Question

which graph shows the solution set of $\frac{x - 1}{x - 3}<0$?

Explanation:

Step1: Find critical points

Set the numerator and denominator equal to zero. For $x - 1=0$, $x = 1$; for $x - 3=0$, $x = 3$. These are the critical - points.

Step2: Test intervals

We have three intervals to test: $(-\infty,1)$, $(1,3)$ and $(3,\infty)$.

• For $x=0$ (in the interval $(-\infty,1)$), $\frac{0 - 1}{0 - 3}=\frac{-1}{-3}=\frac{1}{3}>0$.
• For $x = 2$ (in the interval $(1,3)$), $\frac{2-1}{2 - 3}=\frac{1}{-1}=-1<0$.
• For $x = 4$ (in the interval $(3,\infty)$), $\frac{4 - 1}{4 - 3}=\frac{3}{1}=3>0$.

The inequality $\frac{x - 1}{x - 3}<0$ is satisfied when $1

Answer:

The graph with an open - circle at $x = 1$, an open - circle at $x = 3$, and a line segment between $x = 1$ and $x = 3$ on the number - line.