Answer
See explanation.
Work Step by Step
$f^{\prime}(x)<0$ the $f^{\prime}(x)>0$ on the left and on right of $x=-\sqrt{7},$ so $x=-\sqrt{7}$ is $a$ relative maximum.
$\pm \sqrt{7}=x$
Using the first derivative test
$f^{\prime}(x)<0$ on the left and $f^{\prime}(x)>0 $ on the right of $x=\sqrt{7}, $ so $x=\sqrt{7}$ is a relative minimum
