# calculus

posted by 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.

• calculus - ,

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.