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

1 Answer
Jul 6, 2015

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:
enter image source here
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]}