Assume $\arctan$ is differentiable. Prove that $\arctan(x) =1-x$ has a solution in $(0,1)$. Assume $\arctan$ is differentiable. Prove that $\arctan(x) =1-x$ has a solution in $(0,1)$. The hint given is to use the intermediate value theorem (IVT).

I do not understand how to show there is a solution using the intermediate value theorem. Could I get help to better understand how to apply IVT in this case?


Consider the function

$$ f(x)= \arctan(x)+x-1 $$

and see this $f(0)=-1$ and $f(1) =\pi/4$ which implies there is a point $x\in (0,1)$ where $f(x)=0$ (by IVT) that gives you
$$ \arctan(x)+x-1 = 0 $$
has a real root.
