calculus
posted by dana on .
(a) Find the critical points of f(x) = x^(1/3)x^(1/5)
(b) Determine the maximum and minimum values of f on the interval [0; 2] and where they occur.

I will assume that you know that critical points would be points such as the x and y intercepts, the max and min points, and any points of inflection.
I think the problem here lies in trying to solve different equations containing fractional exponents.
the yintercept is easy, let x = 0
y = 0
for the xintercept let y = 0
so x^(1/2  x^(1/5) = 0
x^(1/3) = x^(1/5)
raise it to the 15th power, (15 is the LCM of 3 and 5)
x^5 = x^3
x^5  x^3  0
x^3(x^2  1) = 0
x = 0 or x = ± 1
After you get dy/dx = (1/3)x^(2/3)  (1/5)x^(4/5)
you will have to set this equal to zero as well
so the same sort of thing as above
x^(2/3) = x^(4/5) , after dividing out the 1/3
so 1/x^(2/3) = 1/x^(4/5)
x^(4/5) = x^(2/3) after crossmultiplying
raise to the 15th
x^12 = x^10
x^10(x^2  1) = 0
x = 0 or x = ± 1 (well, isn't that interesting)
etc.