# Find f(0) and f'(0)

1. Oct 9, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
If f(x) is a continuous and differentiable function and f(1/n)=0, for all nεN, then
a)f(x)=0 for all x ε (0,1]
b) f(0)=f'(0)=0
c) f'(0)=f"(0)
d)f(0)=0 and f'(0) need not be zero.

2. Relevant equations

3. The attempt at a solution
I would say d but the correct answer is b. Why should I believe that f'(0)=0 if f(0)=0?

2. Oct 9, 2013

### Office_Shredder

Staff Emeritus
If I instead told you that f(1/n) = 1/n, then f(0) = 0 still but the actual values of the sequence probably suggest the function looks a bit different near zero (can you guess what f'(0) will be?). You need to use the actual values of the sequence to show that the derivative is zero.

3. Oct 9, 2013

### UltrafastPED

Clearly f(0) = 0 because the limit as n->infinity is zero ...

Also it is zero at an unlimited number of points near zero; for every finite interval specified, no matter how small, there are an infinite number of zeros in that interval.

But it was given that the function was differentiable as well as continuous; so form the derivative as
f'(0) = lim n-> inf [f(0+1/n) - f(0)]/[1/n] ... but the numerator is zero, and so is f'(0).

Hence (b) is correct.

4. Oct 10, 2013

### HallsofIvy

Staff Emeritus
Another way of looking at it: 1/n is a sequence of points converging to 0. Since f is continuous, f(0)= lim f(1/n)= 0.

As UltrafastPED said, [f(0+ 1/n)- f(0)]/(1/n)= 0. Either the limit, as n goes to 0, does not exist or it is 0. Since we are told that the function is differentiable, the limit exists, so is 0, so f'(0)= 0.