# How do you use the second derivative test to find min/max/pt of inflection of y= x^5-5x?

Jul 6, 2015

Maximum at $x = - 1 \mathmr{and} y = 4$
Minimum at $x = 1 \mathmr{and} y = - 4$
Inflection point at $x = 0 \mathmr{and} y = 0$.

#### Explanation:

Start by taking the first derivative:
$y ' = 5 {x}^{4} - 5$
set it equal to zero:
$5 {x}^{4} - 5 = 0$
$x = \pm 1$ corresponding to $y = \pm 4$; these will be the $x$-coordinates of maxima or minima!

Now we find the second derivative:
$y ' ' = 20 {x}^{3}$
we analyze the sign of the second derivative by setting it bigger than zero:
$20 {x}^{3} > 0$
so this is true when $x > 0$
we have an inflection point at $x = 0$ where your function changes concavity;
Graphically:

With this we discover that :
$x = - 1 \mathmr{and} y = 4$ is a maximum and
$x = 1 \mathmr{and} y = - 4$ is a minimum.

The graph of your fnction looks like:
graph{x^5-5x [-10, 10, -5, 5]}