Question

What are the critical points for `f(x)=x^(2/5)(x-10)`?

Answer 1

Critical numbers for function`f` are numbers in the domain of `f` at which `f'(x)=0` or `f'(x)` does not exist. For this function the critical numbers are `0` and `20/7`.

Explanation:

For `f(x)=x^(2/5)(x-10)`, I prefer to rewrite the function before differentiating.

`f(x) = x^(7/5)-10x^(2/5)`

So,
`f'(x) =7/5x^(2/5) - 10*2/5x^(-3/5)`

`= (7x^(2/5))/5 - 4/x^(3/5)`

`= (7x-20)/(5x^(3/5))`

`f'(x) = 0` at `x=20/7` which is in the domain of `f`/

`f'(x)` does not exist at `5x^(3/5) = 0` which happens at `x=0`. Note that `0` is in the domain of the original function, `f`, so it is a critical number for `f`.

The critical numbers for `f` are: `0` and `20/7`.

Additional Notes

`f"(x)` changes from positive to negative as we move past `0`, fo `f(0) = 0` is a local maximum value of `f`

`f'(x)` changes from negative to positive as we move past `20/7`, so `f(20/7)` is a local minimum value of `f`.

It was not asked, but here is the graph of the function: `f(x)=x^(2/5)(x-10)`,

graph{y =x^(2/5)(x-10) [-21.1, 36.62, -24.43, 4.43]}