Question

How do you find critical points for `G(x)= ^3sqrt(x²-x)`?

Answer 1

Hey there :)

The critical points of `G(x) = (x^2-x)^(1/3)` occur at `x=1/2, x=0` and `x=1`.

In order to find the critical points, you must take the derivative of `G(x)`. Using the chain rule, you should obtain `G'(x)=(2x-1)/(3(x^2-x)^(2/3))` ( Try it! )

Critical points of a function are any points in the domain of the function where the value of the derivative of the function is 0 or undefined. (See http://en.wikipedia.org/wiki/Critical_point_%28mathematics%29 for more info!)

So for your function, we are looking for the `x` values such that `G'(x) = 0`. This is `(2x-1)/(3(x^2-x)^(2/3)) = 0`. Now the only way this is possible is if `2x-1 = 0`. This gives `x = 1/2`.

But we must also see where `G'(x)` is undefined!
Now we can't have a zero denominator, so `G'(x)` is undefined if `x^2-x = 0`. Factoring, we get `x(x-1) = 0`. This gives `x = 0` and `x=1`.