Knowing which theorem of calculus to use to prove number/nature of solutions How does one go about showing that the equation $$ arctanx = x^2  $$ has at least one solution and then in turn show that the equation has exactly one positive solution? 
I figured for part a) "show at least one solution" it could be possible to rewrite
the equation as $$ x^2 - arctanx = 0 $$ and apply the intermediate value theorem. and for part b) "exactly one positive solution"  to examine the behaviour of the second derivative to get (concave up or down, could that give an indication of if the graph only has one positive solution?) but I'm not exactly sure if these are the correct approaches. Any suggestions as to the correct method would be much appreciated. 
 A: I am presuming that by $\arctan x$ you just mean one branch of that function that whose range is $[\pi,\pi]$ (There are more branches obtained by adding $2\pi k$ to that branch). Obviously $x=0$ is a solution. As $x$ grows from $0$ it obeys $\arctan x > x$ for small positive $x$. (this can easily be deduced from $y > \tan y$ from elementary trigonometry by taking $x=\tan y$). So f(x)=$x^2-\arctan x$ is negative for small positive values less than 1 since $x^2<x<\arctan x$ But this branch of $\arctan x$ is bounded above by $\pi$ and $x^2$ is unbounded so eventually they have to meet at the only positive solution $x_1$ of $f(x)=0$. For $x>x_1$ obviously $x^2>\arctan x$ and $f(x)$ is positive. This the only solution since the derivative of $x^2$ is $2x$ and is growing with $x$ so the function $x^2$ will always stay on one side of its tangent line at $x_1$, whereas the derivative of $\arctan x$ is $1/(1+x^2)$ is decreasing with increasing $x$ and thus its tangents will always stay on one side of its tangent at the point of intersection. We need to show that $2x>1/(1+x^2)$ This can be done by considering the function $2x^3+2x-1>0$ that is monotonously increasing. It has a zero at $x_0=0.42385$ where $f(x)<0$ so $x_0<x_1$ and the 2 functions stay away from each other for $x>x_1$
Note that if we consider more branches $\arctan x$ there will be more positive solutions to the right of $x_1$
A: An idea: calculate the second derivative of $f(x)=x^2-\arctan(x)$ and show that is it positive. Then the first derivative is increasing. It is increasing from negative to positive values (check it), therefore the function itself is first decreasing (where $f'<0$) and then increasing (where $f'>0$). Thus, the equation $f(x)=a$ can have at most two solutions (at most one solution for each interval of monotonicity). One solution for $f(x)=0$ is $x=0$.
A: It is actually much easier to show it if we put the equation in the form:
$$x=\tan x^2$$
Then for small positive $x$, such that $x<1$, $\tan x^2<x^2<x$ and the function $f(x)=\tan x^2-x$ is negative. But $f'(x)=2x/(1+x^2)-1$ is always positive show $f(x)$ is monotonously increasing function so at some point $x_1$, $f(x)=0$. For $x>x_1$ we only need to consider the fact teh derivative of $\tan x^2=1/(1+x^2)$ is larger than $1$ and thus the curve $y=\tan x^2$ and the line $y=x$ will not meet again. 
A: Let $f(x)=x^2-arctanx$. Determine $f'(x)$. Then you should see that the equation $f'(x)=0$ has two solutions. Hence, by the mean value theorem, the equation $f(x)=0$ has at most three solutions !
Edit: this answer is not correct ! FRED
