(3,1.5)\). Differentiate: \(f'(x)=-x+2\), so the tangent slope at \(x=3\) is \(-1\). 4. The normal slope is \(1\). Through \(P\), the normal is \(y-1.5=x-3\), or \(y=x-1.5\). 5. Its intercepts are \((0,-1.5)\) and \((1.5,0)\). The triangle area is \(\frac{1}{2}(1.5)(1.5)=1.125=\frac{9}{8}\) square units.

a) \(a=-0.5\), \(b=2\) b) \(n: y=x-1.5\) c) \(\frac{9}{8}\) square units

- Compute both derivatives before matching any panels. - Use the formulas you obtained to identify the line and parabola representing the derivatives. - Confirm each pairing by comparing derivative zeros with horizontal tangents of the original graph.

1. Differentiate term by term: \(f'(x)=x\) and \(g'(x)=-0.6x^2+1.5\). 2. Graph (3) is the upward-opening parabola with vertex \((0,-2)\), so it represents \(f\). Graph (1) is the line \(y=x\), so it represents \(f'\). 3. Graph (2) is the odd cubic with negative leading coefficient, so it represents \(g\). Graph (4) is the downward-opening parabola with y-intercept \(1.5\), so it represents \(g'\). 4. The zero of graph (1) at \(x=0\) matches the horizontal tangent of graph (3) there. The zeros of graph (4) align with the turning points of graph (2).

a) \(f'(x)=x\); \(g'(x)=-0.6x^2+1.5\) b) (1) \(f'\); (2) \(g\); (3) \(f\); (4) \(g'\) c) The derivative zeros align with horizontal tangents of their corresponding function graphs.

- Set the function equal to zero to find the x-intercepts. - Use the derivative for each tangent slope. - Solve the two tangent equations together to locate \(S\). - Use \(\overline{PQ}\) as the triangle's base.

1. Set \(f(x)=0\): \(-0.25x^2+4=0\), so \(x=\pm4\). Thus, \(P=(-4,0)\) and \(Q=(4,0)\). 2. Differentiate: \(f'(x)=-0.5x\). At \(P\), the slope is \(2\), so \(t_1:y=2x+8\). At \(Q\), the slope is \(-2\), so \(t_2:y=-2x+8\). 3. Solving the tangent equations together gives \(S=(0,8)\). 4. Triangle \(PQS\) has base \(8\) and height \(8\), so its area is \(32\) square units.

a) \(P=(-4,0)\) and \(Q=(4,0)\) b) \(t_1:y=2x+8\) and \(t_2:y=-2x+8\) c) \(32\) square units

- Differentiate \(f\) before trying to match the panels. - Use the computed derivative formula to identify the linear panel. - Then use vertical and horizontal translations to distinguish \(g\) and \(h\). - Check that a zero of the derivative aligns with a horizontal tangent of \(f\).

1. Differentiate term by term: \(f'(x)=2x-2\). 2. The graph of \(f\) is an upward-opening parabola with zeros at \(0\) and \(2\) and vertex \((1,-1)\), so it is graph II. 3. The derivative \(f'(x)=2x-2\) is a line with zero at \(x=1\), so it is graph IV. Its zero aligns with the horizontal tangent at the vertex of graph II. 4. The graph of \(g=f+3\) is shifted up \(3\) units and has vertex \((1,2)\), so it is graph I. 5. The graph of \(h(x)=f(x+2)\) is shifted left \(2\) units and has vertex \((-1,-1)\), so it is graph III.

a) \(f'(x)=2x-2\) b) Graph I: \(g\); Graph II: \(f\); Graph III: \(h\); Graph IV: \(f'\) c) Graph IV crosses the x-axis at \(x=1\), the x-coordinate of the horizontal tangent at the vertex of graph II.

- Set the two function expressions equal to locate the intersections. - Differentiate both polynomials to obtain tangent slopes. - Use the acute angle-between-lines formula and round only the final angle.

1. Set the functions equal: \(x^2-1=7-x^2\), giving \(x=\pm2\). The intersection points are \((2,3)\) and \((-2,3)\). 2. Differentiate: \(f'(x)=2x\) and \(g'(x)=-2x\). At \(x=2\), the tangent slopes are \(4\) and \(-4\). 3. Thus, \(\tan(\phi)=\left|\frac{4-(-4)}{1+4(-4)}\right|=\frac{8}{15}\), so \(\phi\approx28.1^\circ\). 4. By symmetry, the acute angle at the other intersection is also \(28.1^\circ\).

a) \((2,3)\) and \((-2,3)\) b) The angle at each intersection is approximately \(28.1^\circ\).

- Set the two function expressions equal to find all intersections. - Differentiate both functions to obtain the tangent slopes at each common input. - Use the acute angle-between-lines relation once the slopes are known.

1. Set \(\frac{1}{4}x^3=x\). Then \(x(x^2-4)=0\), so the intersection points are \((-2,-2)\), \((0,0)\), and \((2,2)\). 2. The tangent slopes are given by \(f'(x)=\frac{3}{4}x^2\) and \(g'(x)=1\). 3. At \((0,0)\), the slopes are \(0\) and \(1\), so the acute angle is \(45^\circ\). 4. At \((-2,-2)\) and \((2,2)\), the slopes are \(3\) and \(1\). Thus, \(\tan\theta=\frac{1}{2}\), so \(\theta=\arctan\!\left(\frac{1}{2}\right)\approx26.6^\circ\).

a) \((-2,-2)\), \((0,0)\), and \((2,2)\) b) At \((0,0)\), the acute angle is \(45^\circ\). At the other two points, it is \(\arctan\!\left(\frac{1}{2}\right)\approx26.6^\circ\).

- Use the derivative at \(x=u\) to find the tangent slope. - Write the tangent equation in point-slope form. - Set \(x=0\) to find the y-intercept. - The projection \(P_1\) has x-coordinate \(0\) and the same y-coordinate as \(P\).

1. Differentiate: \(f'(x)=2x\). 2. At \(x=u\), the tangent line is \(y=2u(x-u)+u^2=2ux-u^2\). 3. At the y-axis, \(x=0\), so \(y=-u^2\). Therefore, the y-intercept is \(Q(0,-u^2)\). 4. The projection point is \(P_1(0,u^2)\). 5. The distance from \(Q\) to the origin is \(|-u^2|=u^2\), and the distance from \(P_1\) to the origin is \(|u^2|=u^2\). 6. The distances are equal, so the origin is the midpoint of \(\overline{P_1Q}\).

a) \(t: y=2ux-u^2\), so its y-intercept is \(Q(0,-u^2)\). b) Both distances equal \(u^2\). The origin is the midpoint of \(\overline{P_1Q}\).

- Use the power rule to find the derivative. - The derivative value is the tangent slope. - Parallel lines have equal slopes. - Substitute each resulting input into the original function to find the full point.

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 the polynomial term by term. - A horizontal tangent has slope \(0\). - A tangent parallel to \(y=x\) must have slope \(1\). - Substitute each solved input into \(f\) to recover the point.

1. Differentiate term by term: \(f'(x)=-x+3\). 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 \(R=(2,4)\)

- Differentiate \(f\) and \(g\) first. - Compare the degree and symmetry of each derivative with the graphs. - Find the zeros of each derivative and match them to the \(x\)-intercepts shown.

1. Differentiate each function using the power rule: \(f'(x) = x^2 - 1\) and \(g'(x) = x^3 - 4x\). 2. The graph of \(f'\) is an upward-opening parabola, symmetric about the \(y\)-axis, with vertex \((0, -1)\) and zeros at \(x=-1\) and \(x=1\). This is graph A. 3. Since \(g'(x)=x(x-2)(x+2)\), its graph is an odd cubic with zeros at \(x=-2\), \(x=0\), and \(x=2\). This is graph B.

Graph A represents \(f'\), and graph B represents \(g'\).

- Use the slope formula between \(P\) and the parameterized point \(Q_a\). - The derivative at \(a\) gives the tangent slope. - Write the tangent line in terms of \(a\), then impose the condition that it passes through \(P\).

1. Since \(Q_a=\left(a,\frac{1}{2}a^2+2\right)\), the secant slope is \(m_a=\frac{a^2+4}{2a}\). 2. Differentiate: \(f'(x)=x\). 3. The tangent at \(Q_a\) is \(y=a(x-a)+\frac{1}{2}a^2+2=ax-\frac{1}{2}a^2+2\). 4. For this line to pass through \(P=(0,0)\), \(0=-\frac{1}{2}a^2+2\). Thus, \(a=2\) or \(a=-2\).

a) \(m_a=\frac{a^2+4}{2a}\) b) \(a=2\) and \(a=-2\)

- Rewrite the reciprocal using a negative exponent before applying the Power Rule. - Use point-slope form at the general input \(a\). - Substitute the coordinates of \(P\) into the tangent equation.

1. Rewrite \(g(x)=2x^{-1}\) and differentiate: \(g'(x)=-\frac{2}{x^2}\). 2. The tangent at \(x=a\) is \(y=-\frac{2}{a^2}(x-a)+\frac{2}{a}\), which simplifies to \(y=-\frac{2}{a^2}x+\frac{4}{a}\). 3. Since the line passes through \(P=(0,4)\), \(4=\frac{4}{a}\), so \(a=1\).

a) \(t_a:y=-\frac{2}{a^2}x+\frac{4}{a}\) b) \(a=1\)

- Use the factored form of a quadratic with the two given zeros. - Differentiate that form after expanding it. - Write the tangent line through the origin using its derivative slope. - Use the given intersection point of the tangents to determine the remaining coefficient.

1. Since the zeros are \(0\) and \(4\), write \(f(x)=ax(x-4)=ax^2-4ax\). 2. Differentiate: \(f'(x)=2ax-4a\). At the origin, the tangent slope is \(-4a\), so the tangent line there is \(y=-4ax\). 3. Since \(S=(2,-6)\) lies on this tangent, \(-6=-8a\), so \(a=\frac{3}{4}\). 4. Therefore, \(f(x)=\frac{3}{4}x^2-3x\).

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

- Write a general tangent line at an unknown input \(u\). - The tangent slope at \(u\) is \(f'(u)\). - Require the tangent line to pass through the given external point. - Use the condition \(x>0\) when choosing a solution.

1. Differentiate: \(f'(x)=\frac{1}{4}x\). 2. Let the point of tangency have x-coordinate \(u\). The tangent line is \(y=f'(u)(x-u)+f(u)\). 3. Substitute the expressions for \(f'(u)\) and \(f(u)\): \(y=\frac{1}{4}u(x-u)+\frac{1}{8}u^2+2=\frac{1}{4}ux-\frac{1}{8}u^2+2\). 4. The tangent passes through \((0,0)\), so \(0=-\frac{1}{8}u^2+2\). Thus, \(u^2=16\). Because \(u>0\), \(u=4\). 5. Find the point and slope: \(f(4)=4\) and \(f'(4)=1\). Therefore, \(B=(4,4)\) and the ray has slope \(1\).

\(B=(4,4)\) and \(m=1\)

- A tangent at input \(u\) has slope \(h'(u)\). - Write the tangent line at the unknown point of tangency. - Require that line to pass through \(P=(2,0)\). - The resulting equation may have more than one solution for \(u\).

1. Differentiate: \(h'(x)=2x-4\). 2. Let the point of tangency be \(B=(u,h(u))\). A tangent line at \(u\) is \(y=(2u-4)(x-u)+u^2-4u+5\). 3. Require the line to pass through \(P=(2,0)\): \(0=(2u-4)(2-u)+u^2-4u+5\). 4. Simplifying gives \(u^2-4u+3=0\), so \(u=1\) or \(u=3\). 5. The points of tangency are \((1,h(1))=(1,2)\) and \((3,h(3))=(3,2)\). 6. At \(u=1\), the slope is \(-2\), giving \(y=-2(x-2)=-2x+4\). At \(u=3\), the slope is \(2\), giving \(y=2(x-2)=2x-4\).

The points of tangency are \((1,2)\) and \((3,2)\). The tangent lines are \(y=-2x+4\) and \(y=2x-4\), respectively.

- First find the two x-intercepts of the parabola. - Use the derivative to write both tangent lines. - Their intersection is the third vertex of the triangle. - Use the x-axis segment as the base when computing the area.

1. Factor \(f_k(x)=x(x-k)\). The x-intercepts are \(O=(0,0)\) and \(A=(k,0)\). 2. Differentiate: \(f_k'(x)=2x-k\). The tangent slopes are \(f_k'(0)=-k\) and \(f_k'(k)=k\). 3. The tangent lines are \(y=-kx\) and \(y=k(x-k)=kx-k^2\). 4. Set the equations equal: \(-kx=kx-k^2\). Since \(k>0\), \(x=\frac{k}{2}\). Then \(y=-\frac{k^2}{2}\), so \(S=(\frac{k}{2},-\frac{k^2}{2})\). 5. The base \(OA\) has length \(k\), and the height is \(\frac{k^2}{2}\). Thus, the area is \(\frac{1}{2}\cdot k\cdot\frac{k^2}{2}=\frac{k^3}{4}\). 6. Solve \(\frac{k^3}{4}=16\). This gives \(k^3=64\), so \(k=4\).

\(k=4\)

- Use the slope formula for two given points. - The derivative at \(a\) gives the tangent slope at \(Q_a\). - Check \(a=0\) separately before using a slope formula that divides by \(a\). - If the tangent passes through \(P\), compare its slope with the slope of \(PQ_a\). - Solve the resulting equation for \(a\).

1. The slope through \(P=(0,-5)\) and \(Q_a=(a,a^2-1)\) is \(m_a=\frac{(a^2-1)-(-5)}{a-0}=\frac{a^2+4}{a}\). 2. Differentiate: \(f'(x)=2x\), so the tangent slope at \(Q_a\) is \(f'(a)=2a\). 3. If \(a=0\), the tangent line is \(y=-1\), which does not pass through \(P\). For \(a\ne 0\), a tangent line through both \(Q_a\) and \(P\) must have slope \(m_a\). Thus, \(2a=\frac{a^2+4}{a}\). 4. Since \(a\ne 0\), multiply by \(a\): \(2a^2=a^2+4\). Therefore, \(a^2=4\), so \(a=-2\) or \(a=2\).

a) \(m_a=\frac{a^2+4}{a}\) b) \(a=-2\) and \(a=2\)

- Use the slope formula for two points. - The derivative at \(a\) gives the tangent slope. - Check \(a=0\) separately before using a slope formula that divides by \(a\). - If the tangent passes through \(R\), compare its slope with the slope of \(RS_a\). - Solve the resulting equation for \(a\).

1. Use the slope formula: \(m_a=\frac{(\frac{1}{2}a^3+4)-(-4)}{a-0}=\frac{\frac{1}{2}a^3+8}{a}=\frac{a^3+16}{2a}\). 2. Differentiate: \(g'(x)=\frac{3}{2}x^2\), so the tangent slope at \(x=a\) is \(\frac{3}{2}a^2\). 3. If \(a=0\), the tangent line is \(y=4\), which does not pass through \(R\). For \(a\ne 0\), a tangent line through both \(S_a\) and \(R\) must have slope \(m_a\): \(\frac{3}{2}a^2=\frac{a^3+16}{2a}\). 4. Since \(a\ne 0\), multiply by \(2a\): \(3a^3=a^3+16\). Thus, \(2a^3=16\), so \(a=2\).

a) \(m_a=\frac{a^3+16}{2a}\) b) \(a=2\)

- Treat \(k\) as a constant when differentiating and simplifying. - Use the point-slope form of a tangent line at \(x=x_0\). - For the triangle, find the length of the side on the x-axis and the vertical distance from the third vertex to the x-axis. - Lengths and areas must be positive.

1. Solve \(x^3 - kx = 0\): \(x(x^2-k)=0\), so the zeros are \(x=0\) and \(x=\pm\sqrt{k}\). Therefore, \(x_P=\sqrt{k}\). 2. Differentiate: \(f_k'(x)=3x^2-k\). 3. At the origin, \(f_k'(0)=-k\). Since the tangent passes through \((0, 0)\), \(t_0: y=-kx\). 4. At \(P(\sqrt{k}, 0)\), the slope is \(f_k'(\sqrt{k})=3k-k=2k\). Thus \(t_P: y=2k(x-\sqrt{k})\). 5. Find the intersection of the tangent lines by solving \(-kx=2k(x-\sqrt{k})\). Since \(k>0\), \(x=\frac{2}{3}\sqrt{k}\), and then \(y=-\frac{2}{3}k\sqrt{k}\). 6. The triangle has base \(\sqrt{k}\) and height \(\frac{2}{3}k\sqrt{k}\). Therefore, \(A(k)=\frac{1}{2}\cdot\sqrt{k}\cdot\frac{2}{3}k\sqrt{k}=\frac{1}{3}k^2\).

a) \(x_P=\sqrt{k}\) b) \(t_0: y=-kx\); \(t_P: y=2k(x-\sqrt{k})\) c) The tangent lines intersect at \(\left(\frac{2}{3}\sqrt{k}, -\frac{2}{3}k\sqrt{k}\right)\), giving \(A(k)=\frac{1}{3}k^2\).

- 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 \(n=5\)

- Use the normal-line formula at a general input \(a\). - Differentiate the square-root function. - Set \(y=0\) to find the x-intercept. - A constant difference does not depend on \(a\).

1. Differentiate: \(f'(x)=\frac{1}{2\sqrt{x}}\). 2. At \(x=a\), the normal slope is \(m_n=-\frac{1}{f'(a)}=-2\sqrt{a}\). 3. The normal line is \(y-\sqrt{a}=-2\sqrt{a}(x-a)\). 4. At the x-intercept, \(y=0\): \(0=-2\sqrt{a}(x-a)+\sqrt{a}\). Since \(a>0\), divide by \(\sqrt{a}\) to get \(2(x-a)=1\). 5. Thus, \(x=a+0.5\), and the horizontal distance is \((a+0.5)-a=0.5\).

a) \(n_a: y=-2\sqrt{a}(x-a)+\sqrt{a}\) b) \(S_x=(a+0.5,0)\), so the horizontal distance is always \(0.5\).

- Write the tangent line at an unknown input \(u\). - Require that line to pass through the external point. - Use the midpoint formula after finding both points of tangency. - Points on the same vertical line have distance equal to the difference of their y-coordinates.

1. A tangent at input \(u\) has equation \(y=f'(u)(x-u)+f(u)\). Since \(f'(x)=2x\), this becomes \(y=2u(x-u)+u^2-2=2ux-u^2-2\). 2. Require the tangent to pass through \(P=(0,-3)\): \(-3=-u^2-2\). Thus, \(u^2=1\), so \(u=-1\) or \(u=1\). 3. The points of tangency are \((-1,-1)\) and \((1,-1)\). 4. For \(u=-1\), the tangent is \(y=-2x-3\). For \(u=1\), the tangent is \(y=2x-3\). 5. The midpoint is \(M=(0,-1)\). Since \(M\) and \(P\) have the same x-coordinate, their distance is \(|-1-(-3)|=2\).

a) \(B_1=(-1,-1)\) and \(B_2=(1,-1)\) b) \(y=-2x-3\) and \(y=2x-3\) c) \(M=(0,-1)\), and the distance from \(M\) to \(P\) is \(2\).

- Relate the tangent and normal slopes at a general point. - Write the normal line at \(x=u\). - Substitute the coordinates of \(A\). - Check separately the case where the tangent is horizontal.

1. Since \(P\) lies on the graph, \(v=u^2\). Also, \(f'(x)=2x\). 2. For \(u\ne0\), the normal slope is \(-\frac{1}{2u}\), so the normal line is \(y-u^2=-\frac{1}{2u}(x-u)\). 3. Substitute \(A(0,1.5)\): \(1.5-u^2=\frac{1}{2}\). Thus, \(u^2=1\), so \(u=1\) or \(u=-1\). These give \((1,1)\) and \((-1,1)\). 4. Check \(u=0\) separately. The tangent at \((0,0)\) is horizontal, so the normal is the y-axis, which contains \(A\). Therefore, \((0,0)\) is also a solution.

\((1,1)\), \((-1,1)\), and \((0,0)\)

- Apply the power rule to \(x^{-1}\). - Convert the tangent slope to the normal slope. - Write the normal line at a general input and substitute the origin. - Find all real solutions of \(a^4=1\).

1. Differentiate: \(f'(x)=-\frac{1}{x^2}\). 2. At \(x=a\), the normal slope is \(m_n=-\frac{1}{f'(a)}=a^2\). 3. The normal line through \(P\left(a,\frac{1}{a}\right)\) is \(y-\frac{1}{a}=a^2(x-a)\). 4. Substitute the origin: \(-\frac{1}{a}=-a^3\), so \(a^4=1\). 5. The real solutions are \(a=1\) and \(a=-1\). The corresponding points are \((1,1)\) and \((-1,-1)\).

\(a=1\) gives \(P=(1,1)\), and \(a=-1\) gives \(P=(-1,-1)\).

- Differentiate the polynomial and write the tangent equation at a general input \(a\). - Substitute the coordinates of \(P\) into that general line. - Factor the resulting equation in \(a\).

1. Differentiate term by term: \(f'(x)=1.5x^2-1.5\). 2. The tangent at \(x=a\) is \(t_a(x)=f'(a)(x-a)+f(a)\), which simplifies to \(t_a(x)=(1.5a^2-1.5)x-a^3\). 3. Requiring the tangent to pass through \(P=(2,-3)\) gives \(-3=2(1.5a^2-1.5)-a^3\), or \(a^2(3-a)=0\). 4. Thus, \(a=0\) or \(a=3\). The tangent lines are \(t_0(x)=-1.5x\) and \(t_3(x)=12x-27\).

a) \(t_a(x)=(1.5a^2-1.5)x-a^3\) b) \(a=0\) gives \(t_0(x)=-1.5x\), and \(a=3\) gives \(t_3(x)=12x-27\).

- Apply the power rule after writing the reciprocal with a negative exponent. - Find the tangent line's x- and y-intercepts. - The intercepts give the legs of the right triangle. - Check whether \(a\) remains after simplifying the area formula.

1. Differentiate: \(f'(x)=-\frac{1}{x^2}\). 2. At \(x=a\), the tangent line is \(y=-\frac{1}{a^2}(x-a)+\frac{1}{a}=-\frac{1}{a^2}x+\frac{2}{a}\). 3. Setting \(x=0\) gives the y-intercept \(S_y\left(0,\frac{2}{a}\right)\). 4. Setting \(y=0\) gives \(x=2a\), so the x-intercept is \(S_x(2a,0)\). 5. The triangle has base \(2a\) and height \(\frac{2}{a}\). Its area is \(\frac{1}{2}(2a)\left(\frac{2}{a}\right)=2\), so it is independent of \(a\).

a) \(y=-\frac{1}{a^2}x+\frac{2}{a}\) b) The intercepts are \(S_x(2a,0)\) and \(S_y\left(0,\frac{2}{a}\right)\), and the area is always \(2\).