Find both the maximum value and the minimum value of 3x⁴ - 8x³ + 12x² - 48x + 25 on the interval [0, 3].
Maximum value = 25 at x = 0; Minimum value = -39 at x = 2
Explanation
Following the method demonstrated in Example 27 of the textbook, we find the derivative, identify critical points within the interval, and evaluate the function at all critical points and endpoints to determine absolute maximum and minimum values.
Solution Steps
Step 1: Let f(x) = 3x⁴ - 8x³ + 12x² - 48x + 25. Find the derivative: f'(x) = 12x³ - 24x² + 24x - 48 = 12(x³ - 2x² + 2x - 4)
Step 2: Find critical points by setting f'(x) = 0: 12(x³ - 2x² + 2x - 4) = 0. Factorizing: x³ - 2x² + 2x - 4 = x²(x - 2) + 2(x - 2) = (x² + 2)(x - 2) = 0. This gives x = 2 (since x² + 2 = 0 has no real solution).
Step 3: The critical point x = 2 lies in the interval [0, 3]. Now evaluate f(x) at critical point and endpoints:
Step 4: f(0) = 3(0) - 8(0) + 12(0) - 48(0) + 25 = 25
Step 5: f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39
Step 6: f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16
Step 7: Comparing all values: f(0) = 25, f(2) = -39, f(3) = 16. Therefore, absolute maximum value is 25 (at x = 0) and absolute minimum value is -39 (at x = 2).