Calculus 8th Edition

This function is not one-to-one since $f(x)=0$ for both $x=2$ and $x=-2$ ($2^4-16=16-16=0$ and $(-2)^4-16=16-16=0$) so two different $x$-es: $x=-2$ and $x=2$ both map to the same value $y=0$ which means that there is no one-to-one correspondence between values of the argument and values of the function.