Find the domain of $f(x)$ $$f(x)=\sqrt{\cos(\sin x)}+\arcsin \left(\frac{1+x^2}{2x}\right)$$
My Approach:
$\cos(\sin x) \geq 0 ; -1\leq \frac{1+x^2}{2x} \leq 1$
The first inequality holds for all $x$ and, for second one we have:
$(x+1)^2 \geq 0$ and, $(x-1)^2 \leq 0$ which is true only for $x=1$
My Problem:
The answer mentioned in the book is $x=1,-1$
I want to know how $-1$ is in the domain.
The condition for the arc sine is $$\left\lvert\frac{1+x^2}{2x}\right\rvert\le 1\iff x\ne 0 \,\text{ and }\,1+x^2\le 2\lvert x\rvert\iff x\ne 0 \,\text{ and }\,(1-\lvert x\rvert)^2\le 0$$ which is satisfied if and only if $\;\lvert x\rvert=1$.
We have $\forall x>0$ $$1+x^2\geq 2x $$ $$\implies \frac {1+x^2}{2x}\geq 1$$
the equality holds for $x=1$
and $\forall x <0$ $$1+x^2\geq -2x $$ $$\implies \frac {1+x^2}{2x}\leq -1$$
the equality holds for $x=-1$
the domain is then $\{-1,1\} $.
you have $$-1\le \frac{1+x^2}{2x}\le 1$$ in the case that $$x>0$$ we get $$0\le (1+x)^2$$ which is true. and $$(x-1)^2\le 0$$ which gives $$x=1$$ int the case that $$x<0$$ we get $$-2x\geq 1+x^2\geq 2x$$ which gives $$0\le (x+1)^2$$ thus $$x=-1$$ and $$(x-1)^2\geq 0$$
