1. ## Differentiability of functions

Determine whether the following two functions are differentiable at the given value
a) f(x) = xsin(1/x), f(0)=0: at given value 0
b)
f(x) = 1-x if x<1
x(1-x) if x>= 1 at given value 1

I think a is differentiable. I found the difference quotient and found that as
h->0 the limit of the quotient -> 0 also.

But I can't seem to get started on b!

2. for b

find the left and right limits

$\displaystyle \lim_{x\to 1^-} \frac{f(x)-f(1)}{x-1}$

$\displaystyle \lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1}$

See they are equal so it is differenciable on 1

3. is it really that simple! I've been looking at that for ages!!!

4. Do you understand that this means you were wrong about (a) being differentiable?

The derivative at x= 0 would be given by $\displaystyle \lim_{h\to 0}\frac{h sin(1/h)- 0}{h}= \lim_{h\to 0} sin(1/h)$ and that limit does NOT exist.

It is also true that, while derivatives are not necessarily continuous, they do have the "intermediate value property". That means that if a function is differentiable, the limits of the two derivatives on each side must be equal.

For (b), the f'= -1 for x< 1 and f'= 1- 2x for x< 1. The limits as x goes to 1 are the same so the function is differentiable at x= 1.