1. Show it's twice differentiable

f(x) = 0 x=0
e^(-1/x^2) x/=0

how?

to find
f''(0) = limit( (f'(x)-f'(0))/x,x, goes to 0), but we dont know f'(0)?

2. First thing first, do you know to find the derivative? To be differential at a point, it must be continuous, so we need to make sure the limit as f approaches zero is equal to zero. We also need to find the limit as f' approaches zero, and if that equals zero, then we are good to go.

3. Originally Posted by colby2152
First thing first, do you know to find the derivative? To be differential at a point, it must be continuous, so we need to make sure the limit as f approaches zero is equal to zero. We also need to find the limit as f' approaches zero, and if that equals zero, then we are good to go.

sure, i show it's continuous
it's differentiable elsewhere. so need to show when x=0, its dirivative exists
i show it by limit( (f(0+h)-f(0))/h),h goes to zero), this works.
then to show it's twice diff, i used the same method, but, we dont know f'(0), do we?

4. Originally Posted by szpengchao
sure, i show it's continuous
it's differentiable elsewhere. so need to show when x=0, its dirivative exists
i show it by limit( (f(0+h)-f(0))/h),h goes to zero), this works.
then to show it's twice diff, i used the same method, but, we dont know f'(0), do we?
The derivative, f'(x) is equal to:

$f'(x) = lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

To see if it is continuous at zero, take the limit as f'(x) approaches zero. It must be equal to zero.