Question

Find the intervals on which the function f defind by f(x)=x^4-8x^2+16 is concave upward or concave downward.

Answer 1

Given

$f(x)=x^4-8x^2+16$

Since

$f'(x) = 4x^3-16x\\ f''(x)= 12x^2-16$

Solve

$f''(x)=0$

we get

$x=\frac{2\sqrt{3}}{3},\ \ \ \ \:x=-\frac{2\sqrt{3}}{3}$

Now , we check the sign of $f''(x)$ , since

$f''(-2)>0,\ \ \ f''(0)<0,\ \ f''(2)>0$

Then $f(x)$ is concave up on $(-\infty , -2\sqrt{3}/3)\cup (2\sqrt{3}/3,\infty)$ and

concave down on $(-2\sqrt{3}/3,2\sqrt{3}/3)$