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

- The exponent can be negative in the power rule. - Check whether the original function has a value at the excluded input.

1. By the power rule, \(f'(x)=-2x^{-3}=-\frac{2}{x^3}\). 2. The original function \(1/x^2\) is undefined at \(0\), so it cannot have a derivative there.

\(f'(x)=-\frac{2}{x^3}\), defined only for \(x\ne0\).

- A fractional exponent follows the same power-rule pattern. - How can you rewrite a negative power as a fraction?

1. Use the fractional exponent in the power rule: \(g'(x)=\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{-\frac{2}{3}}\). 2. Rewrite it as \(g'(x)=\frac{1}{3x^{\frac{2}{3}}}\).

\(g'(x)=\frac{1}{3x^{\frac{2}{3}}}\)

- 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: \(y=-3x+3\).

\(y=-3x+3\)

- 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: \(y=4x-8\).

\(y=4x-8\)

- A smooth connection requires the two graphs to share a point and a slope. - Find the point on the hill at \(x=2\). - Use the derivative to find the hill's slope at that point. - Write the line using the point and slope.

1. Find the connection point: \(f(2)=-0.2(2)^2+2(2)+1=4.2\), so the point is \((2,4.2)\). 2. Differentiate: \(f'(x)=-0.4x+2\). 3. Find the slope at the connection point: \(f'(2)=-0.4(2)+2=1.2\). 4. A smooth connection requires the line to have the same point and slope. Using point-slope form, \(y-4.2=1.2(x-2)\). 5. Simplify: \(y=1.2x+1.8\).

\(y=1.2x+1.8\)

- 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: \(y=-5x+7\).

\(y=-5x+7\)

- 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) \(h'(z)=3cz^2-\frac{1}{c}\)

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

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

- 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: \(f'(x)=6\left(\frac{1}{3}\right)x^{-\frac{2}{3}}=2x^{-\frac{2}{3}}\).

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

- 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) \(y=-x\)

- 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.

- 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 \(x=2\)

- 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) \(11\)

- Differentiate using the power rule. - Compare each derivative directly with the corresponding perimeter or surface-area formula. - Find the constant factor that multiplies the boundary measure to produce the derivative. - Check whether the factor is the same in both cases.

1. For the square, \(A'(x)=2x\). Since \(P(x)=4x\), \(A'(x)=\frac{1}{2}P(x)\). 2. For the cube, \(V'(x)=3x^2\). Since \(S(x)=6x^2\), \(V'(x)=\frac{1}{2}S(x)\). 3. In both cases, the ratio of the derivative to the boundary measure is \(\frac{1}{2}\).

a) \(A'(x)=2x=\frac{1}{2}P(x)\) b) \(V'(x)=3x^2=\frac{1}{2}S(x)\) c) The ratio is \(\frac{1}{2}\), or \(1:2\).

- 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 \(y=-3x+6.25\).

\(y=-3x+6.25\)

- 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.

- 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 while keeping the exponent as an unknown integer. - What happens to a power of \(1\)?

1. The power rule gives \(f'(x)=nx^{n-1}\). 2. At \(x=1\), this is \(f'(1)=n\). 3. The given slope therefore means \(n=5\).

\(n=5\)

- Start the limit method with \(f(0+h)-f(0)\). - The power-rule method produces a derivative function before you substitute \(0\). - Compare the final slopes, not just the algebraic forms.

1. At \(x=0\), the difference quotient is \(\frac{h^3-0}{h}=h^2\) for \(h\ne0\). Its limit is \(0\). 2. By the power rule, \(f'(x)=3x^2\), so \(f'(0)=0\). 3. Both methods agree.

Limit definition: \(\lim_{h\to0}h^2=0\). Power rule: \(f'(x)=3x^2\), so \(f'(0)=0\).

- Compare the student's result with the full power rule, not just its coefficient. - Which part of the student's result would you change, and why?

1. The coefficient \(3\) is correct, but the student left the original exponent unchanged. 2. The exponent must decrease by one, so \(f'(x)=3x^2\).

The power was not reduced. The correct derivative is \(f'(x)=3x^2\).

- 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

- What equation identifies the point where the object reaches the ground? - The derivative at the impact point gives the tangent slope. - Use the magnitude of the slope when finding the acute angle with the horizontal. - Round the angle as requested.

1. Find where the shot put reaches the ground by solving \(h(x)=0\). The solutions are \(x=-2\) and \(x=10\), so the physically relevant impact point is \(x=10\). 2. Differentiate: \(h'(x)=-0.2x+0.8\). 3. Find the slope at impact: \(h'(10)=-0.2(10)+0.8=-1.2\). 4. The acute angle \(\alpha\) with the horizontal satisfies \(\tan(\alpha)=|-1.2|=1.2\). Thus, \(\alpha=\arctan(1.2)\approx 50.2^\circ\).

\(50.2^\circ\)

- What does the derivative tell you about the graph at a point? - For a smooth transition to the ground, which two quantities must match? - Use the relationship between slope and angle.

1. Differentiate: \(f'(x)=\frac{1}{4}x-1\). 2. At \(x=0\), the slope is \(f'(0)=-1\). The acute angle \(\alpha\) satisfies \(\tan(\alpha)=|-1|=1\), so \(\alpha=45^\circ\). 3. At \(x=4\), \(f(4)=\frac{1}{8}\cdot 4^2-4+2=0\), so the ramp reaches the ground. 4. Also, \(f'(4)=\frac{1}{4}\cdot 4-1=0\), which matches the slope of the horizontal line \(y=0\). Since both the height and slope agree, the ramp joins the ground smoothly at \((4,0)\).

a) \(45^\circ\) b) \(f(4)=0\) and \(f'(4)=0\), so the ramp and the ground line \(y=0\) have the same point and slope at \((4,0)\).

- A tangent continuation has the same value and slope as the curve at the transition point. - Differentiate the given function. - Use point-slope form for the tangent line. - Verify that the line passes through the point on the curve.

1. Find the transition point: \(g(2)=2^3-4\cdot 2^2+5\cdot 2=2\), so the point is \((2,2)\). 2. Differentiate: \(g'(x)=3x^2-8x+5\). 3. Find the slope at \(x=2\): \(g'(2)=3\cdot 2^2-8\cdot 2+5=1\). 4. Use point-slope form: \(y-2=1(x-2)\). 5. Simplify: \(y=x\).

\(y=x\)

- 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 \(k>0\), \(k=\frac{1}{4}\).

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

- Find the point where the straight continuation begins. - A smooth connection requires the line to have the same slope as the curve. - Write the tangent line at \(x=10\). - Set the line's output equal to the target height.

1. Find the transition point: \(f(10)=0.01(10)^3-0.2(10)^2+1.5(10)+2=7\), so the point is \((10,7)\). 2. Differentiate: \(f'(x)=0.03x^2-0.4x+1.5\). 3. Find the slope at the transition: \(f'(10)=0.03(100)-0.4(10)+1.5=0.5\). 4. A smooth straight continuation is the tangent line: \(y-7=0.5(x-10)\), so \(y=0.5x+2\). 5. Set the height equal to \(12\): \(12=0.5x+2\). Solving gives \(x=20\).

\(x=20\,\text{m}\)

- A smooth transition means the linear model has the same value and slope at the joining time. - Find the temperature and rate of change at \(t=10\). - Write the linear model for times after \(10\) minutes. - Convert the final time to the requested unit.

1. Find the temperature at the transition: \(T(10)=0.1(10)^2-4(10)+80=50\). 2. Differentiate: \(T'(t)=0.2t-4\). The rate at \(t=10\) is \(T'(10)=-2\) degrees Celsius per minute. 3. A smooth linear continuation has the same value and rate at \(t=10\): \(L(t)-50=-2(t-10)\), so \(L(t)=-2t+70\). 4. Solve \(25=-2t+70\). This gives \(t=22.5\) minutes. 5. Convert to seconds: \(22.5\cdot 60=1350\) seconds.

\(1350\,\text{s}\)

- The derivative gives the local slope of the trail. - How are slope, percent grade, and angle of inclination related? - A tangent line requires both the point and the slope. - Use point-slope form for the tangent line.

1. Differentiate: \(h'(x)=-0.006x^2+0.12x\). 2. At \(x=5\), \(h'(5)=-0.006(25)+0.12(5)=0.45\). The percent grade is \(0.45\cdot 100\%=45\%\). 3. At \(x=15\), \(h'(15)=0.45\). Therefore, \(\alpha=\arctan(0.45)\approx 24.23^\circ\). 4. At \(x=25\), the slope is \(h'(25)=-0.75\), and \(h(25)=16.25\). The tangent line is \(y-16.25=-0.75(x-25)\), so \(y=-0.75x+35\).

a) \(45\%\) b) \(24.23^\circ\), approximately c) \(y=-0.75x+35\)

- A line perpendicular to the curve at a point is the normal line. - Use the derivative to find the curve's tangent slope. - Tangent and normal slopes are negative reciprocals. - Intersect the normal line with \(y=7\).

1. Differentiate: \(f'(x)=0.4x\). At \(x=5\), the tangent slope is \(f'(5)=2\). 2. The support follows the normal line, whose slope is \(-\frac{1}{2}\). 3. Through \(P(5,3)\), the normal line is \(y-3=-0.5(x-5)\), or \(y=-0.5x+5.5\). 4. Set \(y=7\): \(7=-0.5x+5.5\), so \(x=-3\). 5. Therefore, \(Q=(-3,7)\).

\(Q=(-3,7)\)

- 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) \(a=2\)

- 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. \(y=-x+3.5\)

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

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

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

- Rewrite 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) \(g'(x)=-\frac{4}{3}x^{-\frac{7}{3}}\); \(g'(1)=-\frac{4}{3}\approx-1.333\)

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

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

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

- Which condition determines the unknown coefficient in the function? - What does the derivative tell you about the tangent at \(P\)? - How can a line equation be used to find its x-intercept?

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 \(\left(\frac{4}{3},0\right)\).

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

- 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 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 \(m=0\).

- 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 \(x>0\)

- Use symmetry to locate the right edge from the total width. - Convert a percent grade to a decimal slope. - The derivative at the edge gives a condition on \(a\). - Evaluate the model at the edge to find the depth.

1. Because the bowl is \(12\,\text{m}\) wide and symmetric, the right edge is at \(x=6\). 2. A grade of \(120\%\) corresponds to slope \(1.2\), so \(f'(6)=1.2\). 3. Since \(f'(x)=2ax\), the condition gives \(12a=1.2\), so \(a=0.1\). Thus \(f(x)=0.1x^2\). 4. The depth is the height of the edge above the origin: \(f(6)=0.1(6)^2=3.6\,\text{m}\).

a) \(f(x)=0.1x^2\) b) \(3.6\,\text{m}\)

- 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 \(x\).

- 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 \(x_2\ne 0\)

- Write the tangent-line equation at a general input \(x_0\). - Use the derivative for the tangent slope. - Substitute the sensor coordinates into the tangent-line equation. - Solve the resulting quadratic equation. - Check each solution against the stated domain.

1. Differentiate: \(f'(x)=-0.5x\). 2. The tangent line at \(x=x_0\) is \(t(x)=f'(x_0)(x-x_0)+f(x_0)\). Substitution and simplification give \(t(x)=-0.5x_0x+0.25x_0^2+4\). 3. Since the line must pass through \(S(4,1)\), set \(t(4)=1\): \(1=-2x_0+0.25x_0^2+4\). 4. Rearranging gives \(x_0^2-8x_0+12=0\), so \((x_0-2)(x_0-6)=0\). Thus, \(x_0=2\) or \(x_0=6\). 5. Only \(x_0=2\) lies in the rail's domain \([0,4]\).

The laser must be at \(x_0=2\).

- Find the vertex of the parabola. - Translate “stay at or above the roadway” into an inequality for the minimum value. - Differentiate to find the slope at the tower. - Use the tangent relationship between slope and angle.

1. The parabola's lowest point occurs halfway between the towers, at \(x=50\). Its height is \(f_a(50)=30-2500a\). 2. To stay at or above the roadway, the minimum height must satisfy \(30-2500a\ge0\). Therefore, \(a\le0.012\). Together with \(a>0\), this gives \(0<a\le0.012\). 3. If the minimum height is \(5\,\text{m}\), then \(30-2500a=5\), so \(a=0.01\). 4. For \(a=0.01\), \(f_a'(x)=0.02x-1\). At \(x=0\), the slope is \(-1\). The acute angle relative to the horizontal is \(\arctan(1)=45^\circ\).

a) \(0<a\le0.012\) b) \(a=0.01\) c) \(45^\circ\)

- Perpendicular lines have slopes that are negative reciprocals. - Use the derivative to find the walking path's tangent slope at \(A\). - Write the connector line through \(A\). - Set the connector and creek equations equal to find their intersection.

1. Differentiate: \(f'(x)=-x+2\). At \(x=3\), the tangent slope is \(f'(3)=-1\). 2. The perpendicular connector has slope \(1\). 3. Through \(A(3,3)\), its equation is \(y-3=x-3\), or \(y=x\). 4. Intersect this line with the creek: \(x=-2x+6\). Thus, \(3x=6\), so \(x=2\) and \(y=2\). 5. Therefore, the bridge location is \(B=(2,2)\).

The connector path is \(y=x\), and the bridge is at \(B=(2,2)\).

- First find the exact takeoff point on the ramp. - Use the derivative to find the ramp's slope at takeoff. - Model the initial path with the tangent line. - Substitute the wall's x-coordinate into the tangent-line equation.

1. Find the takeoff point: \(f(2)=-0.1\cdot2^3+0.6\cdot2^2=1.6\), so the point is \((2,1.6)\). 2. Differentiate: \(f'(x)=-0.3x^2+1.2x\). 3. Find the slope at takeoff: \(f'(2)=-0.3\cdot2^2+1.2\cdot2=1.2\). 4. The tangent line is \(y-1.6=1.2(x-2)\), or \(y=1.2x-0.8\). 5. At the wall, \(y=1.2\cdot6-0.8=6.4\). Therefore, \(h=6.4\,\text{m}\).

\(h=6.4\,\text{m}\)

- 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) \(Q=(-1,2)\)

- 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) \(a=1\)

- 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 \(m=1\)

- 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) \(a=2\)

- 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 \(n=5\)

- The upper edges lie on the x-axis, so use their coordinates in the model. - A \(50\%\) grade corresponds to derivative value \(0.5\). - The parabola's lowest point occurs at its vertex. - Report depth as a positive distance.

1. The top edges are at \(x=\pm10\), and \(f(10)=0\). 2. Since \(f'(x)=2ax\) and \(f'(10)=0.8\), \(20a=0.8\), so \(a=0.04\). 3. Use \(f(10)=0\): \(0.04(10)^2+c=0\), giving \(c=-4\). Thus \(f(x)=0.04x^2-4\). 4. The lowest point is \(f(0)=-4\), so the maximum depth is \(4\,\text{m}\). 5. A \(50\%\) slope is \(0.5\). Solve \(f'(x)=0.08x=0.5\), giving \(x=6.25\).

a) \(f(x)=0.04x^2-4\) b) \(4\,\text{m}\) c) \(6.25\,\text{m}\) from the center

- 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 \(P=(-1,-1)\).

- 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 \(2\).