(1 point) Consider the function \( f(x)=x^{2} e^{3 x} \). This function has two critical numbers \( A

Ask by Burton Garrett. in the United States

Mar 13,2025

To find the critical numbers of the function \( f(x) = x^2 e^{3x} \), we first need to compute its derivative and set it to zero. The product rule gives us: \[ f'(x) = 2x e^{3x} + x^2 (3e^{3x}) = e^{3x} (2x + 3x^2) \] Setting \( f'(x) = 0 \) leads us to solve: \[ e^{3x} (2x + 3x^2) = 0 \] Since \( e^{3x} \neq 0 \) for all \( x \), we need to solve: \[ 2x + 3x^2 = 0 \] Factoring gives: \[ x(3x + 2) = 0 \] This results in critical points: \[ x = 0 \quad \text{and} \quad 3x + 2 = 0 \implies x = -\frac{2}{3} \] Thus, the critical numbers are: \[ A = -\frac{2}{3} < B = 0 \] Next, we evaluate the sign of \( f'(x) \) over the intervals: 1. For the interval \( (-\infty, -\frac{2}{3}) \), plug in \( x = -1 \): \[ f'(-1) = e^{-3}(2(-1) + 3(-1)^2) = e^{-3}(2)(1) > 0 \quad \text{(Positive)} \] 2. For the interval \( (-\frac{2}{3}, 0) \), plug in \( x = -\frac{1}{2} \): \[ f'(-\frac{1}{2}) = e^{3/2}(2(-\frac{1}{2}) + 3(-\frac{1}{2})^2) = e^{3/2}(-1 + \frac{3}{4}) < 0 \quad \text{(Negative)} \] 3. For the interval \( (0, \infty) \), plug in \( x = 1 \): \[ f'(1) = e^3(2(1) + 3(1)^2) = e^3(5) > 0 \quad \text{(Positive)} \] Now we fill in the blanks: \[ (-\infty, A) \quad + \quad (A, B) \quad - \quad (B, \infty) \quad + \] Therefore, we conclude that \( f(x) \) has a local MAX at \( A \) and a local MIN at \( B \). Final answers: \[ A = -\frac{2}{3}, \quad B = 0 \] \[ (-\infty, A) : +, \quad (A, B) : -, \quad (B, \infty) : + \] Thus we conclude that \( f(x) \) has a local MAX at \( A \) and a local MIN at \( B \).