Question

What is the inverse function of `h(x)= 3-(x+4)/(x-7)` and how do you evaluate `h^-1(9)`?

Answer 1

`h^(-1)(y)=(7y-25)/(y-2)` where `y!=2`.
`h^(-1)(9)=38/7`

Explanation:

`h(x)= 3-(x+4)/(x-7)` having `x!=7`
I find the common denominator and sum all together:
`h(x)=(3x-21)/(x-7)-(x+4)/(x-7)`
`h(x)=(2x-25)/(x-7)`
Now I simplify making the division of the 2 polynomials and I obtain:
`"quotient"=2, "reminder"=-11`
So I can write the function as:
`h(x)=2(x-7)/(x-7)-11/(x-7)`
`h(x)=2-11/(x-7)`.

Now, the question is how to find the inverse function? Firstly, I try to isolate x:
`(x-7)(h(x)-2)=-11`
`(x-7)=-11/(h(x)-2)`
`x=-11/(h(x)-2)+7`
Therefore we rewrite better the function:
`h^(-1)(y)=-11/(y-2)+7=(7y-14-11)/(y-2)=(7y-25)/(y-2)`.
So we can state that:
`h^(-1)(y)=(7y-25)/(y-2)` where `y!=2`.

If we want to find `h^(-1)(9)`:
`h^(-1)(9)=(7*9-25)/(9-2)=(63-25)/7=38/7`

Answer 2

The inverse function is `h^(-1)(x) = (7x-25)/(x-2)`.
`h^(-1)(x) = 38/7`

Explanation:

Since the original function is not that complex, you can determine its inverse function faster by solving the function for `x` and switching the result for `h(x)`.

`h(x) = 3 - (x+4)/(x-7)`

`h(x) = (3(x-7)- (x+4))/(x-7) = (3x - 21 -x - 4)/(x-7)`

`h(x) = (2x - 25)/(x-7)`

Solve this form of the equation for `x` to get

`h^(x) * (x-7) = 2x-25`

`x * h(x) - 7 * h(x) * 7 = 2x-25`

`x * h(x) - 2x = 7 * h(x) - 25`

`x( h(x) - 2) = 7 * h(x) - 25`

`x = (7 * h(x) - 25)/(h(x) - 2)`

Once you isolate `x`, simply switch `h(x)` for `x` to get the inverse function

`h^(-1)(x) = (7x - 25)/(x-2)`

To evaluate `h^(-1)(9)`, simply substitute `x` with `9` to get

`h^(-1)(9) = (7 * 9 - 25)/(9-2) = 38/7`