Find the domain of $f(x)$ $=$ $\sqrt{1 + ln(1- x)}$

Question

My solution

Since the argument of ln must be greater than zero:

$1 - x > 0$

$ - x > -1$

$ x < 1 $

Then, because $f(x)$ must be a real number, we have:

$1 + ln(1-x) \ge 0$

$ln(1-x) \ge -1$

For the next result I used the follwing property: $e^{ln (x)}$ $=$ $x$

$1-x \ge e^{-1}$

Which simplifies to: $x \le 1 - e^{-1}$

Thus, option B it's the answer. Right? If so, then how can we verify that $- \infty \lt x $?

Answer 1

You need $1+\ln(1-x)\geq 0$ in order to plug it into the square root function. Then, \begin{equation*} 1+\ln(1-x)\geq 0 \iff \ln(1-x)\geq -1 \iff 1-x\geq e^{-1} \end{equation*} which implies that the domain of your function is $\{x\in\mathbb{R} \mid x\leq 1-e^{-1}\}=(-\infty,1-e^{-1}]$. This equivalence of sets is the crux of what was pointed out in the first two comments.

Answer 2

Let $f(x)$ be $\sqrt{1+ln(1-x)}$. Let $$u=1+ln(1-x)$$, so we have $$f(x)=\sqrt{u}$$, and we know that the square root function is only valid if $$u \ge 0$$, so we have $$1+ ln(1-x) \ge 0$$, which becomes $$ln(1-x) \ge -1$$, which means $$1-x \ge \frac1e$$, thus, $$x \le 1 - \frac1e$$, or $$-\infty \lt x \le 1 - \frac1e$$