Question

How do you find the critical points for `f(x) = (x^2-10x)^4` and the local max and min?

Answer 1

three critical points: `x=0,5,10`
Relative minimum at `x=0, 10`
Relative maximum at `x=5`

Explanation:

First, we need to find `f'(x)`.
Using rules for taking derivative, we get following:
`f'(x) = 4*(x^2-10x)^3*(2x-10)`

Critical points occur when the slope is zero meaning `f'(x)=0`.

`f'(x) = 4*(x^2-10x)^3*(2x-10)=0`

`(2x-10)=0` => `x=5`
`(x^2-10x)=0` => `x=0, 10`

So, three critical points would be `x=0,5,10`.

To find relative maximum and minimum, we can do a sign test.

For `x in`(-∞, 0) => `f'(x)` is negative.
For `x in`(0, 5) => `f'(x)` is positive.
For `x in`(5, 10) => `f'(x)` is negative.
For `x in`(10, ∞) => `f'(x)` is positive.

At `x=0`, function has relative minimum because `f'(x)` changes signs from negative to positive.

At `x=5`, function has relative maximum because `f'(x)` changes signs from positive to negative.

At `x=10`, function has relative minimum because `f'(x)` changes signs from negative to positive.