Prove that the equation $x^{2}-x\sin(x)-\cos(x)=0$ has only one root in the closed interval $(0,\infty)$. Here's the graph
(http://www.wolframalpha.com/input/?i=%28x%5E2%29-xsenx-cosx%3D0).
The part I'm having trouble with is proving that the root is unique. 
I can use the intermediate value theorem to find the interval where the root is, but from that I'm lost. 
I know you can look at the first or second derivative, but I don't know how to use that when the sign of sin and cos varies.
Thanks.
 A: Let $f(x) = x^2 - x\sin x - \cos x$, for $x \ge 0$. Then $f$ is continuous with $f(0) = -1 < 0$ and $f(\pi) = \pi^2 + 1 > 0$. So by the intermediate value theorem, $f$ has a zero in $(0,\pi)$. Since 
$$f'(x) = 2x - x\cos x = x(2 - \cos x) \ge x(2 - 1) = x > 0$$
on $(0,\infty)$, $f$ is strictly increasing on $(0,\infty)$. Therefore, $f$ has a unique root in $(0,\infty)$.
A: $f(x)=x^2-x \sin x -\cos(x)$ and $f'(x)=2x-x \cos x=x(2-\cos x)$. Clearly $2-\cos x>0$ and for $x>0$ we have that $f'(x)> 0$ $x>0$, therefore $f(x)$ is strctly increasing. now since $f(0)<0$ and $f(\infty)>0$, $f(x)$ just has one root on $x>0$.
This is a graphs of $f(x)$

and here is $f'(x)$

A: Given $f\left(x\right)= x^2-x\sin \left(x\right)-\cos \left(x\right)$. You are studying  $x^2-x\sin \left(x\right)-\cos \left(x\right)=0$ so when the function intersects x-axis. Study its monotony by studying its derivative $$f'\left(x\right)=2x-x\cos \left(x\right)$$ You will find that the derivative is positive for $x>0$ and $f\left(0\right)=-1$ so $f\left(x\right)$ will intersect x-axis only once in $\left(0,+∞\right)$.
