Q.Show that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3x-7$ is bijective, and hence find its inverse.

A function is bijective when it is both one-one and onto; once that is established, its inverse is found by solving $y=f(x)$ for $x$.

One-one. Suppose $f(x_1)=f(x_2)$:

So distinct inputs give distinct outputs — $f$ is one-one.

Onto. Take any $y\in\mathbb{R}$ and solve $f(x)=y$:

Since $y$ is real, $\dfrac{y+7}{3}$ is a real number lying in the domain, so every codomain element has a pre-image — $f$ is onto. Being both one-one and onto, $f$ is bijective.

Inverse. From $y=3x-7$ we solved $x=\dfrac{y+7}{3}$. Relabelling the variable,

Verification (dual check):

$f^{-1}\big(f(x)\big)=\frac{(3x-7)+7}{3}=\frac{3x}{3}=x.$ …

A frequent slip is dividing only part of the numerator by $3$, writing $x=y+\frac{7}{3}$ instead of $x=\frac{y+7}{3}$ — the entire right-hand side must be divided. Another is c …