# Inverse function

1. Apr 24, 2010

### thereddevils

The condition for the inverse function, f^(-1) to happen is function , f is one-one .

S0 consider this function , f(x)=x^2-5 , which is NOT a one-one function , and

f^(-1)=y

x=f(y)

x=y^2-5

y^2=x+5

$$f^{-1}(x)=\pm\sqrt{x+5}$$

Seems that the inverse function of f exists without satisfying that condition .

2. Apr 24, 2010

### Cyosis

That inverse function you found is not even a function. For every x there are two values of y. For it to be a function there can only be one value of y for every x.

Secondly f(x) in general is not a function, but just a number in its codomain. Your function f is given by $f:A\rightarrow B, f(x)=x^2-5$. With A its domain and B its codomain. Depending on A and B f can have an inverse.

example:

$$f:[0,1] \rightarrow [-5,-4]; f(x)=x^2-5$$

this function has an inverse.

$$f:[-1,1] \rightarrow [-5,-4]; f(x)=x^2-5$$

this one does not.

3. Apr 24, 2010

### thereddevils

thanks ! I just realised it can have inverse when its broken into half .