Find the absolute maximum and minimum values of $f(x) = 2x^3-9x^2 + 12x-5$ in [0, 3]. Also find the points of maxima and minima.

Absolute maximum: 4 at $x=3$; Absolute minimum: −5 at $x=0$

The correct answer is Option (2) → Absolute maximum: 4 at $x=3$; Absolute minimum: −5 at $x=0$

Given $f(x) = 2x^3-9x^2 + 12x-5$   ...(i)

It is differentiable for all x in [0, 3].

Differentiating (i) w.r.t. x, we get

$f'(x) = 2.3x^2 - 9.2x+12=6(x^2 - 3x + 2)$.

Now $f'(x) = 0 ⇒6(x^2 - 3x + 2) = 0⇒ x^2 - 3x + 2 = 0$

$⇒(x-1) (x-2)= 0⇒x= 1, 2$.

Also 1, 2 both are in [0, 3], therefore, 1 and 2 both are turning points.

Further, $f(1) = 2.1^3-9.1^2 + 12.1-52-9+12-5 = 0,$

$f(2) = 2.2^3-9.2^2 + 12.2-5 = 16-36 + 24-5=-1,$

$f(0) = -5$

and $f(3)= 2.3^3-9.3^2+ 12.3-5= 54-81 +36-5= 4$.

Therefore, the absolute maximum value = 4 and the absolute minimum value = -5. The point of maxima is 3 and the point of minima is 0.