# Finding the particular anti derivative when given f'(x) and TWO f(value)s

1. Nov 26, 2011

### skyturnred

1. The problem statement, all variables and given/known data

Here is the question directly from the book:

f'(x)=x^(-1/3), f(1)=1, f(-1)=-1, find f(x).

2. Relevant equations

3. The attempt at a solution

So there are about ten of these types, and I am very confused. I am confused because if you solve for f(x) with f(a) you get a different value for c than if you solved for f(b).

So finding f(x) I get f(x)=(3/2)x^(2/3) + c
If I make f(x)=1 and x=1, I get the value of c to be -1/2
But if I make f(x)=-1 and x=-1, I get the value of c to be -5/2..

2. Nov 26, 2011

### micromass

Staff Emeritus
Notice that f(0) is not defined. So you actually have two branches. So your function can be described as

$$f(x)=\left\{\begin{array}{c}(3/2)x^{2/3} + c_1 ~\text{if}~x>0\\ (3/2)x^{2/3} + c_2 ~\text{if}~x<0 \end{array}\right.$$

3. Nov 26, 2011

### skyturnred

So whenever f'(0)=DNE, the function f(x) can be written as a piecewise function?

4. Nov 26, 2011

### micromass

Staff Emeritus
Indeed!!

Do notice that you proved that the anti-derivative of a function is unique up to a constant. But you only proved that for functions that exists everywhere on an interval. If a function does not exist in a point, then the result isn't true anymore!!

5. Nov 27, 2011

### skyturnred

OK, thanks again! You helped tremendously.