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