Question

8x^3+2x^2-8 estimate the relative maxima and relative minima

Answer 1

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 $-\dfrac{431}{54}$.