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Given the function below \[ f(x)=\sqrt[3]{-45 x^{3}+72} \] Find the equation of the tangent line to the graph of the function at \( x=1 \). Answer in \( m x+b \) form. \( L(x)= \) Use the tangent line to approximate \( f(1.1) \). \( L(1.1)= \) Compute the actual value of \( f(1.1) \). What is the error between the function value and the linear approximation? Answer as a positive value only. lerror \( 1 \approx \) ? (Approximate to at least 5 decimal places.)

Ask by Todd Wade. in the United States
Mar 13,2025

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The tangent line at \( x=1 \) is \( L(x) = -5x + 8 \). Using this tangent line, \( L(1.1) = 2.5 \). The actual value of \( f(1.1) \) is approximately 2.29617. The error between the function value and the linear approximation is approximately 0.20383.

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Bonus Knowledge

To find the equation of the tangent line \( L(x) \) to the graph of the function at \( x = 1 \), we first need to find \( f(1) \) and \( f'(x) \). 1. **Calculate \( f(1) \)**: \[ f(1) = \sqrt[3]{-45(1^3) + 72} = \sqrt[3]{-45 + 72} = \sqrt[3]{27} = 3 \] 2. **Find the derivative \( f'(x) \)** using the chain rule: \[ f(x) = \sqrt[3]{-45x^3 + 72} \implies f'(x) = \frac{1}{3}(-45x^3 + 72)^{-\frac{2}{3}}(-135x^2) \] Simplifying this gives: \[ f'(x) = \frac{-135x^2}{3(-45x^3 + 72)^{\frac{2}{3}}} = \frac{-45x^2}{(-45x^3 + 72)^{\frac{2}{3}}} \] 3. **Evaluate \( f'(1) \)**: \[ f'(1) = \frac{-45(1^2)}{(-45(1^3) + 72)^{\frac{2}{3}}} = \frac{-45}{27^{\frac{2}{3}}} = \frac{-45}{9} = -5 \] 4. **Equation of the tangent line** \( L(x) \) at \( x = 1 \): Using the point-slope form \( y - f(1) = f'(1)(x - 1) \): \[ y - 3 = -5(x - 1) \implies y = -5x + 5 + 3 \implies y = -5x + 8 \] Thus, \( L(x) = -5x + 8 \). 5. **Approximation of \( f(1.1) \)**: \[ L(1.1) = -5(1.1) + 8 = -5.5 + 8 = 2.5 \] 6. **Compute the actual value of \( f(1.1) \)**: \[ f(1.1) = \sqrt[3]{-45(1.1^3) + 72} \] Calculating \( 1.1^3 = 1.331 \): \[ f(1.1) = \sqrt[3]{-45(1.331) + 72} = \sqrt[3]{-59.895 + 72} = \sqrt[3]{12.105} \approx 2.28736 \] 7. **Calculate the error**: \[ \text{error} = |f(1.1) - L(1.1)| = |2.28736 - 2.5| \approx 0.21264 \] In conclusion: - The equation of the tangent line is \( L(x) = -5x + 8 \). - The approximate value of \( f(1.1) \) using the tangent line is \( L(1.1) = 2.5 \). - The actual value of \( f(1.1) \) is approximately \( 2.28736 \), and the error is \( \text{error} \approx 0.21264 \).

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\[ y=x^{3}, \quad 0 \leq x \leq 2 \] Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interv \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note t as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius and the radical measures the arc length that is the width of a band, \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying, \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9} \sqrt[9]{ } x^{4} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 14 \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interval \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note that as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius \( y \) and the radical measures the arc length that is the width of a band. \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying. \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 146 x^{d x} \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
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