# Thread: Proof that f is differentiable at 0

1. ## Proof that f is differentiable at 0

So the question states " If f satisfies |f(x)|</= |x|^(7/2) for all x, prove that f is differentiable at 0".
We just learned about Mean Value Theorem, Intermediate Value Theorem, Triangle inequality, Rolle's Theorem, and Extreme Value Theorem.
I don't even know where to start with this question. Any help would be greatly appreciated.
((I apologize in advance for the format that this is typed in, I am not that great at typing math stuff. (It is supposed to be less than or equal to; and it is the absolute value of x to the power of (7/2))).

2. ## Re: Proof that f is differentiable at 0

??? The statement "If f satisfies $|f(x)|\le |x|^{7/2}$ for all x then f is differentiable at x= 0" cannot be proven because it is obviously untrue. f(x)= |x| satisfies $|f(x)|\le |x|^{7/2}$ for all x, but is not differentiable at x= 0.

3. ## Re: Proof that f is differentiable at 0

Halls - for any positive x with x <1, $|x|\leq |x^{7/2}|$ is false!

Hints:
1. First show f(0)=0.
2. Now show the absolute value of the difference quotient satisfies:
$$0\leq\left|{f(0+h)-f(0)\over h}\right|\leq|h^{5/2}|$$
3. Conclude by the "squeeze" theorem that in fact f'(0)=0.