## College Algebra 7th Edition

$h(x)$ is one-to-one.
The function $h(x)=x^3+8$ is one-to-one because it passes the horizontal line test (a shifted $x^3$ graph). We can show this algebraically: We start with the assumption that: $x_1\ne x_2$ Then: ${x_1}^3\ne {x_2}^3$ (Since cubing a unique number results in a unique value.) ${x_1}^3+8\ne {x_2}^3+8$ So the function would not have the same $y$ value for two different $x$ values (one-to-one).