Question

Let \( f(x) = x^3 - \dfrac{9}{2}x^2 + 6x - 2 \) be a function defined on the closed interval \( [0, 3] \). Then, the global maximum value of \( f(x) \) is

• A. 4.5
• B. 0.5
• C. 2.5
• D. 3.0

Hint:

To find global extrema on a closed interval, evaluate \( f(x) \) at endpoints and at points where \( f'(x) = 0 \).

Answer 1

The Correct Option is C

Given: \[ f(x) = x^3 - \frac{9}{2}x^2 + 6x - 2 \] Take derivative: \[ f'(x) = 3x^2 - 9x + 6 \] Solve \( f'(x) = 0 \): \[ 3x^2 - 9x + 6 = 0 \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow x = 1, 2 \] Evaluate \( f(x) \) at endpoints and critical points: \[ f(0) = -2,\quad f(1) = 1 - 4.5 + 6 - 2 = 0.5,\quad f(2) = 8 - 18 + 12 - 2 = 0,\quad f(3) = 27 - 40.5 + 18 - 2 = 2.5 \] Maximum = 2.5