# Question 9ff5f

Aug 19, 2017

$x \in \left(- 1 , 0\right)$

#### Explanation:

Right from the start, you know that your solution interval cannot include $x = 1$ because that is the one value of $x$ that will make the denominator undefined.

So you know that you need to have $x \ne 1$.

Now, let's separate this compound inequality into two inequalities

$- 1 < \frac{x + 1}{x - 1} \text{ " and " } \frac{x + 1}{x - 1} < 0$

Remember, the solution interval must satisfy both inequalities!

For the first inequality, you know that

$- 1 < \frac{x + 1}{x - 1}$

Now, it is absolutely vital to keep in mind that we can have $2$ possible cases here

$\textcolor{w h i t e}{a}$

• $\underline{x - 1 < 0}$

In this case, the first inequality can be rewritten as

$- 1 \cdot \left(x - 1\right) \textcolor{red}{>} x + 1$

Notice the fact that we must flip the sign of the inequality because we're multiplying on both sides by a negative number, since $x - 1 < 0$.

This simplifies to

$- x + 1 > x + 1$

$- 2 x > 0 \implies x < 0 \text{ } \textcolor{b l u e}{\left(1\right)}$

$\textcolor{w h i t e}{a}$

• $\underline{x - 1 > 0}$

In this case, the first inequality can be written as

$- 1 \cdot \left(x - 1\right) < \left(x + 1\right)$

This simplifies to

$- x + 1 < x + 1$

$- 2 x < 0 \implies x > 0$

However, keep in mind that in this case, we need

$x - 1 > 0 \implies x > 1$

so you can say that

$x > 1 \text{ } \textcolor{b l u e}{\left(2\right)}$

So in order for the first inequality to be true, you need

 overbrace((-oo", " 0))^(color(blue)("from (1)")) " " or " " overbrace((1", " oo))^(color(blue)("from (2)"))" "color(darkorange)("( * )")

$\textcolor{w h i t e}{\frac{a}{a}}$
For the second inequality, you know that

$\frac{x + 1}{x - 1} < 0$

The same approach applies here as well--you have two possible cases to look at.

• $\underline{x - 1 < 0}$

In this case, you have

$x + 1 \textcolor{red}{>} 0 \cdot \left(x - 1\right)$

This simplifies to

$x + 1 > 0 \implies x > - 1$

Since you also need

$x - 1 < 0 \implies x < 1$

you can say that you have

$- 1 < x < 1 \text{ } \textcolor{\mathrm{da} r k g r e e n}{\left(1\right)}$

$\textcolor{w h i t e}{a}$

• $\underline{x - 1 > 0}$

In this case, the second inequality can be rewritten as

$x + 1 < 0 \cdot \left(x - 1\right)$

This simplifies to

$x + 1 < 0 \implies x < - 1$

But since you also need

$x - 1 > 0 \implies x > 1$

you can say that you have

$1 < x < - 1 \implies x \in \left\{\emptyset\right\} \text{ } \textcolor{\mathrm{da} r k g r e e n}{\left(2\right)}$

According to $\textcolor{\mathrm{da} r k g r e e n}{\left(1\right)}$ and $\textcolor{\mathrm{da} r k g r e e n}{\left(2\right)}$, the second inequality is only satisfied by

x in (-1, 1)" "color(darkorange)("(* *)")#

$\textcolor{w h i t e}{\frac{a}{a}}$
Finally, to find the solution interval for the compound inequality, you need to intersect the $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{( * )}}$ and $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(* *)}}$ solution intervals.

You have

$\left(- \infty , 0\right) \text{ " or " " (1, oo)" " and " } \left(- 1 , 1\right)$

the equivalent of

$\left(- \infty , 0\right) \text{ " uu " " (1, oo)" " nn " } \left(- 1 , 1\right)$

which gets you

$x \in \left(- 1 , 0\right)$

To double-check the result, you can either graph the two inequalities separately and intersect their solution intervals, or graph the compound inequality and look at the solution interval.

The graph for $- 1 < \frac{x + 1}{x - 1}$ looks like this The graph for $\frac{x + 1}{x - 1} < 0$ looks like this and the graph for the compound inequality looks like this 