Question

What are the critical points for `f(x) = xsqrt(16 - x^2)`?

Answer 1

The critical numbers are `+-2sqrt2`, and `+-4`.

Explanation:

Critical numbers for a function, `f`, are numbers in the domain `f` at which `f'(x) = 0` or `f'(x)` does not exist.

For, `f(x) = xsqrt(16 - x^2)`, the derivative is:

`f'(x) = sqrt(16-x^2) + x 1/(2sqrt(16-x^2)) *(-2x)`

`= sqrt(16-x^2)/1 - x^2/sqrt(16-x^2)`

`= (16-2x^2)/sqrt(16-x^2)`

Derivative is 0
`f'(x) = 0` when `16-2x^2 = 0`.
Which happens at `x=+-sqrt8 = +-2sqrt2`.

Both `sqrt8` and `-sqrt8` are in the domain of `f`, so they are both critical numbers.

Derivative does not exist
`f'(x)` does not exist when `16-x^2 =0`, which happens at `+-4`.
Both `4` and `-4` are in the domain of `f`, so they are both critical numbers.