The absolute maximum value of function f(x)=x^3-3x+2 in [0,2] is:

The correct answer is Option (3) → 4 ##

Given, $f(x) = x^3 - 3x + 2$

On differentiating, we get

$f'(x) = 3x^2 - 3$

For critical points, put $f'(x) = 0$:

$3x^2 - 3 = 0$

$\Rightarrow x^2 = 1$

$\Rightarrow x = \pm 1$

Since $x = -1$ is outside the given interval $[0, 2]$, we consider $x = 1$.

Now, value of $f(x)$ at critical points and endpoints:

$f(0) = 0^3 - 3(0) + 2 = 2$

$f(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$

$f(2) = 2^3 - 3(2) + 2 = 8 - 6 + 2 = 4$

The maximum value is 4.