Question

The absolute maximum of $y=x^3-3x+2$ in $0\le x\le 2$ is:

• A. $4$
• B. $6$
• C. $3$
• D. $0$

Answer 1

Answer: (A)

$4$

$f\left(x\right)=x^3-3x+2$
$f^{\prime }\left(x\right)=3x^2-3$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=\pm 1$
Since , $x=-1$ does not lies in the given range of x.
$f\left(1\right)=0$
$f\left(0\right)=2$ and $f\left(2\right)=4$
So, absolute maximum value is 4.