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

Jul 23, 2015

You can use three easy steps:

#### Explanation:

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

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);
$y ' = 0$
$2 x = 0$
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:
$2 x > 0$ 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!