How to find the domain of $\sqrt{\frac{1}{1-x}-1}$ using algebra?

Question

Textbook answer: [0, 1)

At first, I determined the restriction of the fraction, which is that $x \neq 1$. Lastly, I determined the restriction of the radical.

$\sqrt{\frac{1}{1-x}-1}\geq 0$ $\Rightarrow\frac{1}{1-x}-1\geq 0$ $\Rightarrow \frac{1}{1-x}\geq 1$ $\Rightarrow 1\geq 1-x$ $\Rightarrow x\geq 0$

So why is the domain [0, 1) and not [0, 1) $\cap$ (1, $\infty$)? Does algebra not work for this case or is my algebra/reasoning wrong?

Answer 1

Note that when $1-x <0$, that is, $x >1$, we have to reverse the inequality giving us: $$\frac {1}{1-x}\geq 1 \implies 1 \leq 1-x \implies x \leq 0$$ which is impossible.

Note that when $1-x >0$, that is $x<1$, we get: $$\frac {1}{1-x} \geq 1 \implies 1 \geq 1-x \implies x \geq 0$$

Thus the domain is $$[0,1) $$

We cannot multiply straightaway by $(1-x) $ because we do not know it's sign.

Answer 2

$$ \frac{1}{1-x}-1\geq0 $$ or $$ \frac{x}{1-x}\geq0 $$ or $$ 0\leq x<1. $$

You can not do, which you did because if $1-x$ is negative then the sign of the inequality is changed.

Answer 3

You must take care when multiplying by $1-x$ as you do not know whether it is positive or negative. We will treat these cases separately.

The given function is valid when the interior function to the square root is nonnegative. That is, the points $x$ in the domain satisfy $$\frac{1}{1-x}-1\geq 0\iff \frac{1}{1-x}\geq 1\iff \begin{cases}1\geq 1-x\qquad 1-x> 0\\1\leq 1-x\qquad 1-x< 0\end{cases}\iff\begin{cases}x\geq 0\qquad x< 1\\x\leq 0\qquad x>1\end{cases}$$ Of course, the second case (i.e. $x\leq 0$ and $x>1$) is impossible. Hence, the domain is the set of points satisfying $0\leq x<1$.

Answer 4

The big mistake in your solution is that you cleared fractions in the non-linear inequality $\frac{1}{1-x} \geq 1$ (this inequality is non-linear because $x$ appears in the denominator of a fraction). You should never clear fractions in a non-linear inequality. Instead, to solve this type of inequality, you make one side $0$, write the non-zero side as a single fraction, and then use a sign chart.

So, let's subtract $1$ from both sides to make one side $0$. The inequality becomes $$\frac{1}{1-x} - 1 \geq 0.$$ Now that one side is $0$, let's make the non-zero side into one single fraction. To do this, multiply the term $1$ by $\frac{1-x}{1-x}$ to get: $$\frac{1}{1-x} - \frac{1-x}{1-x} \geq 0.$$ Now just subtract the fractions to get $$ \frac{1 - (1-x)}{1-x} \geq 0, $$ i.e., $$\frac{x}{1-x} \geq 0. $$ Now, hopefully you remember how to solve using a sign chart. You first find the critical points, i.e., the $x$-values that make the numerator and denominator $0$. $x = 0$ makes the numerator $0$, and $x = 1$ makes the denominator $0$, so $0$ and $1$ are our critical points. Draw a number line, mark $0$ and $1$ on it, and plug test points from each region of the number line into $\frac{x}{1-x}$ to figure out the sign.

Since we want $\frac{x}{1-x} \geq 0$, the region we will choose is the positive region, which turns out to be $[0,1)$. Notice that we include the critical point $0$ since it makes the numerator $0$, which makes the whole fraction $0$. But we don't include $1$, even though it makes the denominator $0$, because this makes the whole fraction undefined.