Question

The function `g(x)=(6x-8)/(5-x)` is one-to-one. How do you find the inverse of `g`? What is the domain and range of `g` and `g^(-1)`?

Answer 1

The inverse of a function is found by switching the x and y values.

Explanation:

Before we perform the calculation to find `g^(-1)(x)`, let's state the domain and range of `g(x)`

First, read the part of my answer about asymptotes at the bottom of the page. Following that, you'll notice that there will be a horizontal asymptote at `y = 6/-1 = -6`, so the range will be `y !=-6`, and there will be a vertical asymptote at `x = 5`, making the domain `x !=5`

`y = (6x - 8)/(5 - x)`

`x = (6y - 8)/(5 - y)`

`x(5 - y) = 6y - 8`

`5x - xy - 6y + 8 = 0`

`5x + 8 = xy + 6y`

`5x + 8 = y(x + 6)`

`(5x + 8)/(x + 6) = y`

`g^(-1)(x) = (5x + 8)/(x + 6)`

Now, for the domain and range of `g^(-1)`:

Let's look at the rules for vertical, horizontal and oblique asymptotes, since this is a rational function.

Vertical asymptotes:

Vertical asymptotes occur when the denominator is 0, because division by 0 in mathematics is undefined.

The vertical asymptotes of rational function `y = (a + b)/(x + n)` can be found by setting the denominator to 0 and solving for x.

`x + 6 = 0`

`x = -6`

Therefore, there is a vertical asymptote at `x = -6`

Hence, our domain is `x != -6`

Horizontal asymptotes:

The placement of horizontal asymptotes will depend on the degree of the function. The degree of a function is the highest exponent in the equation. For example, in the polynomial function `y = 3x^4 + 2x^3 - 5x^2 + 9x - 1`, the degree is 4, since the highest exponent is a fourth power.

Back to the problem at hand, here are the scenarios when we will have a horizontal asymptote:

`•` The degree of the denominator is higher than that of the numerator. In this case, there will be a horizontal asymptote at `x = 0`

`•` The degrees of the denominator and the numerator are equal. In this case, the horizontal asymptote will be at the ratio of the coefficients of the highest degree. For example, in `ƒ(x) = (ax + b)/(nx - c)`, there will be an asymptote at `x = a/n`, a and n being any real numbers, `n !=0`.

Looking at our problem, we see that the highest degree in both the numerator and the denominator is 1 (`x^1`).

`y = (5x + 8)/(x + 6)`

The coefficients of x (the highest degrees) are 5 in the numerator and 1 in the denominator.

Therefore, we will have a horizontal asymptote at `y = 5`

Hence, we can deduce that our range will be `y != 5`

Oblique asymptotes:

Oblique asymptotes exist only when the degree of the numerator is higher than that of the denominator. This is obviously not the case in this situation. Since the answer is getting pretty long, I'm not going to get into the workings of this special type of asymptote.

Hopefully this helps!