Chapter 4.9, Problem 31WE

To determine

To find and graph the solution of given inequality

Answer to Problem 31WE

The solution of the given inequality is $\left\{x:x=0\text{ or }x=2\right\}$ and the graph is represented.

Explanation of Solution

Given information:

The inequality is $x^2\left(x^2+4\right)\le 4x^3$

Assume $x^2\left(x^2+4\right)\le 4x^3$ as $x^2\left(x^2+4\right)=4x^3$ and solve it by factoring

The solution of the given inequality can be obtained as,

  $\begin{array}{l}{x}^2\left(x^2+4\right)\le 4x^3\left[\text{Original equation}\right]\\ x^2\left(x^2+4\right)=4x^3\\ x^2\left(x^2+4\right)-4x^3=0\\ x^2\left[\left(x^2+4\right)-4x\right]=0\\ x^2\left(x^2-4x+4\right)=0\\ x^2{\left(x-2\right)}^2=0\\ x=0,x=2\end{array}$

If $x^2\left(x^2+4\right)=4x^3$ means $x^2{\left(x-2\right)}^2=0$ then $x^2\left(x^2+4\right)\le 4x^3$ means $x^2{\left(x-2\right)}^2\le 0$

Since $x^2{\left(x-2\right)}^2\ge 0$ for all real values of $x$ and from above it can be said that $x^2{\left(x-2\right)}^2=0$ when $x=0\text{ or }x=2$ .

Therefore for any real values of $x$ that is neither 0 nor 2 then $x^2{\left(x-2\right)}^2>0$ . However, the inequality is $x^2{\left(x-2\right)}^2\le 0$ . This can only be true when $x^2{\left(x-2\right)}^2=0$ .

Notice that $x^2{\left(x-2\right)}^2=0$ is a product of two squares, so it is always greater than 0. The only time it’s not greater than 0 is when it is exactly 0.

Therefore, the solution set of given inequality is $\left\{x:x=0\text{ or }x=2\right\}$

Now, the graph of the solution set can be represented as,