# What is the relative minimum, relative maximum, and points of inflection of f(x) = x^4 - 4x^2?

Nov 24, 2016

$f ' \left(x\right) = 4 {x}^{3} - 8 x$
$f ' \left(x\right) = 4 x \left({x}^{2} - 2\right)$

$4 x \left({x}^{2} - 2\right) = 0$

$x = 0 , x = \pm \sqrt{2}$

The multiplicity of each of these zeros is odd, therefore, there will be a minimum or maximum at each value.

For:
$x > \sqrt{2}$
$y > 0$
Positive slope

For:
$0 < x < \sqrt{2}$
$y < 0$
Negative slope

For:
$- \sqrt{2} < x < 0$
$y > 0$
Positive slope

For :
$x < - \sqrt{2}$
$y < 0$
Negative slope

$\therefore x = \sqrt{2} \implies$ Local minimum
$\therefore x = 0 \implies$ Local maximum
$\therefore x = - \sqrt{2} \implies$ Local minimum

$f ' ' \left(x\right) = 12 {x}^{2} - 8$
$f ' ' \left(x\right) = 4 \left(3 {x}^{2} - 2\right)$

$3 {x}^{2} - 2 = 0$
$x = \pm \sqrt{\frac{2}{3}}$

For:
$x > \sqrt{\frac{2}{3}}$
$y > 0$
Concave up

For:
$- \sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}}$
$y < 0$
Concave down

For:
$x < - \sqrt{\frac{2}{3}}$
$y > 0$
Concave up

$\therefore$ There are points of inflection at:
$x = \pm \sqrt{\frac{2}{3}}$