Answer
$\left( -\infty ,-3 \right]\bigcup \left\{ 0 \right\}\bigcup \left[ 2,\infty \right)$ or $\left\{ x\left| x \right.\le -3\text{ or }x=0\text{ or }x\ge 2 \right\}$
Work Step by Step
The inequality is,
${{x}^{4}}+{{x}^{3}}\ge 6{{x}^{2}}$
Rearrange the inequality.
${{x}^{4}}+{{x}^{3}}-6{{x}^{2}}\ge 0$
From the given graph it is clear that the inequality $p\left( x \right)\ge 0$ is possible in the interval $\left( -\infty ,-3 \right]$,$\left\{ 0 \right\}$ and $\left[ 2,\infty \right)$. Because the graph is moving below the $x$-axis in the interval $\left( -3,0 \right)$ and $\left( 0,2 \right)$, it means the value of $p\left( x \right)\le 0$.
Thus, the solution of the inequality ${{x}^{4}}+{{x}^{3}}\ge 6{{x}^{2}}$, from the provided graph, is $\left( -\infty ,-3 \right]\bigcup \left\{ 0 \right\}\bigcup \left[ 2,\infty \right)$ or $\left\{ x\left| x \right.\le -3\text{ or }x=0\text{ or }x\ge 2 \right\}$.
