#### Alfredo Dawlabany

##### New member

- Joined
- Aug 22, 2017

- Messages
- 45

Hi

Is the inverse of a function always a function ?

I asked myself this question because of an exercise that I was solving but couldn't find an answer.

We have \(\displaystyle f(x)=\sqrt{1+sinx}\) defined over \(\displaystyle \left [\frac{-\pi}{2} ,\frac{\pi}{2} \right ]\)

Let's see its curve on the graph and also the one of its inverse:

The inverse's curve doesn't seem to be a function to me (maybe I'm missing some information in my mind).

A function is a map (every x has a unique y-value), while on the inverse's curve some x-values have 2 y-values.

By doing some calculations, I get \(\displaystyle f^{-1}(x)\,=\,\arcsin(x^{2}-1)\)

Its curve (the one I calculated) is like that (the grey one below):

The part defined over negative x of this curve shocked me.

Can someone explain to me how this can happen ?

And why shouldn't the grey curve be like the purple one ?

Is the inverse of a function always a function ?

I asked myself this question because of an exercise that I was solving but couldn't find an answer.

We have \(\displaystyle f(x)=\sqrt{1+sinx}\) defined over \(\displaystyle \left [\frac{-\pi}{2} ,\frac{\pi}{2} \right ]\)

Let's see its curve on the graph and also the one of its inverse:

The inverse's curve doesn't seem to be a function to me (maybe I'm missing some information in my mind).

A function is a map (every x has a unique y-value), while on the inverse's curve some x-values have 2 y-values.

By doing some calculations, I get \(\displaystyle f^{-1}(x)\,=\,\arcsin(x^{2}-1)\)

Its curve (the one I calculated) is like that (the grey one below):

The part defined over negative x of this curve shocked me.

Can someone explain to me how this can happen ?

And why shouldn't the grey curve be like the purple one ?

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