\(f(x)=x^3+x^2-5x-2\)
\(\implies f'(x)=3x^2+2x-5=(3x+5)(x-1)\)
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 (\(f''(x)<0\)), then a maximum occurs; if it's convex (\(f''(x)>0\)), then a minimum occurs.
You have
\(f''(x)=6x+2\)
and so
\(f''\left(-\dfrac53\right)=-8<0\implies f(x)\text{ is concave at }x=-\dfrac53\)
\(f''(1)=8>0\implies f(x)\text{ is convex at }x=1\)
This means a maximum of \(f\left(-\dfrac53\right)=\dfrac{121}{27}\) and a minimum of \(f(1)=-5\).
