$f\left(x\right)=x^3+x^2+4x+\sin x$

$\Rightarrow f^{\text{'}}\left(x\right)=3x^2+2x+4+\cos x$

$f^{\text{'}}\left(x\right)=3\left[{\left(x+\frac{1}{3}\right)}^2+\frac{11}{9}\right]-\left(-\cos x\right)>0$ $\mathrm{as}\mathrm{ }3\mathrm{ }{\left[{\left(x+\frac{1}{3}\right)}^2+\frac{11}{9}\right]}_{\min }=\frac{11}{3}$

and $-\cos x$ has the maximum value $1.$

$\Rightarrow f\left(x\right)$ is strictly increasing and hence it is one-one

Also, $\underset{x\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)\Rightarrow \mathrm{\infty }$ and $\underset{x\to -\mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)\Rightarrow -\mathrm{\infty }.$

Thus, the range of $f\left(x\right)$ is $R,$ hence it is onto.