Question

Solve the non-linear inequality. Express the solution using interval notation and graph the solution set.

$\frac{x}{x+1}>3x$

Short answer

The interval notation is $\left(-\infty ,-1\right)\cup \left(-\frac{2}{3},0\right)$.

The graph of the solution is:

Step-by-step solution

Step 1. Solve the equation.

$\begin{array}{c}\frac{x}{x+1}>3x\\ \frac{x}{x+1}-3x>0\\ \frac{x-3x^2-3x}{x+1}>0\\ \frac{-2x-3x^2}{x+1}>0\\ \frac{-x\left(3x+2\right)}{x+1}>0\\ \frac{x\left(3x+2\right)}{x+1}<0\end{array}$

Set the numerator and denominator equal to zero.

$\begin{array}{c}3x+2=0\Rightarrow x=-\frac{2}{3}\\ x+1=0\Rightarrow x=-1\\ x=0\end{array}$

The zeros are $x=-1,-\frac{2}{3},0$.

The zeros divide the real line into four intervals, $(-\infty ,-1),\left(-1,-\frac{2}{3}\right),\left(-\frac{2}{3},0\right),(0,\infty )$

Step 2. Determining the sign of each factor.

To determine the sign of each factor on each of the intervals that we found, we use test values. We choose a number inside each interval and check the sign of each factor.

Step 3. Making the table.

The value of inequality should be less than zero.

So, the required solution set is $\left(-\infty ,-1\right)\cup \left(-\frac{2}{3},0\right)$.

Step 4. Graph the solution.

The required graph is