Question

What are the critical points of `f(x) = x*sqrt(8-x^2)`?

Answer 1

x= `+-2, +-2sqrt(2)`

Explanation:

To find the critical points of a function, find where the first derivative equals 0.
For `f(x)` stated above, `f'(x) = (8-2x^2)/sqrt(8-x^2)`

You need to use the product rule for derivatives: `u * dv + v * du`
where in this case, `u = x`, and `v = sqrt(8-x^2)`

`du` is the derivative of `u` and the derivative of `x= 1` so `du = 1`.

`dv` is the derivative of `v`; the derivative of `sqrt(8-x^2)` requires the use of the Chain Rule because you are taking the derivative of the function of a function (the square root of `(8-x^2)`).
It is easier to represent `sqrt((8-x^2))` using exponents:
`sqrt((8-x^2)) = (8-x^2)^(1/2)`.

Now use the Chain Rule to find the derivative of `v`:
`d/dx((8-x^2)^(1/2)) = 1/2(8-x^2)^(-1/2)*-2x`
(Don't forget to find the derivative of the inside function --
that's where the `-2x` comes from!)

Clean this up so `dv= -x/sqrt(8-x^2)`

OKAY, put it all together now:

`u*dv + v*du =`
`x*-x/sqrt(8-x^2) + sqrt(8-x^2) * 1`
`= -x^2/sqrt(8-x^2) + sqrt(8-x^2)`

Combine these two fractions into one by getting common denominators and you get `f'(x) = (8-2x^2)/sqrt(8-x^2)`

Critical values occur when `f'` = 0 or undefined (i.e. the
numerator = 0 and/or denominator = 0).
Setting the numerator = 0 and solving results in `x=+-2`
Setting the denominator = 0 and solving results in `x=+-2sqrt(2)`

If you use Symbolab, you can quickly get these answers, I hope my explanation helped! :)