Question

Find the intervals on which the function f defined by f( x )=x⁴-8x²+16is concave upward or concave downward.

Answer 1

$f'=4x^3-16x$

$f''=12x^2-16$

$12x^2-16>0\iff 3x^2-4>0\implies x\in (-\infty,-\frac{2\sqrt3 }{3})\bigcup(\frac{2\sqrt3 }{3},+\infty)$, then the function is concave upward.

$f''<0\iff x\in (-\frac{2\sqrt3}{3},\frac{2\sqrt3}{3})$ , then the function is concave downward.