Given, $f\left(x\right)=x^3+4x+5$
Now, $f^{\prime }\left(x\right)=3x^2+4>0,\forall x\in R$
$\therefore f\left(x\right)$ is strictly increasing function
$\therefore$ It is one-one
Clearly, $f\left(x\right)$ is a continuous function and also increasing on $R$
$\underset{\mathrm{x}\to -\infty }{\lim }f\left(x\right)=-\infty$ and $\underset{\mathrm{x}\to \infty }{\lim }f\left(x\right)=\infty$
$\therefore f\left(x\right)$ takes every value between $-\infty$ and $\infty .$
Thus, $f\left(x\right)$ is an onto function.