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, always check endpoints and critical points where the derivative is zero.

Answer 1

The Correct Option is C

We are asked to find the global maximum of the function: \[ f(x) = x^3 - \frac{9}{2}x^2 + 6x - 2 \text{on the interval } [0, 3] \]
To find the global maximum on a closed interval, evaluate the function at: - Endpoints: $x = 0$ and $x = 3$
- Critical points inside the interval (where $f'(x) = 0$)
Step 1: Compute the derivative of $f(x)$:
\[ f'(x) = 3x^2 - 9x + 6 \]
Step 2: Set $f'(x) = 0$ and solve:
\[ 3x^2 - 9x + 6 = 0 \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow (x - 1)(x - 2) = 0 \Rightarrow x = 1, 2 \]
These are critical points in the interval $[0, 3]$. Now evaluate $f(x)$ at $x = 0$, $1$, $2$, and $3$.
Step 3: Evaluate function values:
\[ f(0) = 0 - 0 + 0 - 2 = -2\] \[f(1) = 1 - \frac{9}{2} + 6 - 2 = 1 - 4.5 + 6 - 2 = 0.5\] \[f(2) = 8 - \frac{9}{2}(4) + 12 - 2 = 8 - 18 + 12 - 2 = 0\] \[f(3) = 27 - \frac{9}{2}(9) + 18 - 2 = 27 - 40.5 + 18 - 2 = 2.5\] Step 4: Compare all values:
\[ f(0) = -2, f(1) = 0.5, f(2) = 0, f(3) = 2.5 \] The global maximum value of $f(x)$ on $[0, 3]$ is $\boxed{2.5}$, which occurs at $x = 3$.