Question

For what positive numbers will the cube of a number be less than the number?

Short answer

The cube of a number will be less than the number in the set of$\left\{x\right|0<x<1\}.$

Step-by-step solution

Step 1. Given Information 

We have to find out for what positive numbers will the cube of a number be less than the number.

Step 2. Form an inequality expression 

To find the positive numbers, let the number bex.

According to the question,

$$\begin{gathered} x^3<x \\ {x}^3-x<0 \end{gathered}$$

Step 3. Solve the inequality expression 

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

Now, let's find the real zeros

$$\begin{gathered} x^3-x=0 \\ x\left(x^2-1\right)=0 \\ x=0and{x}^2-1=0 \\ \left(x+1\right)\left(x-1\right)=0 \\ x=1,-1 \end{gathered}$$

Step 4. Check the points 

From the real zeros, we get three intervals which are $\left(-\infty ,-1\right),\left(-1,0\right),\left(0,1\right),\left(1,\infty \right).$

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

$$\begin{gathered} f\left(-3\right)={\left(-3\right)}^3-\left(-3\right) \\ f\left(-3\right)=-27+3 \\ f(-3)=-24 \end{gathered}$$

It is negative.

Let $x=-\frac{1}{2}$, substitute in $f\left(x\right)=x^3-x$

$$\begin{gathered} f\left(-\frac{1}{2}\right)={\left(-\frac{1}{2}\right)}^3-\left(-\frac{1}{2}\right) \\ f\left(-\frac{1}{2}\right)=\frac{3}{8} \end{gathered}$$

It is positive.

Let $x=\frac{1}{2}$, substitute in $f\left(x\right)=x^3-x$

$$\begin{gathered} f\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^3-\left(\frac{1}{2}\right) \\ f\left(\frac{1}{2}\right)=-\frac{3}{8} \end{gathered}$$

It is negative.

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

$$\begin{gathered} f\left(3\right)={\left(3\right)}^3-\left(3\right) \\ f\left(3\right)=27-3 \\ f\left(3\right)=24 \end{gathered}$$

It is positive.

Thus, the set is $\left(-\infty ,-1\right)and\left(0,1\right)$because$f\left(x\right)<0$but we need positive numbers the solution will be$\left\{x\right|0<x<1\}.$