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 the global maximum or minimum of a function on a closed interval, evaluate the function at the critical points and endpoints, then compare the values.

Answer 1

The Correct Option is C

We are given the function: \[ f(x) = x^3 - \dfrac{9}{2} x^2 + 6x - 2 \] and we need to find the global maximum value of $f(x)$ on the interval $[0, 3]$.
1. First, find the first derivative $f'(x)$ to determine the critical points: \[ f'(x) = 3x^2 - 9x + 6 \] 2. Set $f'(x) = 0$ to find the critical points: \[ 3x^2 - 9x + 6 = 0 \] Divide the equation by 3: \[ x^2 - 3x + 2 = 0 \] Factor the quadratic equation: \[ (x - 1)(x - 2) = 0 \] So, $x = 1$ and $x = 2$ are the critical points. 3. Evaluate $f(x)$ at the critical points and the endpoints of the interval: - At $x = 0$: \[ f(0) = 0^3 - \dfrac{9}{2} \times 0^2 + 6 \times 0 - 2 = -2 \] - At $x = 3$: \[ f(3) = 3^3 - \dfrac{9}{2} \times 3^2 + 6 \times 3 - 2 = 27 - \dfrac{9}{2} \times 9 + 18 - 2 = 27 - 40.5 + 18 - 2 = 2.5 \] - At $x = 1$: \[ f(1) = 1^3 - \dfrac{9}{2} \times 1^2 + 6 \times 1 - 2 = 1 - \dfrac{9}{2} + 6 - 2 = 1 - 4.5 + 6 - 2 = 0.5 \] - At $x = 2$: \[ f(2) = 2^3 - \dfrac{9}{2} \times 2^2 + 6 \times 2 - 2 = 8 - \dfrac{9}{2} \times 4 + 12 - 2 = 8 - 18 + 12 - 2 = 0 \] 4. The global maximum value occurs at $x = 3$, and the corresponding value of $f(x)$ is $2.5$.
Thus, the global maximum value of $f(x)$ is $2.5$.