How do use the first derivative test to determine the local extrema #x^2+1#?

1 Answer
Jul 23, 2015

Answer:

You can use three easy steps:

Explanation:

First you evaluate the derivative of your function:
#y'=2x#

Second you set it equal to zero (so you get the #x# coordinate(s) of points which have zero derivative, i.e., they neither go 'up" nor "down"...so they are max or min);
#y'=0#
#2x=0#
or #x=0# which is a max or min...

Third we need to understand if it is a max or min so we set the derivative bigger than zero:
#y'>0# this will give you the interval where the function is going up (derivative bigger than zero) or down (derivative less than zero):

you get that #y'>0# when:
#2x>0# or #x>0#
this means that the derivative is bigger than zero (function going up) when #x# is bigger than zero and is smaller than zero (going down) when #x# is less than zero! so #x=0# is a MINIMUM!
Your minimum has coordinates #x=0, y=1# (substituting #x=0# into your function).

Graphically:
graph{x^2+1 [-10, 10, -5, 5]}
you can see that for #x<0# the function goes down and for #x>0# up!