Chapter 2.8, Problem 67E

Solution

To solve: The problem without using a calculator.

The solution set is equal to $\left(0,1\right)$ .

Given information:

Consider the given inequality.

  $x^2<x$

Calculation:

Rewrite the given inequality:

  $\begin{array}{c}{x}^2-x<0\\ x\left(x-1\right)<0\end{array}$

To set numerator and the denominator equal to zero. These are the values that make the quotient zero or undefined and solve for $x$ :

  $\begin{array}{c}x\left(x-1\right)=0\\ x=0,1\end{array}$

So, the boundary points are $0,1$ .

Now, locate these boundary points on the number line (or sign chart) and dividing the number line into intervals and check the function $f\left(x\right)$ is positive or negative at each interval below:

  

The boundary points divide the number line into three test intervals. Excluding the boundary points (because of the given less than sign), the intervals are below:

  $\left(-\infty ,0\right),\left(0,1\right),\left(1,\infty \right)$

Now, take one representative number within each test interval and substitute that number into the inequality below:

  $\begin{array}{l}\left(-\infty ,0\right),\left(0,1\right)\left(1,\infty \right)\\ {\left(-1\right)}^2<-1{\left(\frac{1}{2}\right)}^2<\frac{1}{2}{\left(2\right)}^2<2\\ 1<-1\text{False }\frac{1}{4}<\frac{1}{2}\text{True}4<2\text{False}\end{array}$

Therefore, the interval $\left(0,1\right)$ belongs to the solution set and the intervals $\left(-\infty ,0\right)$ and $\left(1,\infty \right)$ does not belong to the solution set.

Hence, the solution set is $\left(0,1\right)$ .