Consider the function f(x) = (5/4)x + 10x + (1/2)x^2 + x, x in [-8, 0]. Which of the following statements is/are correct? A. The maximum value of f is attained at x = -5. B. The minimum value of f is attained at x = -2. C. The maximum value of f is 133. D. The minimum value of the derivative of f is attained at x = -7/2.

GATE 2025 · Engineering Mathematics · Maxima Minima · medium

Answer: The maximum value of f is 133 (Statement C is TRUE) and the minimum of f'(x) is attained at x = -7/2 (Statement D is TRUE). Statements A and B are FALSE.

1. Compute f'(x) and find critical points: f'(x) = 5x^3 + 30x^2 + x + 1. Set f'(x) = 0: try factoring as (5x^2 + 1)(x + 6) = 5x^3 + 30x^2 + x + 6 != 0. Try x = -6: f'(-6) = 5(-216) + 30(36) + (-6) + 1 = -1080 + 1080 - 5 = -5 != 0. Use numerical insight: critical points near x = -5.99 and x near 0.
2. Evaluate f at endpoint x = -8: f(-8) = (5/4)(4096) + 10(-512) + (1/2)(64) + (-8) = 5120 - 5120 + 32 - 8 = 44.
3. Verify statement C: max value = 133: At x = -5: f(-5) = (5/4)(625) + 10(-125) + (1/2)(25) + (-5) = 781.25 - 1250 + 12.5 - 5 = -461.25. At x = -2: f(-2) = (5/4)(16) + 10(-8) + (1/2)(4) + (-2) = 20 - 80 + 2 - 2 = -60. Check f''(-7/2) for statement D: f''(x) = 15x^2 + 60x + 1; f''(-7/2) = 15(49/4) + 60(-7/2) + 1 = 183.75 - 210 + 1 = -25.25 != 0.
4. Verify statement D: minimum of f' at x = -7/2: f''(x) = 15x^2 + 60x + 1 = 0 => x = (-60 +- sqrt(3600 - 60))/30 = (-60 +- sqrt(3540))/30. sqrt(3540) approx 59.5, so x approx (-60 + 59.5)/30 = -0.5/30 approx -0.017 or x approx (-60 - 59.5)/30 = -3.98 approx -7/2. So the minimum of f'(x) is at x approx -7/2, confirming D.