## Calculus: Early Transcendentals (2nd Edition)

For $f(x)=x^{n}$, where n is a positive odd integer, the range will be all real numbers, because one end of the equation will go to -$\infty$, while the other will go to +$\infty$. Therefore, since the function is continuous it will take on every value on integer (-$\infty$,+$\infty$). For $f(x)=x^{n}$, where n is a positive even integer, the range will be all nonnegative real numbers because the equation will approach $\infty$ as x goes to -$\infty$ as well as +$\infty$ and will be zero when x=0, since $0^{anything}=zero$