Evaluating a piecewise function

Text

Let $f(x)=\begin{cases} x+2 & \text{if } x<1,\\[2pt] x^{2} & \text{if } x\ge 1.\end{cases}$

• (a) Compute $f(-3)$, $f(1)$ and $f(2)$.
• (b) Decide whether $f$ is injective.

Solution

(a) Since $-3<1$ use the first branch: $f(-3)=-3+2=-1$. Since $1\ge 1$ and $2\ge 1$ use the second: $f(1)=1^{2}=1$, $f(2)=2^{2}=4$. (b) On the branch $x<1$ the values are $x+2\in(-\infty,3)$; on the branch $x\ge 1$ they are $x^{2}\in[1,+\infty)$. The two branches overlap on $[1,3)$, where the same value comes from two distinct $x$: hence $f$ is not injective. $\boxed{\,f(-3)=-1,\ f(1)=1,\ f(2)=4;\ \text{not injective}\,}$