To find the sign of
$$f(x)=\log(x-1)-\arctan(x-2)$$
I tried searching solutions for
$$\log(x-1)\ge\arctan(x-2)$$
but failed, so I guessed I could use its derivative to find in which intervals the function is monotone, so
$$f'(x)=\frac{1}{x-1}+\frac{1}{(x-2)^2+1}=\frac{x^2-5x+6}{(x-1)(x^2-4x+5)}$$
and
$$f'(x)\ge0\iff x\in(1,2]\cup[3,\infty)$$
which means that f(x) is monotone in $(1,2]$, $[2,3]$ and $[3,\infty)$.
The point for that is so I can apply the Intermediate value theorem, since $f(x)$ is continuous $\forall x\in (1,\infty)$, by finding the images of the boundaries of these intervals. If the lower boundary's image is negative and the upper one's image is positive, or viceversa, there is a root. Now, by blind luck:
$$\lim_{x\to1^+}f(x)=\log(0^+)-\arctan(1^+-2)=-\infty$$ $$f(2)=\log(1)-\arctan(0)=0$$
and that gives us both a root and the fact that $\forall x\in (1,2]\;f(x)\le0$. Great, now
$$f(3)=\log(2)-\frac{\pi}{4}\lt0$$
So no roots here, since $\forall x \in[2,3]\; f(x)\le0$, but at least we know that $\forall x \in(1,3]\; f(x)\le0$. Now here comes the parts I can't solve: given that
$$f(4)=\log(3)-\arctan(2)<0$$ $$\lim_{x\to \infty}f(x)=\log(\infty-1)-arctan(\infty-2)=\infty$$
Which means there is a root in $[4,\infty)$, but how do I find it? I can't seam t find a way to actually have a finite value for this root. Many thanks to whoever replies! (A smiley face will do)(Sorry for any errors).