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

1 Answer
Feb 28, 2018

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! :)