Question

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

Answer 1

Maximum at `x=-1 and y=4`
Minimum at `x=1 and y=-4`
Inflection point at `x=0 and y=0`.

Explanation:

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

Now we find the second derivative:
`y''=20x^3`
we analyze the sign of the second derivative by setting it bigger than zero:
`20x^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 and y=4` is a maximum and
`x=1 and y=-4` is a minimum.

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