What-is-the-domain-of-f-x-arcosh-1-x-2-1-x-2-

Question Number 99055 by Ar Brandon last updated on 18/Jun/20

What is the domain of

f (x) = arcosh (\frac{1 + x^{2}}{1 - x^{2}}) ?

Commented by mathmax by abdo last updated on 18/Jun/20

f (x) = \ln (\frac{1 + x^{2}}{1 - x^{2}} + \sqrt{{(\frac{1 + x^{2}}{1 - x^{2}})}^{2} - 1})

so x \in D_{f} \Leftrightarrow x \neq \bar{+} 1 and ∣ \frac{1 + x^{2}}{1 - x^{2}} ∣⩾ 1 \Rightarrow x \neq \bar{+} 1 and \frac{1 + x^{2}}{1 - x^{2}} ⩾ 1 or \frac{1 + x^{2}}{1 - x^{2}} ⩽ - 1

\frac{1 + x^{2}}{1 - x^{2}} ⩾ 1 \Rightarrow \frac{1 + x^{2}}{1 - x^{2}} - 1 ⩾ 0 \Rightarrow \frac{1 + x^{2} - 1 + x^{2}}{1 - x^{2}} ⩾ 0 \Rightarrow \frac{{2 x}^{2}}{1 - x^{2}} ⩾ 0 \Rightarrow - 1 < x < 1

\frac{1 + x^{2}}{1 - x^{2}} ⩽ - 1 \Rightarrow \frac{1 + x^{2}}{1 - x^{2}} + 1 ⩽ 0 \Rightarrow \frac{1 + x^{2} + 1 - x^{2}}{1 - x^{2}} ⩽ 0 \Rightarrow \frac{2}{1 - x^{2}} ⩽ 0 \Rightarrow 1 - x^{2} ⩽ 0 \Rightarrow

x^{2} - 1 ⩾ 0 \Rightarrow x ⩾ 1 or x ⩽ - 1 \Rightarrow D_{f} =] - 1, 1 [\cup] - \infty, - 1 [\cup] 1, + \infty [= R - {1, - 1}

Commented by Ar Brandon last updated on 19/Jun/20

Thanks Sir . But according to the graph \dots

I {don}^{'} t understand the difference .

Any explanations please ?

Commented by abdomathmax last updated on 19/Jun/20

all i have in my store is this \dots

Commented by Ar Brandon last updated on 19/Jun/20

When we substitute x = 2 in the equation it becomes

undefined . Why ? I {don}^{'} t understand .

Answered by Ar Brandon last updated on 19/Jun/20