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Find the domain of: $$ f(x)= \sqrt[e^{\frac{1}{x^2-1}}] {\arccos \frac{x-1}{x+2}}$$ (it's not really visible but it's the $e^{\frac{1}{x^2+1}}$-th root)

I started solving this problem like this:

condition #1

$$-1\le \frac{x-1}{x+2} \le 1$$

  1. case $$\frac{x-1}{x+2} \ge -1$$ $$\frac{2x+1}{x+2}\ge 0$$ $$\Rightarrow x\in (-\infty,-2]\cup[-\frac12,\infty)$$

  2. case $$\frac{x-1}{x+2} \le 1$$ $$\frac{-3}{x+2} \le 0$$ $$\Rightarrow x\in(-2,\infty)$$

Intersection of these two cases is $x \in [-\frac 12, \infty)$.

condition #2

$$x^2+1 \neq 0$$ $$x \neq \pm 1$$

But, what am I supposed to do with the whole $e^{\frac{1}{x^2+1}}$-th root?

i dont know much about algebra
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1 Answers1

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Once you ensure that $$ A(x)=\frac{1}{x^2-1} $$ is defined, then also $e^{A(x)}$ is defined (and positive), so you can take any nonnegative number to the power $1/(e^{A(x)})$. This requires $x\ne-1$ and $x\ne1$.

Since the arccosine takes its values on $[0,\pi]$, the condition becomes $$ -1\le\frac{x-1}{x+2}\le1 $$ This becomes $$ \begin{cases} \dfrac{2x+1}{x+2}\ge0 \\[6px] \dfrac{-3}{x+2}\le0 \end{cases} $$ The top inequality is satisfied over $(-\infty,-2)\cup[-1/2,\infty)$, the bottom inequality over $(-1/2,\infty)$. Hence the domain is $$ [-1/2,\infty)\setminus\{1\} $$

egreg
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