Derivatives! It's Calculus!?

Question

Multipart question, in stuck on parts 'B' and 'D', though I'll show all of them. Thanks in advance.

f(x) = (x-2)^(2/3)

In words, f of x equals the quantity of x minus two, raised to the two-thirds power.

A) Does f ' (2) exist?

B) Show that the only local extreme value of f occurs at x = 2.

C) Does the result in B contradict the Extreme Value Theorem?

D) Repeat parts A and B for f(x) = (x - a)^(2/3)

Change the 'in' towards the beginning, to 'I'm'.

Don't know how I missed that...

Answer 1

f(x) = (x-2)^3/2 then f'(x) = (3/2)*(x-2)^-(1/3). Derivative is infinite at x=2, so the function is discontinuous at that point.

The function is imaginary for x<2, so values only exist for X>=2. it is a monotonic function above that value so there will be only one minimum and one maximum in any given interval, the minimum occurring at the low end of the interval, the maximum at the end. The minimum at x = 2 is the lowest value the function can be so it is a local extreme.

For part D, the function is imaginary for x>a, monotonic for x>=a.

I hope this helps a bit, I'm nor sure this is what you are after.