College Algebra 7th Edition

We are given: $r(x)=2+\sqrt{x+3}$ This function is one-to-one because it passes the horizontal line test (it is a shifted square root function). We can show this algebraically. We start with the assumption: $x_1 \neq x_2$ $x_1+3\neq x_2+3$ $\sqrt{x_1+3}\neq \sqrt{x_2+3}$ $2+\sqrt{x_1+3}\neq 2+\sqrt{x_2+3}$ Since square rooting a unique value produces a unique value. Thus the function will not have the same $y$ value for two different $x$ values. It is therefore one-to-one.