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 \( \quad(x, y)= \) \( \square \) )

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.