Question

f(x)=(x^2 -4)^ 2/3 critical numbers and point of inflection

Answer 1

For the critical points:
Taking the derivative, you get f'(x) = (2/3)*(x^2 - 4)^(-1/3)*2x. Setting this equal to 0 and dividing out the (x^2 - 4)^(-1/3), you will see that x = 0.

For the point of inflection:
Take the derivative again, either by product or quotient rule and you'll get:
(12(x^2-4)^(1/3) - 8*x^2*(x^2-4)^(-2/3))/(9*(x^2-4)^(2/3))=0. Get rid of the bottom by multiplying it to the other side then divide by (x^2-4)^(1/3) to get:
12-(8x^2)/(x^2-4) = 0. You can simplify further to eventually get x^2 = 12.
Therefore the points of inflection are x = +-sqrt(12)