If $f(x)=x^{2}$ and $g(x)=x sin x+cos x$; Find out the no of intersecting point

Solution without calculus

If $x ge frac{pi}{2}$, we have $g(x) le sqrt{x^2 + 1} < x^2$, and thus $f(x) = g(x)$ has no real solution on $[frac{pi}{2}, infty)$. Also, $0$ is not a solution of $f(x) = g(x)$.

Since $f(0) < g(0)$ and $f(frac{pi}{2}) > g(frac{pi}{2})$, by continuity, $f(x)=g(x)$ has at least one real solution on $(0, frac{pi}{2})$. Let us prove that $f(x)=g(x)$ has exactly one real solution on $(0, frac{pi}{2})$. Suppose there exist $0 < s < t < frac{pi}{2}$ such that $f(s) = g(s)$ and $f(t) = g(t)$. We have begin{align} frac{s^2}{sqrt{s^2+1}} &= frac{ssin s + cos s}{sqrt{s^2+1}} = cos (s - arctan s), \ frac{t^2}{sqrt{t^2+1}} &= frac{tsin t + cos t}{sqrt{t^2+1}} = cos (t - arctan t). end{align} Since $frac{t^2}{sqrt{t^2+1}} - frac{s^2}{sqrt{s^2+1}} = sqrt{t^2+1} - frac{1}{sqrt{t^2 + 1}} - sqrt{s^2+1} + frac{1}{sqrt{s^2 + 1}} > 0$, we have $$cos (s - arctan s) < cos (t - arctan t)$$ which, when combined with $t - arctan t in (0, frac{pi}{2})$ and $s - arctan s in (0, frac{pi}{2})$, results in $$s - arctan s > t - arctan t.$$ Thus, we have $arctan t - arctan s > t - s$ and $tan (arctan t - arctan s) > tan (t - s)$ that is $frac{t - s}{1 + ts} > tan (t-s)$ which, when combined with $tan (t-s) > t-s$, results in $frac{t - s}{1 + ts} > t-s$ which is impossible. The desired result follows.

Thus, $f(x) = g(x)$ has exactly one real solution on $[0, infty)$. Since $f(x)$ and $g(x)$ are both even, $f(x) = g(x)$ has exactly two real solutions on $(-infty, infty)$. We are done.

Of course all statements that use calculus can be proved without using calculus. But this may be very cumbersome. Let's try it here. Here is a picture of the function weare talking about.

We can define the functions $sin t$ and $cos t$ as projection of the unit vector to the x- and y-axis. The angle t is the lenth of the arc.

The following properties we can immediately conclude from the unique circle diagram:

$$sin(t)lt t,; forall t$$ $$cos (t) ; text{is decreasing in}; (0,pi)$$ Also the following formula can be deduced or found in Wiki

$$sin(x)-sin(y)=2cosfrac{x+y}2sinfrac{x-y}2$$

For the function $$f(x)=x^2$$ and $$g(x)=x sin x+cos x$$ we prove the following

1. In the interval $(0,infty)$ the function $f$ is increasing
2. In the interval $(0,pi)$ $f$ is increasing faster than $g$.
3. If $x>pi$ then $f(x)>g(x)$

$$ f(x+h)>f(x), forall h>0, x,x+hin (0,infty)tag 1$$ $$ f(x+h)-f(x)>g(x+h)-g(x), forall h>0, x,x+hin (0,pi)tag 2$$ $$f(x)>g(x), forall x>pi tag 3$$

We have $$ f(x+h)-f(x)=2xh+h^2>0,forall x,h>0$$ and so $(1)$ follows.

From $$ g(x+h)-g(x)\=(x+h)sin(x+h)-xsin x + cos(x+h)-cos(x) = $$ $$(x+h)sin(x+h)-xsin x \ = x(sin(x+h)-sin(x))+h sin(x+h) = x( 2 cos(x+frac h 2)sin(frac h 2) ) + h sin(x+h)\ <x(2 frac h 2)+h(x+h)\ =2xh+h^2\ =f(x+h)-f(x) $$

We further have $$ cos(x+h)-cos(x)<0$$ and so we proved $(2)$.

Finally $(3)$ we get from $$g(x)<xcdot 1 +1 < 4+1< 3^2 < x^2=f(x), ; forall x>pi$$

We have at least one intersection in $(0,pi)$ because $f(0)<g(0)$ and $f(1)>g(1)$ and both function are continues. We have only one point of intersection in $(0,pi)$ because $f$ grows faster than $g$ in $(0,pi)$ and no further point of intersection in $[pi,infty)$ because $f(x)>g(x)$ in $[pi,infty)$.

$f$ is even so for symmetry reasons we have exactly two points of intersection.

A solution:

(we want to show that the equation $x^2-xsin{x} -cos{x}=0$ has exactly two roots)

Let $g(x)=x^2-xsin{x} -cos{x}$
$g$ is continuous (as the sum of the polynomial $x^2$, $-cos{x}$ and $xsin{x}$,which is continuous as multiplication of continuous functions). $g(-pi)=pi^2+1>0,g(0)=-1<0, g(pi)=pi^2+1>0$ Because g, is continuous applying Bolzano's theorem: g(x)=0 has at least one solution for $xin(-pi,0)$ and one for $xin(0,pi)$ Therefore g(x)=0 has at least two solutions.

Let's assume g(x)=0 has more than two solutions, then g(x)=0 has at least three solutions $x_1<x_2<x_3$
$g$ is also differentiable, because of Rolle's theorem $g'(x)=0$ has at least two different solutions. $g'(x)=0iff x(2+cos{x})=0iff x=0$ so it holds $g'(x)=0$ has only one solution, which leads to contradiction. So g(x)=0 has exactly two solutions $square$

I don't know if this qualifies as "without calculus." You know that $f(x)$ is monotonously decreasing in $left. right] infty,0 left] right.$ and monotonously increasing in $left[ right. 0,infty left[ right.$ (you know because you can take the derivative, although in this simple case you don't need to; this is why I'm saying I don't know if this qualifies). You also know that $f(x)$ grows faster than $g(x).$ Finally, you also know that $f(0)=0$ and $g(0)=1$. Putting all this together yields two solutions.