Question

For what positive numbers will the cube of a number exceed four times its square?

Short answer

The cube of a number will exceed four times its square for every positive value of x in the set of$\left\{x\right|x>4\}orintervalnotation(4,\infty ).$

Step-by-step solution

Step 1. Given Information

We have to find out for what positive numbers will the cube of a number exceed four times its square.

Step 2. Form an inequality expression

To find the positive numbers, let the number be x.

According to the question,

$$\begin{gathered} x^3>4x^2 \\ {x}^3-4x^2>0 \end{gathered}$$

Step 3. Solve the inequality expression

Let $f\left(x\right)=x^3-4x^2$

Now, let's find the real zeros

$$\begin{gathered} x^3-4x^2=0 \\ {x}^2\left(x-4\right)=0 \\ {x}^2=0andx-4=0 \\ x=0x=4 \end{gathered}$$

Step 4. Check the points

From the real zeros, we get three intervals which are $\left(-\infty ,0\right),\left(0,4\right),\left(4,\infty \right)$.

Let $x=-2$, substitute in $f\left(x\right)=x^3-4x^2$

$$\begin{gathered} f\left(-2\right)={\left(-2\right)}^3-4{\left(-2\right)}^2 \\ f\left(-2\right)=-8-16 \\ f\left(-2\right)=-24 \end{gathered}$$

It is negative.

Let $x=2$, substitute in $f\left(x\right)=x^3-4x^2$

$$\begin{gathered} f\left(2\right)={\left(2\right)}^3-4{\left(2\right)}^2 \\ f\left(2\right)=8-16 \\ f\left(2\right)=-8 \end{gathered}$$

It is negative.

Let $x=6$, substitute in $f\left(x\right)=x^3-4x^2$

$$\begin{gathered} f\left(6\right)={\left(6\right)}^3-4{\left(6\right)}^2 \\ f\left(6\right)=72 \end{gathered}$$

It is positive.

Thus, the set is $\left\{x\right|x>4\}orintervalnotation\left(4,\infty \right)$where the cube of a number will exceed four times its square.