Show that $x^2 = \sin (x)$ has exactly one positive solution I want to show that $x^2 = \sin(x)$ has exactly one positive solution.
We know that $x^2 - \sin(x)$ only has roots in the segment $[-1,1]$ and we also know that $x=0$ is a root.
How do I show that there exists exactly one root on $(0,\infty)$?
 A: You can use intermediate zero theorem to show that exists a solution $x_0$ in $(\frac{\pi}{4},\frac{\pi}{2})$ in fact
$$f\left(\frac{\pi}{4}\right)<0\\f\left(\frac{\pi}{2}\right)>0$$
then use the fact that $x^2-\sin(x)>0 \space \forall x \ge x_0$ showing that $f$ is incrasing in $[\frac{\pi}{4}, +\infty)$.
$f'(x)=2x-\cos(x)$, so $f'(\frac{\pi}{4})=\frac{\pi}{2}-\frac{\sqrt2}{2}>0$ 
A: We take the second derivative.
$$f''(x) = 2+\sin (x) >0$$
Which means it is concave in all its domain (or convex? They never agree). Point is, it looks like a smiley face. Then, it has at most two intersections with the $y$-axis. Hence, at most two solutions.
And those solutions are $0$ and approximately $0.88$.
A: hint
$f $ is continuous and differentiable at $(0,+\infty). $
Assume that there exist $x_1>x_2>0$ such that
$$f (x_1)=f (x_2)=0=f (0) $$
then by Double Rolle Theorem, 
$$\exists c_1, c_2 \;\;:\;\; f'(c_1)=f'(c_2)=0$$
and
$$\exists c\in (c_1,c_2) \;\;:\;\; f''(c)=0$$
but
$$f''(x)=2+\sin(x)>0$$
A: Hint As you point out, outside the  $[-1,1]$ interval there cannot be solutions, as for $x>1$, $x^2 > \max \sin(x)$.
The existence of a positive solution can be proved by noting that for sufficienty small $x$, $x^2 < \sin x$, while on the other hand $ \sin (1) < 1^2 =1 $: they must then "cross" somewhere in between, very loosely speaking.
Now, $x^2$ is convex, while $\sin$ is concave over the same $(0,1)$ interval, and that should help proving there is only one positive root.
A: The second derivative is positive, so the first derivative had a single root, which is a global minimum of the original function, which therefore has two roots, one at the origin. By Leonardo's argument, the other root is positive.
A: Since $\sin(x)\leq 1$, the positive solutions of $\sin(x)=x^2$ have to lie in the interval $(0,1)$. Over such interval $x^2$ is an increasing and convex function, going from $0$ to $1$, while $\sin(x)$ is an increasing and concave function, going from $0$ to $\sin(1)<1$. It follows that
$$ \exists! x\in(0,1) : x^2=\sin(x). $$
By applying Newton's method to the function $g(x)=x^2-\sin(x)$ with starting point $x_0=1$, we get that the actual solution is pretty close to 
$$ x_1 = \frac{1-\cos(1)+\sin(1)}{2-\cos(1)}\approx 0.9. $$
A more accurate approximation is given by $x_{6}\approx 0.876726215395$.
