Question

How do you find the inverse of `y=(3x-7)/(x+9)`?

Answer 1

`f^-1(x)=(9x+7)/(3-x)`

Explanation:

let, `y=f(x)=(3x-7)/(x+9)` then we have
`x=f^-1(y)`
`rArry=(3x-7)/(x+9)rArry(x+9)=3x-7rArrxy+9y=3x-7`
`rArr9y+7=x(3-y)rArrx=(9y+7)/(3-y)`
since `x=f^-1(y)rArrf^-1(y)=(9y+7)/(3-y)rArrf^-1(x)=(9x+7)/(3-x)`

Answer 2

`f(x)^-1= (-9x-7)/(3+x)`

Explanation:

The inverse of a function switches the imput value and the output value. One easy way to solve inverse functions is by simply switching where the `x's and y's are`. So...
`f(x) = (3x-7)/(x+9)` turns into `x = (3y-7)/(y+9)`
Then from here on it is basic algebra.
`x = (3y-7)/(y+9)`
`x*(y+9) =( 3y-7)`
`xy+9x = 3y -7`
`3y+xy = -9x-7`
`y(3+x)=-9x-7`
`f(x)^-1= (-9x-7)/(3+x)`

If you need any more of an explanation, I will add them in