Question

The absolute maximum value of $y = x^3 - 3x + 2$ in $0 \leq x \leq 2$ is

• A. 4
• B. 6
• C. 2
• D. 0

Answer 1

Answer: (A)

4

Answer 2

To find the absolute maximum value of the function $y = x^3 - 3x + 2$ in the interval $0 \leq x \leq 2$, we follow these steps:

1. Find the derivative of the function: Let $f(x) = x^3 - 3x + 2$. The first derivative is $f'(x) = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3$.

2. Find the critical points: Set the derivative equal to zero to find the critical points: $3x^2 - 3 = 0$ $3(x^2 - 1) = 0$ $x^2 - 1 = 0$ $x^2 = 1$ $x = \pm 1$

   We consider only the critical points that lie within the given interval $0 \leq x \leq 2$. The critical point $x=1$ is in the interval $[0, 2]$. The critical point $x=-1$ is not in the interval $[0, 2]$.

3. Evaluate the function at the critical points and the endpoints of the interval:

   • At the critical point $x=1$: $f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$.

   • At the left endpoint $x=0$: $f(0) = (0)^3 - 3(0) + 2 = 0 - 0 + 2 = 2$.

   • At the right endpoint $x=2$: $f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4$.

4. Determine the absolute maximum value: Compare the values of the function obtained: $0, 2, 4$. The largest among these values is $4$.

Therefore, the absolute maximum value of $y = x^3 - 3x + 2$ in $0 \leq x \leq 2$ is $4$.