Inverse functions ( 11.c)

[Mrnewbie]

The problem is 11.c) and i ve tried it many times already

This is my attempt at solving 11.c) But i think it is wrong and i dont know where i did wrong

 

[Pedja]

As you have already found: [imath]f \circ g(x)=y(x)=3\left(\frac{x+2}{1-x}\right)^2+2[/imath]
Now substitute [imath]x[/imath] by [imath]y^{-1}(x)[/imath] and [imath]y(x)[/imath] by [imath]x[/imath] . After that plug [imath]x=2[/imath] into equation.

 

[Mrnewbie]

As you have already found: [imath]f \circ g(x)=y(x)=3\left(\frac{x+2}{1-x}\right)^2+2[/imath]
Now substitute [imath]x[/imath] by [imath]y^{-1}(x)[/imath] and [imath]y(x)[/imath] by [imath]x[/imath] . After that plug [imath]x=2[/imath] into equation.

can you elaborate more

 

[Dr.Peterson]

The problem itself is wrong. f o g is not an invertible function!

But what you need to do is to solve your equation for x, which means x must be on only one side, unlike your last line. What you'll find is a quadratic equation in x with y as a parameter (that is, treated as a constant). This will have two solutions, which is my concern.

What Pedja said amounts to saying that you can find the answer without getting a general formula for the inverse, In your work, this would mean letting y=2 before solving for x. This happens to give a single number, so you would not notice that the inverse is not a function.