Find the minimum value of f(x) = x^3 - 3x in [0, 2].

The correct answer is Option (2) → -2

Given $f(x) = x^3-3x, x ∈ [0,2]$.

It is differentiable for all $x ∈ [0,2]$.

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

Now $f'(x) = 0 ⇒ 3x^2-3=0⇒x^2 = 1⇒x=1, -1$.

But $x∈  [0,2]$, so $x = 1$ is the only turning point.

$f(1) = 1-3=-2, f(0) = 0-0 = 0, f(2) = 8-6=2$.

Therefore, minimum value of $f(x) = -2$.