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Find the intervals on which \( f(x) \) is increasing and the intervals on which \( f(x) \) is decreasing. Then sketch the graph. Add horizontal tangent lines. \( f(x)=2-27 x+9 x^{2}-x^{3} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is increasing on (Type your answer using interval notation. Use a comma to separate answers as needed.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is decreasing on (Type your answer using interval notation. Use a comma to separate answers as needed.) B. The function is never decreasing.

Ask by Frank Simpson. in the United States
Mar 12,2025

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**Increasing on:** (none) **Decreasing on:** \((-\infty,3) \cup (3,\infty)\)

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To determine the intervals where the function \( f(x) = 2 - 27x + 9x^2 - x^3 \) is increasing or decreasing, we first need to find the derivative \( f'(x) \): \[ f'(x) = -27 + 18x - 3x^2 \] Next, we set the derivative equal to zero to find the critical points: \[ -3x^2 + 18x - 27 = 0 \] Dividing by -3 gives us: \[ x^2 - 6x + 9 = 0 \] Factoring gives: \[ (x - 3)^2 = 0 \] Thus, we have a double root at \( x = 3 \). Now we check the sign of \( f'(x) \) around this critical point: 1. For \( x < 3 \), let's pick \( x = 0 \): \[ f'(0) = -27 + 0 + 0 = -27 \quad (\text{decreasing}) \] 2. For \( x > 3 \), let's pick \( x = 4 \): \[ f'(4) = -27 + 72 - 48 = -3 \quad (\text{decreasing}) \] Since the derivative is negative both before and after \( x = 3 \), we find that \( f(x) \) is decreasing on the entire domain except at \( x = 3 \) where the function has a horizontal tangent line. Combining this information: A. The function is increasing on \( \text{(no intervals)} \). A. The function is decreasing on \( (-\infty, 3) \cup (3, \infty) \). Now let's sketch the graph. The graph would show a local maximum at \( x = 3 \) with a horizontal tangent line, and it would curve downwards on both sides, confirming it is decreasing everywhere else. --- In the realm of mathematics, the history of calculus, particularly the concept of increasing and decreasing functions, has roots in the works of great mathematicians like Isaac Newton and Gottfried Wilhelm Leibniz, who independently developed calculus in the late 17th century. Their foundations in differentiation and integration laid the groundwork for all future exploration of function behavior! In real-world applications, understanding where a function is increasing or decreasing is crucial in fields like economics for determining profit (increasing) and loss (decreasing), as well as in science where graphing natural phenomena helps us analyze motion — think of how engineers might design roller coasters by studying angles and shapes for a thrilling ride!

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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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