Find the values of ' $x$ ' for which the rate of change of $\frac{x^4}{4}+\frac{x^3}{3}-x$ is more than the rate of change of $\frac{x^4}{4}$.

Given: $f\left(x\right)=\frac{x^4}{4}+\frac{x^3}{3}-x$, $g\left(x\right)=\frac{x^4}{4}$.  

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

The rate of change of $f\left(x\right)$ is given by:

$f^{\text{'}}\left(x\right)=x^3+x^2-1.$

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

The rate of change of $g\left(x\right)$ is given by: 

$g^{\text{'}}\left(x\right)=x^3$.

Considering the problem statement,

$$\begin{gathered} f^{\text{'}}\left(x\right)>g^{\text{'}}\left(x\right) \\ \Rightarrow x^3+x^2-1>x^3 \\ \Rightarrow x^2-1>0 \end{gathered}$$

$\Rightarrow (x-1)(x+1)>0$

Evaluating this inequality on the number line, we get

$\Rightarrow x\in \left(-\infty ,-1\right)\cup \left(1,\infty \right).$

Therefore,

The values of $x$ for which the rate of change of $\frac{x^4}{4}+\frac{x^3}{3}-x$ is more than that of $\frac{x^4}{4}$ are found to be $x\in (-\infty ,-1)\cup (1,\infty )$.

Solution is $\left(4\right).$