롤의 정리와 평균값 정리

We are to find the number of positive solutions to the equation $e^x = 4\cos(2x)$. The figure below contains the graphs of $y=e^x$ and $y= 4\cos(2x)$. The solutions to $e^x = 4\cos(2x)$ are precisely the $x$'s where $y=e^x$ and $y= 4\cos(2x)$ cross.

It sure looks like there is exactly one crossing with $x\ge 0$ and that one crossing is somewhere between $x=0$ and $x=1$. Indeed since $\big[e^x- 4\cos(2x)\big]_{x=0} = -3<0$ and $\big[e^x- 4\cos(2x)\big]_{x=1} > e > 0$ and $f(x)=e^x- 4\cos(2x)$ is continuous, the intermediate value theorem guarantees that there is at least one root with $0<x<1$.

We still have to show that there is no second root — even if our graphs are not accurate.

Recall that the range of the cosine function is $[-1,1]$. If $e^x=4\cos(2x)$, then $e^x \leq 4$, so $x \leq \log(4)\approx 1.39$. So, we only need to search for roots of $f(x)$ on the interval $(0,1.4)$: we are guaranteed there are no roots elsewhere. Over this interval, $2x \in (0,2.8)$, so $\sin(2x)>0$, and thus $f'(x)=e^x+8\sin(2 x)>0$. Since $f'(x)$ has no roots in $(0,1.4)$, we conclude by Rolle's Theorem that $f(x)$ has at most one root in $(0,1.4)$ (and so at most one positive root total). Since we've already found that a root of $f(x)$ exists in $(0,1)$, we conclude $e^x=4\cos(2x)$ has precisely one positive-valued solution.