Q.(Case study, continued.) Let $R=\{(L_1,L_2): L_1 \parallel L_2 \text{ and } L_1: y=x-4\}$, then which of the following can be taken as $L_2$?

Invert $f(x)=3x-5$ by solving for $x$: $f^{-1}(x)=\tfrac13(x+5)$.

Set $y=f(x)=3x-5$. To find the inverse, express $x$ in terms of $y$:

$y=3x-5\Rightarrow 3x=y+5\Rightarrow x=\dfrac{y+5}{3}$.

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Solve $y = 3x-4$ for $x$: $x = \frac{y+4}{3}$, so $f^{-1}(x) = \frac{x+4}{3}$.

Let $y = f(x) = 3x - 4$. Solve for $x$:

$y + 4 = 3x \implies x = \frac{y+4}{3}.$ …

$f(x)=x-4$ is onto $\mathbb R$, so its range is $\mathbb R$.

For $f:\mathbb R\to\mathbb R$, $f(x)=x-4$, and any $y\in\mathbb R$ we can find $x=y+4\in\mathbb R$ with $f(x)=y$. Thus ever …

A line parallel to $y=x-4$ must have slope $1$; only option (a) has slope $1$.

$L_1: y=x-4$ has slope $1$.

Check each option's slope:

1. $2x-2y+5=0 \Rightarrow 2y=2x+5 \Rightarrow y=x+\dfrac52$. Slope $=1$. Parallel to $L_1$.
2. $2x+y=5 \Rightarrow y=-2x+5$. Slope $=-2$. Not parallel. …

$f^{-1}(x) = \dfrac{1}{3}(x+4)$.

Let $y=f(x)=3x-4$. To find the inverse, solve for $x$ in terms of $y$:

$y = 3x-4 \implies 3x = y+4 \implies x = \dfrac{y+4}{3}.$

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$(2*3)*4=18$.

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