(1 point) Consider the function \( f(x)=-2 x^{3}+27 x^{2}-84 x+8 \). This function has two critical numbers \( A

Ask by Powers Chavez. in the United States

Mar 13,2025

To find the critical numbers \( A \) and \( B \) for the function \( f(x) = -2x^3 + 27x^2 - 84x + 8 \), we first need to calculate the derivative \( f'(x) \) and then find where it is equal to zero. ### Step 1: Calculate the derivative The derivative of \( f(x) \) is given by: \[ f'(x) = \frac{d}{dx}(-2x^3 + 27x^2 - 84x + 8) \] Calculating the derivative: \[ f'(x) = -6x^2 + 54x - 84 \] ### Step 2: Set the derivative equal to zero To find the critical points, we set \( f'(x) = 0 \): \[ -6x^2 + 54x - 84 = 0 \] Dividing the entire equation by -6 to simplify: \[ x^2 - 9x + 14 = 0 \] ### Step 3: Solve the quadratic equation We can factor the quadratic: \[ (x - 7)(x - 2) = 0 \] Thus, the critical points are: \[ x = 7 \quad \text{and} \quad x = 2 \] So, we have: \[ A = 2 \quad \text{and} \quad B = 7 \] ### Step 4: Determine the sign of \( f'(x) \) in the intervals Now we will test the sign of \( f'(x) \) in the intervals \( (-\infty, A) \), \( (A, B) \), and \( (B, \infty) \). 1. **Interval \( (-\infty, 2) \)**: - Choose a test point, e.g., \( x = 0 \): \[ f'(0) = -6(0)^2 + 54(0) - 84 = -84 \quad (\text{negative}) \] 2. **Interval \( (2, 7) \)**: - Choose a test point, e.g., \( x = 5 \): \[ f'(5) = -6(5)^2 + 54(5) - 84 = -6(25) + 270 - 84 = -150 + 270 - 84 = 36 \quad (\text{positive}) \] 3. **Interval \( (7, \infty) \)**: - Choose a test point, e.g., \( x = 8 \): \[ f'(8) = -6(8)^2 + 54(8) - 84 = -6(64) + 432 - 84 = -384 + 432 - 84 = 0 \quad (\text{negative}) \] ### Step 5: Determine local maxima and minima - Since \( f'(x) \) changes from negative to positive at \( A = 2 \), there is a local minimum at \( A \). - Since \( f'(x) \) changes from positive to negative at \( B = 7 \), there is a local maximum at \( B \). ### Final Answers Thus, we can fill in the blanks as follows: \[ \begin{array}{l} A = 2 \\ B = 7 \end{array} \] For the intervals: \[ (-\infty, A) \quad \text{is } - \\ (A, B) \quad \text{is } + \\ (B, \infty) \quad \text{is } - \] Thus, \( f(x) \) has a local \( \text{MIN} \) at \( A \) and a local \( \text{MAX} \) at \( B \). Final filled in: \[ \begin{array}{l} A = 2 \\ B = 7 \\ (-\infty, A) \quad - \\ (A, B) \quad + \\ (B, \infty) \quad - \\ \text{local MIN at } A \\ \text{local MAX at } B \end{array} \]