finding how many intersections $\cos x $ and $e^x - 2 $ have

Question

I proved it graphically that they only have 1 intersection, but how do I prove it in other ways. (Non-graphic ways)

How many intersections between $ e ^ x - 2 $ and $\cos x$ (proving not graphically)

Answer 1

In the interval $[0,\pi]$, $\cos(x)$ is decreasing and $e^x-2$ is increasing. Besides,$$e^0-2=-1<1=\cos(0)$$and $e^\pi-2>-1=\cos(\pi)$. So the graphs have one and only one intersection point in that region.

Furthermore, $x<0\implies e^x-2<-1\leqslant\cos x$ and $x>\pi\implies e^x-2>1>\cos x$. So, there are no more intersection points.

Answer 2

$f(x)= e^x -2 - cosx$

$f'(x) = e^x + sinx $

Now, for $x \ge0 , f'(x) \ge 1 +sinx >0$ i.e. $f(x)$ is increasing and $f(0)<0$, & $f(\pi/2)>0$ so $(fx_0)=0$ for some unique $x_0 \in (0,\pi/2) \implies e^{x_0} - 2 = cosx_0$.

For $x < 0 $, notice that $ 1 \le2+cosx \le3$ but $e^x <1$, so no solution exist.

Thus, you have only one solution $x_0 \in (0, \pi/2)$

Answer 3

Hint 1. If $x< 0$ then $e^x-2<1-2=-1\leq \cos(x)$.

Hint 2. $f(x)=e^x-2+\cos(x)$ is a continuous function which is strictly increasing in $[0,+\infty)$ (check the derivative $f'$), $f(0)<0$, and $f(\pi/2)>0$.

Answer 4

Notice that $-1\le\cos(x)\le 1$ and $e^x-2>-2$. You only need to check between the range of $\cos(x)$ for solutions, in the other cases the exponential function is too large or too small. $$e^x-2=-1\to e^x=1\to x=0$$ $$e^x-2=1\to e^x=3\to x=\ln(3)$$ So checking in the range $0\le x\le \ln(3)$

Since $e^x-2$ is increasing throughout this interval, and $\cos(x)$ is decreasing, there will only be one intersection. Since it is irrational, we will use an iteration to find it: $$e^x-2=\cos(x)\to e^x=\cos(x)+2\to x=\ln[\cos(x)+2]$$

So iterate: $$x_{n+1}=\ln[\cos(x_n)+2], x_0=1$$ Approximate solution is $x\approx0.9488147556$

Answer 5

The function $\cos(x)$ is bounded between $-1$ and $1$. The function $f(x)=e^x-2$ is increasing for all $x\in\mathbb{R} $; so all intersections with $\cos(x)$ are in between the two values of $x$, call them $x_1$ and $x_2$, where $f(x_1)=-1$ and $f(x_2)=1$.

Now we find $x_1=0$ and $x_2=\ln(3)=1.09\ldots$.

For $y$ in the interval $[0,\ln(3)]$, all values $\cos(y)$ of the $\cos$-function appear exactly once, i.e. are unique, because $\ln(3)<\pi$. Thus $f(x)$ intersects $\cos(x)$ exactly once (from below) in the interval $[0,\ln(3)]$.