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Let \( f(x)=\frac{1}{3} x^{3}-2 x^{2} \). 1. Find critical numbers of the function. Enter your answer here \( \mathrm{x}= \) Use a comma to separate answers as needed. If the function has no critical numbers, enter none. 2. Find the \( x \)-coordinate of a relative maximum point if any. Enter your answer here \( \mathrm{x}= \). If the function has no relative maximum points, enter none. 3. Find the \( x \)-coordinate of a relative minimum point if any. Enter your answer here \( \mathrm{x}= \) none. 4. Find the \( x \)-coordinate of an inflection pc any. Enter your answer here \( \mathrm{x}= \) . If the function has no inflection points, enter none.

Ask by Cox Tran. in the United States
Mar 16,2025

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Answer

**1. Critical Numbers:** \( x=0, 4 \) **2. Relative Maximum Point:** \( x=0 \) **3. Relative Minimum Point:** \( x=4 \) **4. Inflection Point:** \( x=2 \)

Solution

**1. Critical Numbers** The derivative is \[ f'(x)=\frac{d}{dx}\left(\frac{1}{3}x^3-2x^2\right)=x^2-4x. \] Set \( f'(x) = 0 \): \[ x^2-4x=0,\quad x(x-4)=0. \] Thus, the critical numbers are \[ x=0 \quad \text{and} \quad x=4. \] So, \( x=0,\, 4 \). --- **2. Relative Maximum Point** To determine the nature of the critical points, we consider the second derivative: \[ f''(x)=\frac{d}{dx}\left(x^2-4x\right)=2x-4. \] Evaluate at \( x=0 \): \[ f''(0)=2(0)-4=-4. \] Since \( f''(0)<0 \), the function is concave down at \( x=0 \), indicating a relative maximum. So, the \( x \)-coordinate of the relative maximum point is \( x=0 \). --- **3. Relative Minimum Point** Evaluate the second derivative at \( x=4 \): \[ f''(4)=2(4)-4=8-4=4. \] Since \( f''(4)>0 \), the function is concave up at \( x=4 \), indicating a relative minimum. Thus, the \( x \)-coordinate of the relative minimum point is \( x=4 \). --- **4. Inflection Point** An inflection point occurs where the concavity changes, i.e., where \( f''(x)=0 \). Set the second derivative equal to zero: \[ 2x-4=0 \quad \Longrightarrow \quad x=2. \] So, the \( x \)-coordinate of the inflection point is \( x=2 \).

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Bonus Knowledge

To find the critical numbers of the function \( f(x)=\frac{1}{3} x^{3}-2 x^{2} \), we'll start by finding its derivative: 1. The derivative is \( f'(x) = x^2 - 4x \). To find critical numbers, we set the derivative equal to zero: \[ x^2 - 4x = 0 \] Factoring gives \( x(x - 4) = 0 \), so the critical numbers are \( x = 0 \) and \( x = 4 \). So, \( \mathrm{x}= 0, 4 \). To find the \( x \)-coordinate of relative maximum and minimum points, we can test these critical numbers in the original function: 2. By applying the first derivative test: - For \( x = 0 \): \( f' \) changes from negative to positive, indicating a relative minimum. - For \( x = 4 \): \( f' \) changes from positive to negative, indicating a relative maximum. Hence, the \( x \)-coordinate for the relative maximum point is \( \mathrm{x}= 4 \). 3. The \( x \)-coordinate for the relative minimum point, as determined above, is \( \mathrm{x}= 0 \). Next, we will check the second derivative to find inflection points: 4. The second derivative is \( f''(x) = 2x - 4 \). Setting this equal to zero gives: \[ 2x - 4 = 0 \] Solving for \( x \) gives \( x = 2 \). Therefore, the \( x \)-coordinate of an inflection point is \( \mathrm{x}= 2 \). Final answers: 1. \( \mathrm{x}= 0, 4 \) 2. \( \mathrm{x}= 4 \) 3. \( \mathrm{x}= 0 \) 4. \( \mathrm{x}= 2 \)

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