Domain of a function 5

Question

I was solving this domain:

$$ f(x)= \sqrt\frac{(e^x-2)^x + \arccos(e^x -2)}{x- \sqrt {x^2 -1}}$$

My solution is: $x^2 \geq 1$ and the argument of the first sqrt $\geq 0$ (can't solve this one) and $0 \leq x \leq \log3$ and $x \geq 2$ and the denominator $\neq 0$ Could you help me to reach the right solution? note1: Missing the x- factor

Answer 1

You need that

$$\sqrt {x^2 -1}\neq 0 \quad \land\quad x-\sqrt {x^2 -1} \neq0$$

$$-1 \leq e^x -2 \leq 1 \quad \land\quad e^x -2>0 \implies 0 < e^x -2 \leq 1$$

$$\frac{(e^x-2)^x + \arccos(e^x -2)}{x-\sqrt {x^2 -1}}\ge0$$

Answer 2

The necesaary conditions are :

$x^2>1$ as the denominator of the function cant be 0 ,

$e^x -2 >0$ as the function $a^x$ is defined for all real x iff a>0.

and $-1\leq e^x -2\leq 1$ for $ arccos(e^x-2)$.

and $x- \sqrt{x^2-1}>0$

Solving these conditions you will get the domain, as we can clearly see that $ arccos(e^x-2)$ and $(e^x -2)^x$ are non negative functions.