Question

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

Answer 1

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!