The minimum value of $f(x) = 4x^3 - 48 x + 105 $ in the interval [1, 3] is :

41

The correct answer is Option (2) → 41

Given,

$f(x)=4x^3-48x+105$

$⇒f'(x)=12x^2-48$

Now, for critical points,

$12x^2-48=0$

$x^2-4=0$

$x=±2$

Since the interval is [1, 3], we consider x = 2, as it lies within the interval.

$f(1)=4(1)^3-48(1)+105=4-48+105=61$

$f(2)=4(2)^3-48(2)+105=32-96+105=41$

$f(3)=4(3)^3-48(3)+105=108-144+105=69$

∴ Minimum value is 41.