The minimum value of the function $f(x)=x^3-12 x$ on the interval [0, 3] is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -16
$f(x)=x^3-12x$
$f'(x)=3x^2-12=3(x^2-4)$
$f'(x)=0 \Rightarrow x=\pm 2$
$x=2 \in [0,3]$
$f(0)=0,\;\; f(2)=8-24=-16,\;\; f(3)=27-36=-9$
$\text{Minimum value} = -16$
The minimum value is $-16$.