Question

Calculus(Derivatives):

Answer 1

$f(x)=x^3+x^2-5x-2$
$\implies f'(x)=3x^2+2x-5=(3x+5)(x-1)$

Critical points occur when $f'(x)=0$, which happens for $x=-\dfrac53$ and $x=1$.

Check the sign of the second derivative at each critical point to determine the function's concavity at that point. If it's concave ($f''(x)$), then a maximum occurs; if it's convex ($f''(x)>0$), then a minimum occurs.

You have

$f''(x)=6x+2$

and so

$f''\left(-\dfrac53\right)=-8$
$f''(1)=8>0\implies f(x)\text{ is convex at }x=1$

This means a maximum of $f\left(-\dfrac53\right)=\dfrac{121}{27}$ and a minimum of $f(1)=-5$.