Question

If $f\left(x\right)=x^3-3x^2-9x+16,$ then the absolute minimum value of $f\left(x\right)$ in the interval $[-4,4]$ is

[1 mark]

• A. $-11$
• B. $-60$
• C. $-81$
• D. $-16$

Answer 1

The correct option is B $-60$

$f\left(x\right)=x^3-3x^2-9x+16$
Differentiating w.r.t. $x,$ we get
$f^{\prime }\left(x\right)=3x^2-6x-9$
For critical points, we have
$f^{\prime }\left(x\right)=0$
$\Rightarrow x^2-2x-3=0$
$\Rightarrow x=-1,3$

For finding the absolute minimum value, we evaluate $f$ at the critical points and the end points.
$f\left(3\right)=(3{)}^3-3(3{)}^2-9\left(3\right)+16=-11$
$f(-4)=(-4{)}^3-3(-4{)}^2-9(-4)+16=-60$
$f\left(4\right)=-4$
$f(-1)=21$
$\therefore$ The minimum value of $f\left(x\right)$ in the interval $[-4,4]$ is $-60.$