QuestionDirections: Find the inverse of the following functions. Be sure to show all work and use proper notation.
8. $f(x)=\frac{x-3}{x+2}$
9. $h(x)=\frac{4 x-1}{2 x+3}$ $\begin{array}{l} x=\frac{y-3}{y+2} \\ x(y+2)=y-3 \\ x y+2 x=y-3 \\ 2 x-3=y-x y \\ 2 x-3=y(1-x) \\ y=\frac{2 x-3}{1-x} \end{array}$ $x=\frac{4 y-1}{2 y+3}$ $\begin{array}{l} x(2 y+3)=4 y-1 \\ 2 x y+3 x=4 y-1 \\ 3 x+1=4 y-2 x y \\ 3 x+1=y(4-2 x) \end{array}$ $h(x)^{-1}=\sqrt{y=\frac{3 x+1}{4-2 x}}$
10. $y=\frac{x+4}{x-5}$
11. $g(x)=\frac{2 x+1}{3 x+7}$

Studdy Solution

STEP 1

1. We are given functions for which we need to find the inverse.
2. We need to show all work and use proper notation.
3. The process involves algebraic manipulation to express $y$ in terms of $x$ .

STEP 2

1. Find the inverse of $f(x) = \frac{x-3}{x+2}$ .
2. Find the inverse of $h(x) = \frac{4x-1}{2x+3}$ .
3. Find the inverse of $y = \frac{x+4}{x-5}$ .
4. Find the inverse of $g(x) = \frac{2x+1}{3x+7}$ .

STEP 3

Given $f(x) = \frac{x-3}{x+2}$ , we want to find its inverse.
1. Replace $f(x)$ with $y$ :
$y = \frac{x-3}{x+2}$
2. Swap $x$ and $y$ :
$x = \frac{y-3}{y+2}$
3. Solve for $y$ :
$x(y+2) = y-3$
$xy + 2x = y - 3$
$xy - y = -3 - 2x$
$y(x - 1) = -3 - 2x$
$y = \frac{-3 - 2x}{x - 1}$
Thus, the inverse function is:
$f^{-1}(x) = \frac{-3 - 2x}{x - 1}$

STEP 4

Given $h(x) = \frac{4x-1}{2x+3}$ , we want to find its inverse.
1. Replace $h(x)$ with $y$ :
$y = \frac{4x-1}{2x+3}$
2. Swap $x$ and $y$ :
$x = \frac{4y-1}{2y+3}$
3. Solve for $y$ :
$x(2y+3) = 4y - 1$
$2xy + 3x = 4y - 1$
$2xy - 4y = -3x - 1$
$y(2x - 4) = -3x - 1$
$y = \frac{-3x - 1}{2x - 4}$
Thus, the inverse function is:
$h^{-1}(x) = \frac{-3x - 1}{2x - 4}$

STEP 5

Given $y = \frac{x+4}{x-5}$ , we want to find its inverse.
1. Swap $x$ and $y$ :
$x = \frac{y+4}{y-5}$
2. Solve for $y$ :
$x(y-5) = y + 4$
$xy - 5x = y + 4$
$xy - y = 5x + 4$
$y(x - 1) = 5x + 4$
$y = \frac{5x + 4}{x - 1}$
Thus, the inverse function is:
$y^{-1}(x) = \frac{5x + 4}{x - 1}$

STEP 6

Given $g(x) = \frac{2x+1}{3x+7}$ , we want to find its inverse.
1. Replace $g(x)$ with $y$ :
$y = \frac{2x+1}{3x+7}$
2. Swap $x$ and $y$ :
$x = \frac{2y+1}{3y+7}$
3. Solve for $y$ :
$x(3y+7) = 2y + 1$
$3xy + 7x = 2y + 1$
$3xy - 2y = 1 - 7x$
$y(3x - 2) = 1 - 7x$
$y = \frac{1 - 7x}{3x - 2}$
Thus, the inverse function is:
$g^{-1}(x) = \frac{1 - 7x}{3x - 2}$
The inverse functions are:
1. $f^{-1}(x) = \frac{-3 - 2x}{x - 1}$
2. $h^{-1}(x) = \frac{-3x - 1}{2x - 4}$
3. $y^{-1}(x) = \frac{5x + 4}{x - 1}$
4. $g^{-1}(x) = \frac{1 - 7x}{3x - 2}$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms