The minimum value of the function f(x)=x^2+128/x is :

The correct answer is Option (2) → 48

$f(x)=x^2+\frac{128}{x}$

$f'(x)=2x-\frac{128}{x^2}$

for critical points,

$f'(c)=0$

$⇒2c-\frac{128}{c^2}=0$

$⇒2c^3=128$

$⇒c^3=64$

$⇒c=4$

$f(4)=16+\frac{128}{4}=48$