# Is the inverse of a function always a function ?

#### Alfredo Dawlabany

##### New member
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 ?

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#### stapel

##### Super Moderator
Staff member
Is the inverse of a function always a function?
Obviously not! Easy case: a parabola. Its inverse is a sideways parabola, which is not a function. To be stated in functional terms, it would have to be expressed as two square-root functions.

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:

View attachment 9096

The inverse's curve doesn't seem to be a function to me (maybe I'm missing some information in my mind).
No, it is not. But you're inverting the entire original function (which you did not restrict to the stated domain), rather than a portion. This is why the arcsine function is very carefully defined on only a very limited domain.

#### Alfredo Dawlabany

##### New member
Obviously not! Easy case: a parabola. Its inverse is a sideways parabola, which is not a function. To be stated in functional terms, it would have to be expressed as two square-root functions.

No, it is not. But you're inverting the entire original function (which you did not restrict to the stated domain), rather than a portion. This is why the arcsine function is very carefully defined on only a very limited domain.
Thanks, that makes sense. I don't know why I didn't think about the parabola and it's inverse which is a conic. It's more clear now.