Q.Show that the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x - 2$ is one-one and onto.

One-one: suppose $f(x_1) = f(x_2)$. Then $3x_1-2 = 3x_2-2 \Rightarrow 3x_1=3x_2 \Rightarrow x_1=x_2$. So $f$ is one-one.

Onto: let $y \in \mathbb{R}$ be arbitrary. Solve $3x-2=y \Rightarrow x = \dfrac{y+2}{3}$, which is a real number for every real $y$, and $f\left(\dfrac{y+2}{3}\right) = 3\cdot\dfrac{y+2}{3}-2 = y$. So every $y \in \mathbb{R}$ has a preimage, and $f$ …

Assume $f(x_1)=f(x_2)$ and solve for $x_1=x_2$ to prove injectivity; take arbitrary $y$, solve $f(x)=y$ for $x$, …

• Forgetting to confirm the solved $x=\frac{y+2}{3}$ actually lies in the stated domain $\mathbb{R}$. …