Take the function $$y=\frac{\sqrt{\cosh\left(\frac{1+x}{x^2}\right) - 1}}{e^{\frac{2}{x-1}\log\left|x-1\right|}+1}$$ I have to find the domain of this function. These are the condition that I set up: $$\begin{cases} e^{\frac{2}{x-1}\log\left|x-1\right|}+1\neq 0&(1)\\ x-1\neq0&(2)\\ \left|x-1\right|>0&(3)\\ \cosh\left(\frac{1+x}{x^2}\right) - 1\ge0&(4)\\ x^2\neq0&(5) \end{cases}$$ And these are the results: $$\begin{cases} \forall x \in\mathbb{R}&(1)\\ x\neq1&(3)\\ \forall x \in\mathbb{R}&(4)\\ x\neq0&(5) \end{cases}$$

$(1)$ Denominator

$(2)$ Denominator of the exponent
$(3)$ Argument of the logarithm
$(4)$ Argument of the root
$(5)$ Denominator of the argument of $\cosh$

And this is the definition set of $y$:
$$x\in(-\infty, 0)\cup(0, 1)\cup(1, +\infty)$$

I deleted $(2)$ because it's included in the $(3)$.

The $(1)$ is verified for all $x$ because it's an exponential and because I set up the $(3)$

To solve $(3)$ I made the $\vee$ between $x-1<0$ and $x-1>0$.
The $(4)$ is verfied for all $x$ because the range of $\cosh(x)$ is $[1;+\infty)$, so it's always greater or equal than $1$.

So, is it correct? Or I was wrong something?

  • 3
    $\begingroup$ Looks right; you can do (3) a lot easier by noting that $|a|\geq 0$ for all $a$, and $|a|=0$ if and only if $a=0$; so in order to get $|x-1|\gt 0$ you only need $x-1\neq 0$; that is, (2) and (3) are equivalent. Note that "The (4) is verified for all $x$ because the range of $\cosh(x)$ is $[1,\infty)$" (range, not domain). $\endgroup$ – Arturo Magidin Feb 2 '12 at 20:00
  • $\begingroup$ Yeah, I did :) Thank you Arturo! $\endgroup$ – Overflowh Feb 2 '12 at 23:12

This CW answer intends to remove this question from the Unanswered queue.

As Arturo Magidin remarks, your answer is correct. Cheers!


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