Question

How do use the first derivative test to determine the local extrema `f(x) = x³+3x²-9x+15`?

Answer 1

`x=-3` is a local maximum and `x=1` a local minimum.

Explanation:

First we look for points that might be the extrema by solving `f'(x)=0`:
`f'(x)=3x^2+6x-9=3(x+3)(x-1)`
`3(x+3)(x-1)=0 => x=-3 vv x=1`
`x=-3` and `x=1` are two potential extrema.

Now, we can use the graph of the first derivative to determine whether those points are extrema and which type of extrema:

if when passing through point `x` the first derivative changes it's sign from positive to negative then `x` is a local maximum
and if from negative to positive - a locac minimum.

graph{3(x+3)(x-1) [-10, 10, -5.21, 5.21]}

In this case `x=-3` is a local maximum and `x=1` a local minimum.