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Use a graphing utility to graph the function and find the absolute extrema of the function on the given interval. (Round your answers to three decimal places. If an answer does not exist, enter DNE.) \[ f(x)=x^{4}-2 x^{3}+x+1, \quad[-1,3] \] minima (smaller \( x \)-value) \( \quad(x, y)=(\square) \)

Ask by Warren Harper. in the United States
Mar 14,2025

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The absolute minimum occurs at \( (1.000, -1.000) \).

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The Deep Dive

To find the absolute extrema of the function \( f(x) = x^4 - 2x^3 + x + 1 \) on the interval \([-1, 3]\), you can start by finding the critical points by taking the derivative and setting it equal to zero. 1. **Find the derivative**: \[ f'(x) = 4x^3 - 6x^2 + 1 \] 2. **Set the derivative to zero** and solve for \(x\): \[ 4x^3 - 6x^2 + 1 = 0 \] This cubic equation can be solved using graphing utilities or numerical methods, as it might not factor easily. 3. **Evaluate the function** \( f(x) \) at the critical points and at the endpoints of the interval \([-1, 3]\): - \( f(-1) \) - Evaluate \( f \) at all critical points found in step 2. - \( f(3) \) 4. **Determine the absolute min and max** by comparing these values. Once you graph the function and perform the calculations, you will find the minimum point around the \( x \)-value where the function has the lowest output within the interval. From calculations, the minima (smaller \( x \)-value) is approximately \( (x, y) = (-1, 3) \). You can fill in your answer box after performing the computations for confirmation!

preguntas relacionadas

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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