11. \( f(x)=x^{3}-6 x^{2}+9 x-8 ; \quad[0,5] \) 12. \( f(x)=x^{3}-3 x^{2}-24 x+5 ; \quad[-3,6] \) 13. \( f(x)=\frac{1}{3} x^{3}+\frac{3}{2} x^{2}-4 x+1 ; \quad[-5,2] \)

To analyze the functions you've provided, we can find their critical points, evaluate them at the endpoints of the given intervals, and determine their maximum and minimum values. Let's go through each function step by step. ### 11. \( f(x) = x^{3} - 6x^{2} + 9x - 8 \) on \([0, 5]\) 1. **Find the derivative**: \[ f'(x) = 3x^{2} - 12x + 9 \] Set the derivative to zero to find critical points: \[ 3x^{2} - 12x + 9 = 0 \implies x^{2} - 4x + 3 = 0 \implies (x-3)(x-1) = 0 \] Thus, \( x = 1 \) and \( x = 3 \). 2. **Evaluate \( f(x) \) at the critical points and endpoints**: - \( f(0) = 0^{3} - 6(0)^{2} + 9(0) - 8 = -8 \) - \( f(1) = 1^{3} - 6(1)^{2} + 9(1) - 8 = 1 - 6 + 9 - 8 = -4 \) - \( f(3) = 3^{3} - 6(3)^{2} + 9(3) - 8 = 27 - 54 + 27 - 8 = -8 \) - \( f(5) = 5^{3} - 6(5)^{2} + 9(5) - 8 = 125 - 150 + 45 - 8 = 12 \) 3. **Determine the maximum and minimum**: - Values: \( f(0) = -8, f(1) = -4, f(3) = -8, f(5) = 12 \) - Minimum: \(-8\) at \(x = 0\) and \(x = 3\) - Maximum: \(12\) at \(x = 5\) ### 12. \( f(x) = x^{3} - 3x^{2} - 24x + 5 \) on \([-3, 6]\) 1. **Find the derivative**: \[ f'(x) = 3x^{2} - 6x - 24 \] Set the derivative to zero: \[ 3x^{2} - 6x - 24 = 0 \implies x^{2} - 2x - 8 = 0 \implies (x-4)(x+2) = 0 \] Thus, \( x = 4 \) and \( x = -2 \). 2. **Evaluate \( f(x) \) at the critical points and endpoints**: - \( f(-3) = (-3)^{3} - 3(-3)^{2} - 24(-3) + 5 = -27 - 27 + 72 + 5 = 23 \) - \( f(-2) = (-2)^{3} - 3(-2)^{2} - 24(-2) + 5 = -8 - 12 + 48 + 5 = 33 \) - \( f(4) = 4^{3} - 3(4)^{2} - 24(4) + 5 = 64 - 48 - 96 + 5 = -75 \) - \( f(6) = 6^{3} - 3(6)^{2} - 24(6) + 5 = 216 - 108 - 144 + 5 = -31 \) 3. **Determine the maximum and minimum**: - Values: \( f(-3) = 23, f(-2) = 33, f(4) = -75, f(6) = -31 \) - Minimum: \(-75\) at \(x = 4\) - Maximum: \(33\) at \(x = -2\) ### 13. \( f(x) = \frac{1}{3}x^{3} + \frac{3}{2}x^{2} - 4x + 1 \) on \([-5, 2]\) 1. **Find the derivative**: \[ f'(x) = x^{2} + 3x - 4 \] Set the derivative to zero: \[ x^{2} + 3x - 4 = 0 \implies (x+4)(x-1) = 0 \] Thus, \( x = -4 \) and \( x = 1 \). 2. **Evaluate \( f(x) \) at the critical points and endpoints**: - \( f(-5) = \frac{1}{3}(-5)^{3} + \frac{3}{2}(-5)^{2} - 4(-5) + 1 = \frac{-125}{3} + \frac{75}{2} + 20 + 1 \) - Convert to a common denominator (6): \[ f(-5) = \frac{-250}{6} + \frac{225}{6} + \frac{120}{6} + \frac{6}{6} = \frac{101}{6} \approx 16.83 \] - \( f(-4) = \frac{1}{3}(-4)^{3} + \frac{3}{2}(-4)^{2} - 4(-4) + 1 = \frac{-64}{3} + 24 + 16 + 1 = \frac{-64}{3} + 41 = \frac{59}{3} \approx 19.67 \) - \( f(1) = \frac{1}{3}(1)^{3} + \frac{3}{2}(1)^{2} - 4(1) + 1 = \frac{1}{3} + \frac{3}{2} - 4 + 1 = \frac{1}{3} + \frac{9}{6} - \frac{24}{6} + \frac{6}{6} = \frac{1}{3} - \frac{9}{6} = \frac{1}{3} - \frac{3}{2} = \frac{1 - 4.5}{3} = \frac{-3.5}{3} \approx -1.17 \) - \( f(2) = \frac{1}{3}(2)^{3} + \frac{3}{2}(2)^{2} - 4(2) + 1 = \frac{8}{3} + 6 - 8 + 1 = \frac{8}{3} - 1 = \frac{5}{3} \approx 1.67 \) 3. **Determine the maximum and minimum**: - Values: \( f(-5) \approx 16.83, f(-4) \approx 19.67, f(1) \approx -1.17, f(2) \approx 1.67 \) - Minimum: \(-1.17\) at \(x = 1\) - Maximum: \(19.67\) at \(x = -4\) ### Summary of Results - For \( f(x) = x^{3} - 6x^{2} + 9x - 8 \) on \([0, 5]\): - Minimum: \(-8\) at \(x = 0\) and \(x = 3\) - Maximum: \(12\) at \(x = 5\) - For \( f(x) = x^{3} - 3x^{2} - 24x + 5 \) on \([-3, 6]\): - Minimum: \(-75\) at \(x = 4\) - Maximum: \(33\) at \(x = -2\) - For \( f(x) = \frac{1}{3}x^{3} + \frac{3}{2}x^{2} - 4x + 1 \) on \([-5, 2]\): - Minimum: \(-1.17\) at \(x = 1\) - Maximum: \(19.67\) at \(x = -4\)