Differentiability

Prove that if f(x) = x^alpha, where alpha = 1/n for some n in N (the natural numbers), then y = f(x) is differentiable and f'(x) = alpha x^(alpha - 1). Progress I have made so far: I have managed to prove, (x^n)' = n x^(n - 1) for n in N and x in R both from the definition of differentiation involving the limit and the binomial theorem or equivalently using induction on n. Feel free to use this result although anything else should be made rigorous. It should be possible to prove this by the basic definition of the derivative. Thanks!

Limits of Functions

For which real values alpha does lim {x -> 0+} x^alpha sin(1/x) exist? It is easy to show using the epsilon - delta definition below that this limit exists for all real alpha >= 1. In fact the limit is zero in this case. The case alpha equals zero is also quite simple and the limit does not exist. Consider the two sequences a_n = 2/((4n + 1)pi) and b_n = 2/((4n + 3)pi) These go to zero through positive values as n --> infinity and yet f(a_n) = 1, however f(b_n) = -1, so this limit cannot exist. I AM INTERESTED IN THE CASE alpha < 0 in which case the limit diverges, but this must be shown rigorously...Thanks! Definition [Right Hand Limit] Let I be a nonempty interval with a... click for more

Differentiation

Prove that if f(x) = x^alpha, where alpha = 1/n for some n in N (the natural numbers), then y = f(x) is differentiable and f'(x) = alpha x^(alpha - 1). Progress I have made so far: I have managed to prove, (x^n)' = n x^(n - 1) for n in N and x in R both from the definition of differentiation involving the limit and the binomial theorem or equivalently using induction on n. Feel free to use this result although anything else should be made rigorous. It should be possible to prove this by the basic definition of the derivative. Thanks! Please no fancy inverse function theorem, just the basic definition of the derivative...

Functions: Proof by Induction

Let n be a natural number, and let f(x) = x^n for all x are members of R. 1) If n is even, then f is strictly increasing hence one-to-one, on [0,infinity) and f([0,infinity)) = [0,infinity). 2) If n is odd, then f is strictly increasing, hence one-to-one, on R and f(R) = R. "This needs to be a proof by induction, proving the fact that f is strictly increasing. The proof needs to look at the values of n, not x"

Real Analysis: Mean Value Theorem

Please see the attached file for the fully formatted problem. Let f(x) = ˆ e− 1/x 2 x 6 = 0 0 x = 0 Show that the nth derivative of f(x) exists for all n 2 N . Please justify all steps and be rigorous because it is an analysis problem. (Note: The problem falls under the chapter on Differentiability on R in the section entitled The Mean Value Theorem.)