Posted by **dana** on Saturday, December 18, 2010 at 3:42am.

(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.

- calculus -
**Reiny**, Saturday, December 18, 2010 at 9:04am
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 y-intercept is easy, let x = 0

y = 0

for the x-intercept 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 cross-multiplying

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.

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