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Q: Find the range and domain of the function $$f(x) = \sqrt{1-e^{x+2}}?$$

I've found the domain, which is $x \le -2$ by solving the inequality $1-e^{x+2} \ge 0$.

I've tried to find the range by taking the inverse of $f$, which gives me $f^{-1} = \ln(1-x^2)-2$. Then, since for $\ln(1-x^2)$ to be defined, $1-x^2>0$, so solving this inequality gives the interval $x \in (-1,1)$, which I thought is the range of $f$. However, graphing it out on desmos shows that the range is only $[0,1)$. What am I doing wrong?

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    $\begingroup$ $x\leq 2$ is not the domain. $\endgroup$ Commented Sep 10, 2021 at 1:34
  • $\begingroup$ Corrected. Thanks. $\endgroup$
    – John Fei
    Commented Sep 10, 2021 at 1:38
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    $\begingroup$ because values of $x$ in the domain of $f^{-1}$ have to live inside of $[0,1)$. even though $x$ can be in $(-1,1)$ for the given formula of $f^{-1}$, they must, by necessity, live inside of $[0,1)$ since $\text{im}(f)$ is bounded below by $0$ and above by $1$. $\endgroup$
    – C Squared
    Commented Sep 10, 2021 at 1:45
  • $\begingroup$ @C Squared Thanks again! $\endgroup$
    – John Fei
    Commented Sep 10, 2021 at 2:02
  • $\begingroup$ Just to add some intuition as to why this is occurring: whenever solving for a variable in an equation, if you square both both sides, you may get extraneous solutions. Similarly, if you square both sides of an inequality, you may get extraneous interval solutions. $\endgroup$
    – Bonnaduck
    Commented Sep 13, 2021 at 20:40

2 Answers 2

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$\sqrt{1-e^{x+2}} $ is always $\ge0$ as we have a square root function.

The minimum value it can achieve is $0$ when $ e^{x+2} =1$ or $x=-2$ and the maximum it can achieve is $1$, when $e^{x+2} = 0$ or $x\to-\infty$.

So the range is $[0,1)$.

The mistake in your process is that first you've let,

$x = \sqrt{1-e^{f^{-1}(x)+2}} $. For this to hold we need $x\ge 0$(as we have the square root). If you use this fact in your solution, you'll get the required range $[0,1)$.

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Well since \begin{eqnarray*} f: \mathbf{R}&\longrightarrow& \mathbf{R}\\ x&\longmapsto& f(x)=\sqrt{1-e^{x+2}}. \end{eqnarray*} so the domain of $f$ is given by $${\rm Dom}(f)=\{x\in \mathbb{R}:1-x^{x+2}\geqslant 0\}=]-\infty; -2] $$ and then the image of $f$ is given by $${\rm Im}(f)=f(]-\infty;-2])=[0;1[$$ because $f$ is injective function.

Remark: Note that $$\lim_{x\to-\infty} \sqrt{1-e^{x+2}}=1.$$

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