Compute the derivative:
\(\dfrac{\mathrm d}{\mathrm dx}\bigg[8x^3+2x^2-8\bigg]=24x^2+4x\)
Set equal to zero and find the critical points:
\(24x^2+4x=4x(6x+1)=0\implies x=0,-\dfrac16\)
Compute the second derivative at the critical points to determine concavity. If the second derivative is positive, the function is concave upward at that point, so the function attains a minimum at the critical point. If negative, the critical point is the site of a maximum.
\(\dfrac{\mathrm d^2}{\mathrm dx^2}\bigg[8x^3+2x^2-8\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[24x^2+4x\bigg]=48x+4\)
At \(x=0\), the second derivative takes on the value of \(4\), so the function is concave upward, so the function has a minimum there of \(-8\).
At \(x=-\dfrac16\), the second derivative is \(-4\), so the function is concave downward and has a maximum there of \(-\dfrac{431}{54}\).
