$f\left(x\right)=\sqrt[3]{x^{4}}$

$f\left(x\right)=\frac{x^2}{x^4+16}$

$f\left(x\right)=2x-4$

$f\left(x\right)=\frac{1}{x\sqrt{x}}$

$f\left(x\right)=\left(4x+1\right)\left(6x+3\right)$

$f\left(x\right)=\frac{1}{\left(1+x\right)^2}$

$f\left(x\right)=x^2-4x$

$f\left(x\right)=x^{\frac{4}{3}}$