Question

`f(x)=(5-x)/(x-3)` what would be the inverse function `f^-1(x)` ?

Answer 1

`f^-1(x)=(5+3x)/(x+1)`

Explanation:

Make the function in terms of `y` and equate to `x`:
`x=(5-y)/(y-3)`

`x(y-3)=5-y`

`xy+y=5+3x`

`y(x+1)=5+3x`

`y=(5+3x)/(x+1)`

Answer 2

`f^(-1)(x)=y=(3x+5)/(x+1)`

Full detail of calculation given. In practise you would use shortcuts and use less lines.

Explanation:

Basically the method is that you change the equation into the form of `x="something to do with "y` then where ever there is an `x` you write `y` and wherever there is a `y` you write `x`. So your target is to have just one `x` and for it to be on the LHS of the = and everything else on its right./ Then you do the switch.

Set `f(x)=y=(5-x)/(x-3)`

Multiply both sides by `(x-3)`

`y(x-3)=(5-x)`

Multiply out the LHS bracket

`yx-3y=5-x`

Add `x` to both sides

`yx+x-3y=5`

Add `3y` to both sides

`yx+x=5+3y`

Factor out the `x` on the LHS

`x(y+1)=5+3y`

Divide both sides by `(y_1)`

`x=(3y+5)/(y+1)`

NOW YOU DO THE SWITCH

`f^(-1)(x)=y=(3x+5)/(x+1)`

This should be a reflection about the line `y=x`

`color(blue)("Original equation graph"color(magenta)(" + reflected (inverse function)"`

`color(red)("+ line of reflection.")`