How does this work?

[The Blade Dancer]

What does all this even mean? How do you know that 1 - 1/x is the inverse of 1 / 1-x? How do you "put the output back"?

[Legendaryboy991]

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[v4913]

@the-blade-dancer Two functions are inverses if they are exact copies of each other, except with the x and y switched. So for example x = y^2 and y = x^2 are inverses. Taking the y by itself onto the left side, y = sqrt(x) and y = x^2 are inverses. So similarly, y = 1/(1-x) and y = 1-1/x are inverses because, switching the x and y and simplifying will result in the other equation.

One cool property of a function and its inverse is that you can take a number, put it through the function, put the result through the inverse function, and you'll end up with the original number. It's almost like the inverse function reverses all the steps that the original function did. That's what it means by "putting the output back".

[The Blade Dancer]

@v4913 said in How does this work?:

So similarly, y = 1/(1-x) and y = 1-1/x are inverses because, switching the x and y and simplifying will result in the other equation.

I understand this, but I don't see how they have this inverse relationship. If I flip 1/(1-x) I get 1-x/1, which is not the same as 1-1/x.

[professionalbronco]

@v4913 I believe that \(y=\sqrt{x}\) and \(y=x^2\) are inverses only for the interval \(x\in[0,\infty)\).
@The-Blade-Dancer:
Note: the inverse of a function \(f\) does NOT mean \(\frac{1}{f}\). Think of it this way.

Suppose \(f(x)=\frac{1}{1-x}\) and \(g(x)=1-\frac{1}{x}\). The function \(g\) is the inverse of \(f\) when \(f(g(x))\) and \(g(f(x))\) both equal \(x\). For this particular example, we have

$$f(g(x))=f\left(1-\frac{1}{x}\right)=\frac{1}{1-\left(1-\frac{1}{x}\right)}=\frac{1}{\frac{1}{x}}=x $$

and

$$g(f(x))=g\left(\frac{1}{1-x}\right)=1-\frac{1}{\frac{1}{1-x}}=1-(1-x)=x. $$

Hence, the functions \(f(x)=\frac{1}{1-x}\) and \(g(x)=1-\frac{1}{x}\) are inverses.

Basically, finding \(f(g(x))\) and \(g(f(x))\) is "putting the output back", because we have an expression for \(g(x)\) (or \(f(x)\)) and then we put that output into the next function.

(If you are still confused on how I found \(f(g(x))\) and \(g(f(x))\), remember that the order of operations states that you do parentheses first. I hope this helped!)

[Legendaryboy991]

I believe you just copy and paste

[professionalbronco]

@legendaryboy991 me? I didn't copy and paste...

[energizedpanda]

@legendaryboy991 don't assume things there are very orz people in this world

[Legendaryboy991]

ok ok...