Question

1. directions

drag and drop the correct answer choice to each answer blank.
match each function to its inverse. each answer may be used only once.
$f^{-1}(x) = 3x + 2$ $f^{-1}(x) = \frac{x}{3}$ $f^{-1}(x) = \frac{x + 3}{2}$ $f^{-1}(x) = \frac{x + 2}{3}$

function	inverse
$y = 3x$
$y = 3x - 2$
$y = \frac{x - 2}{3}$

Explanation:

Step1: Find inverse of \( y = 2x - 3 \)

Swap \( x \) and \( y \): \( x = 2y - 3 \). Solve for \( y \): \( 2y = x + 3 \), so \( y=\frac{x + 3}{2} \). Thus, inverse is \( f^{-1}(x)=\frac{x + 3}{2} \).

Step2: Find inverse of \( y = 3x \)

Swap \( x \) and \( y \): \( x = 3y \). Solve for \( y \): \( y=\frac{x}{3} \). Thus, inverse is \( f^{-1}(x)=\frac{x}{3} \).

Step3: Find inverse of \( y = 3x - 2 \)

Swap \( x \) and \( y \): \( x = 3y - 2 \). Solve for \( y \): \( 3y = x + 2 \), so \( y=\frac{x + 2}{3} \). Wait, no: \( 3y=x + 2\Rightarrow y=\frac{x + 2}{3} \)? Wait, no, wait: \( x = 3y - 2\Rightarrow 3y=x + 2\Rightarrow y=\frac{x + 2}{3} \)? Wait, no, let's recheck. Wait, original function \( y = 3x - 2 \). Swap \( x \) and \( y \): \( x = 3y - 2 \). Then \( 3y=x + 2\Rightarrow y=\frac{x + 2}{3} \)? Wait, no, wait the inverse options: one is \( f^{-1}(x)=3x + 2 \). Wait, maybe I made a mistake. Wait, no, let's do \( y = 3x - 2 \). Swap \( x \) and \( y \): \( x = 3y - 2 \). Then \( 3y=x + 2\Rightarrow y=\frac{x + 2}{3} \)? But there's an option \( f^{-1}(x)=3x + 2 \). Wait, maybe another function. Wait, function \( y=\frac{x - 2}{3} \): swap \( x \) and \( y \): \( x=\frac{y - 2}{3}\Rightarrow 3x=y - 2\Rightarrow y = 3x + 2 \). Ah! So step3: function \( y = 3x - 2 \): swap \( x \) and \( y \): \( x = 3y - 2\Rightarrow 3y=x + 2\Rightarrow y=\frac{x + 2}{3} \)? No, wait, no, let's correct step3 and step4.

Wait, step3: function \( y = 3x - 2 \). Swap \( x \) and \( y \): \( x = 3y - 2 \). Then \( 3y = x + 2 \), so \( y=\frac{x + 2}{3} \)? No, that's not matching. Wait, function \( y=\frac{x - 2}{3} \): swap \( x \) and \( y \): \( x=\frac{y - 2}{3} \). Multiply both sides by 3: \( 3x=y - 2 \). Then \( y = 3x + 2 \). So inverse of \( y=\frac{x - 2}{3} \) is \( f^{-1}(x)=3x + 2 \).

Then function \( y = 3x - 2 \): swap \( x \) and \( y \): \( x = 3y - 2 \). Then \( 3y = x + 2 \), so \( y=\frac{x + 2}{3} \). Wait, but the options have \( f^{-1}(x)=\frac{x + 2}{3} \) and \( f^{-1}(x)=3x + 2 \). So let's redo:

Function \( y = 3x - 2 \):

1. Swap \( x \) and \( y \): \( x = 3y - 2 \)
2. Solve for \( y \): \( 3y = x + 2 \Rightarrow y=\frac{x + 2}{3} \). So inverse is \( f^{-1}(x)=\frac{x + 2}{3} \).

Function \( y=\frac{x - 2}{3} \):

1. Swap \( x \) and \( y \): \( x=\frac{y - 2}{3} \)
2. Multiply both sides by 3: \( 3x = y - 2 \)
3. Solve for \( y \): \( y = 3x + 2 \). So inverse is \( f^{-1}(x)=3x + 2 \).

So correcting step3 and step4:

Step3: Find inverse of \( y = 3x - 2 \)

Swap \( x \) and \( y \): \( x = 3y - 2 \). Solve for \( y \): \( 3y = x + 2 \Rightarrow y=\frac{x + 2}{3} \). Thus, inverse is \( f^{-1}(x)=\frac{x + 2}{3} \).

Step4: Find inverse of \( y=\frac{x - 2}{3} \)

Swap \( x \) and \( y \): \( x=\frac{y - 2}{3} \). Multiply by 3: \( 3x = y - 2 \). Solve for \( y \): \( y = 3x + 2 \). Thus, inverse is \( f^{-1}(x)=3x + 2 \).

Answer:

• For \( y = 2x - 3 \): \( f^{-1}(x)=\frac{x + 3}{2} \)
• For \( y = 3x \): \( f^{-1}(x)=\frac{x}{3} \)
• For \( y = 3x - 2 \): \( f^{-1}(x)=\frac{x + 2}{3} \)
• For \( y=\frac{x - 2}{3} \): \( f^{-1}(x)=3x + 2 \)

So the matches are:

Function	Inverse
\( y = 3x \)	\( f^{-1}(x)=\frac{x}{3} \)
\( y = 3x - 2 \)	\( f^{-1}(x)=\frac{x + 2}{3} \)
\( y=\frac{x - 2}{3} \)	\( f^{-1}(x)=3x + 2 \)