# Inverse of a function

1. Jan 29, 2010

### hangover

If a function is bijective, then its inverse exists. Is there any example that inverse of a function exists but the original function is not bijective?

2. Jan 29, 2010

A function has to be bijective in order to have an inverse.

3. Jan 29, 2010

### elibj123

A function can be locally bijective, so it's inverse exists only in some finite interval.

For example $$x^{2}$$ is not a bijective in any interval containing x=0 (since f'(0)=0) but if you restrict yourself to x>0, then you off course have the inverse
$$f(x)=\sqrt{x}$$ or in x<0 the inverse is $$f(x)=-\sqrt{x}$$.

Last edited: Jan 29, 2010
4. Jan 29, 2010

### HallsofIvy

Another example. f(x)= ex, as a function from R to R, is not surjective, so not and does not have an inverse. In particular, there is no f-1(-1). But if I consider it as a function from R to R+, the positive real numbers, then it is bijective and f-1(x)= ln(x).

A function, from A to B, has an inverse if and only if it is bijective.

5. Jan 29, 2010

### hangover

Thanks a lot! It may be a typo in my textbook.

However, why is x^2 not bijective if we define the domain containing zero(like x=>0)? It is a continous function so isn't it bijective at this interval though being not strictly increasing? Thanks

6. Jan 29, 2010

### Staff: Mentor

f(x) = x2 on [0, infinity) is strictly increasing. If a and b are any two numbers in this interval such that a < b, then f(a) < f(b).

7. Jan 29, 2010

### elibj123

I had a little confusion in defining the inverse sorry.

For $$x\geq 0$$ the inverse of $$y=x^{2}$$ is $$x=f^{-1}(y)=\sqrt{y}$$

For $$x\leq 0$$ the inverse of $$y=x^{2}$$ is $$x=f^{-1}(y)=-\sqrt{y}$$

Notice of course that the inverse is defined only over $$y\geq 0$$, since the range of x^2 is only the non-negative real numbers.

Also, the interval where the function can be define as a bijection may inclue x=0, but only as a boundary point.

Last edited: Jan 29, 2010