A function is given. Use the second derivative to determine each of the following. \[ f(x)=\frac{1}{4} x^{4}+\frac{1}{2} x^{3}-3 x^{2}+5 \] Determine intervals on which the function is concave up. (Enter your answer using interval notation.) \[ (-\infty,-2) \cup(1, \infty) \] Determine intervals on which the function is concave down. (Enter your answer using interval notation.) \( \square \) \[ 2 \text {, } \] Locate the points of inflection. (If an answer does not exist, enter DNE.) \[ \begin{array}{ll} (x, y)=\left(\begin{array}{ll} -2,-7 & \\ (x, y)=\left(\begin{array}{ll} 1 & \\ & (\text { smaller } x \text {-value }) \\ (\text { (larger } x \text {-value) } \end{array}\right. \end{array}\right) \end{array} \]

To determine the concavity of the function \( f(x) = \frac{1}{4} x^{4} + \frac{1}{2} x^{3} - 3 x^{2} + 5 \), we will first find the second derivative of the function and then analyze it to find the intervals of concavity and points of inflection. ### Step 1: Find the first derivative \( f'(x) \) The first derivative of \( f(x) \) is calculated as follows: \[ f'(x) = \frac{d}{dx} \left( \frac{1}{4} x^{4} + \frac{1}{2} x^{3} - 3 x^{2} + 5 \right) \] Calculating this gives: \[ f'(x) = x^{3} + \frac{3}{2} x^{2} - 6 x \] ### Step 2: Find the second derivative \( f''(x) \) Now, we differentiate \( f'(x) \) to find the second derivative: \[ f''(x) = \frac{d}{dx} \left( x^{3} + \frac{3}{2} x^{2} - 6 x \right) \] Calculating this gives: \[ f''(x) = 3x^{2} + 3x - 6 \] ### Step 3: Set the second derivative to zero to find critical points To find the points of inflection, we set \( f''(x) = 0 \): \[ 3x^{2} + 3x - 6 = 0 \] Dividing the entire equation by 3 simplifies it to: \[ x^{2} + x - 2 = 0 \] Now we can factor this quadratic: \[ (x + 2)(x - 1) = 0 \] Thus, the critical points are: \[ x = -2 \quad \text{and} \quad x = 1 \] ### Step 4: Determine the concavity To determine the intervals of concavity, we will test the sign of \( f''(x) \) in the intervals defined by the critical points \( x = -2 \) and \( x = 1 \): 1. **Interval \( (-\infty, -2) \)**: - Choose \( x = -3 \): \[ f''(-3) = 3(-3)^{2} + 3(-3) - 6 = 27 - 9 - 6 = 12 \quad (\text{positive}) \] - Concave up. 2. **Interval \( (-2, 1) \)**: - Choose \( x = 0 \): \[ f''(0) = 3(0)^{2} + 3(0) - 6 = -6 \quad (\text{negative}) \] - Concave down. 3. **Interval \( (1, \infty) \)**: - Choose \( x = 2 \): \[ f''(2) = 3(2)^{2} + 3(2) - 6 = 12 + 6 - 6 = 12 \quad (\text{positive}) \] - Concave up. ### Summary of Results - **Concave Up Intervals**: \( (-\infty, -2) \cup (1, \infty) \) - **Concave Down Intervals**: \( (-2, 1) \) ### Step 5: Locate the points of inflection The points of inflection occur at \( x = -2 \) and \( x = 1 \). We will find the corresponding \( y \)-values by substituting these \( x \)-values back into the original function \( f(x) \). 1. For \( x = -2 \): \[ f(-2) = \frac{1}{4}(-2)^{4} + \frac{1}{2}(-2)^{3} - 3(-2)^{2} + 5 = \frac{1}{4}(16) - 4 - 12 + 5 = 4 - 4 - 12 + 5 = -7 \] - Point of inflection: \( (-2, -7) \) 2. For \( x = 1 \): \[ f(1) = \frac{1}{4}(1)^{4} + \frac{1}{2}(1)^{3} - 3(1)^{2} + 5 = \frac{1}{4} + \frac{1}{2} - 3 + 5 = \frac{1}{4} + \frac{2}{4} - \frac{12}{4} + \frac{20}{4} = \frac{11}{4} \] - Point of inflection: \( (1, \frac{11}{4}) \) ### Final Answers - **Concave Down Intervals**: \( (-2, 1) \) - **Points of Inflection**: - \( (x, y) = (-2, -7) \) - \( (x, y) = (1, \frac{11}{4}) \)