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

[skyturnred]

Homework Statement

Here is the question directly from the book:

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

Homework Equations

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..

Thank you so much for your help in advance!

 

[micromass]

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.

 

[skyturnred]

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

f(x)={(3/2)x2/3+c1 if x>0(3/2)x2/3+c2 if x<0

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

 

[micromass]

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

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!

 

[skyturnred]

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!

OK, thanks again! You helped tremendously.