Solution for $\arctan x = x^2$ 
Show that the equation $\arctan x = x^2$ has at least one solution. Then explain why the equation has exactly one positive solution $r$.

So i know that $\arctan x$ is always increasing as $f'(\arctan x) > 0$. I also know that $\arctan x$ goes from negative to positive at some point and thus has one root only, but I am not sure how to apply this to the equation I'm dealing with. 
I guess i could use fixed point iteration to show that the equation has at least one solution, but i don't know how i would go about doing that.
 A: The function $f(x)=\arctan x-x^2$ has
$$
\lim_{x\to-\infty}f(x)=-\infty,
\qquad
f(0)=0,
\qquad
\lim_{x\to\infty}f(x)=-\infty,
$$
and, moreover,
$$
f'(x)=\frac{1}{1+x^2}-2x=\frac{1-2x-2x^3}{1+x^2}
$$
Consider $g(x)=1-2x-2x^3$; since $g'(x)=-2-6x^2<0$, the function is decreasing, so it vanishes at exactly one point $p$; note that $g(0)=1$ and $g(1)=-3$, so $0<p<1$.
Can you finish?

 Since $f'(x)$ has the same sign as $g(x)$, we deduce that $f$ is increasing over $(-\infty,p]$ and decreasing (to $-\infty$) over $[p,\infty)$. Since $p>0$, we have that $f(p)>0$, so for a unique $r>p$ we have $f(r)=0$.

A: Let $f(x) = x^2$ and $g(x) = \arctan(x)$. For the first part: $x=0$.
For the second part we use 2 observations:


*

*$f'(x) = 2x$ and $g'(x) = \frac{1}{1+x^2}$. Thus, as $f(0) = g(0)$, but $f'(0)< g'(0)$, it follows that $f(x) < g(x)$ for small positive $x$.

*On the other hand, $\arctan(x)$ is bounded whereas $x^2$ is not, hence $f(x) > g(x)$ for large positive $x$.
Thus, by the mean value theorem (applied on $f-g$) there must be a positive $x$ for which $f(x)=g(x)$. To show that this point is unique, we can use Rolle's Theorem: 
Assume $f(x) = g(x)$ and $f(y) = g(y)$ for some $x< y$. Then there must be some $\xi\in(x,y)$ such that $f'(\xi)-g'(\xi)=0$. Note that:
$$f'(x) - g'(x) = 2x - \frac{1}{1+x^2} = 0 \iff  2x^3+2x-1 = 0$$
Now if this equation has only $1$ positive solution, then also $f(x)=g(x)$ has only 1 positive solution, since we can apply Rolle for any pair of solutions. E.g. if there were 2 positive solutions $x_0<x_1$ then by Rolle $f'(x)-g'(x) = 0$ must have at least two solutions, one in $(0,x_0)$ and one in $(x_0,x_1)$.
