(1 point) Consider the function \( f(x)=-2 x^{3}+33 x^{2}-168 x+3 \). This function has two critical numbers \( A

Ask by Marshall Graham. in the United States

Mar 13,2025

We start with the function \[ f(x) = -2 x^3 + 33 x^2 - 168 x + 3. \] **Step 1. Find the critical numbers.** The first derivative is \[ f'(x) = \frac{d}{dx}\left(-2x^3 + 33x^2 - 168x + 3\right) = -6x^2 + 66x - 168. \] Factor out \(-6\): \[ f'(x) = -6\left(x^2 - 11x + 28\right). \] Factor the quadratic: \[ x^2 - 11x + 28 = (x-4)(x-7). \] Thus, setting \(f'(x) = 0\): \[ -6(x-4)(x-7) = 0 \quad \Longrightarrow \quad x = 4 \quad \text{or} \quad x = 7. \] Since \(4 < 7\), we have: \[ A = 4 \quad \text{and} \quad B = 7. \] **Step 2. Compute the second derivative.** The second derivative is \[ f''(x) = \frac{d}{dx}\left(-6x^2 + 66x - 168\right) = -12x + 66. \] Now evaluate at \(x=4\) and \(x=7\): \[ f''(4) = -12(4) + 66 = -48 + 66 = 18, \] \[ f''(7) = -12(7) + 66 = -84 + 66 = -18. \] **Step 3. Determine the nature of the critical points.** Since - \(f''(4) = 18 > 0\), the function is concave up at \(x=4\), indicating a **local minimum**. - \(f''(7) = -18 < 0\), the function is concave down at \(x=7\), indicating a **local maximum**. **Final Answers:** \[ A = 4, \quad B = 7, \quad f''(4) = 18, \quad f''(7) = -18. \] Thus, \(f(x)\) has a local minimum at \(x=4\) and a local maximum at \(x=7\).