# What is the Range of the function?

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?

• $x\leq 2$ is not the domain. Commented Sep 10, 2021 at 1:34
• Corrected. Thanks. Commented Sep 10, 2021 at 1:38
• 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 $1$. Commented Sep 10, 2021 at 1:45
• @C Squared Thanks again! Commented Sep 10, 2021 at 2:02
• 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. Commented Sep 13, 2021 at 20:40

$$\sqrt{1-e^{x+2}}$$ 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 $$[0,1)$$.

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