pisrationalhahaha
New member
- Joined
- Aug 22, 2017
- Messages
- 46
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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