Theorem: Let $f$ be continuous on $[a,\,b]$ and assume $f(a)\not=f(b)$. Then for every $\lambda$ such that $f(a)<\lambda<f(b)$, there exists a $c\in(a,\,b)$ such that $f(c)=\lambda$.


Use the Intermediate Value Theorem to prove that the equation $$e^{sin(x)} = 2+cos(x)−sin(x)$$ has at least one positive real solution.


I'm assuming that you only proof that a root of this equation exits, not to give the actual value. However I was able to solve this question graphically and obtained the (multiple) roots: $$x = 2 \pi n+\pi$$
where $n\in\mathbb{Z}$.

How would I use the IVT to answer the original question?

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    $\begingroup$ Let $f(x)=e^{\sin x}-2-\cos x+\sin x$. We have $f(0)=-2<0,f(\frac{\pi}{2})=e-1>0$, so there is a root between 0 and $\frac{\pi}{2}$. $\endgroup$ – almagest May 19 '16 at 11:33
  • $\begingroup$ how do you know to choose $\pi$ and $\pi/2$? $\endgroup$ – UniStuffz May 19 '16 at 11:55
  • $\begingroup$ you start by picking values which are easy to calculate :) $\endgroup$ – almagest May 19 '16 at 12:35

You can't use IVT to verify that those points are solutions. To do that, simply plug them in and observe that they satisfy the equation.

The way IVT is used is to prove that a solution must exist. This is very different than directly finding a solution, as you have done. To use IVT in this problem, first move everything to one side of the equation so that we have $$f(x) = e^{\sin(x)} - 2 - \cos(x) + \sin(x)$$ Now plug in the values $x=\pi/2, 3\pi/2$ and observe that $f(\pi/2) = e - 2 + 1 > 0$, while $f(3\pi/2) = e^{-1} - 2 - 1 < 0$. Therefore, by the IVT, there exists a number $c \in (\pi/2, 3\pi/2)$ such that $f(c) = 0$. The IVT doesn't give us any idea what $c$ is; it just tells us that $c$ exists.

Actually, your solution is much better than using IVT, because it is obviously more useful to know what some of the solutions are, rather than simply knowing that the solutions exist.

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  • $\begingroup$ I've edited my question so that it makes more sense, I want to know how to use the IVT to show that a root exist. Is your method still correct, why do we take x to be numerical values? $\endgroup$ – UniStuffz May 19 '16 at 11:54
  • $\begingroup$ @UniStuffz Yes, what I've written here is still correct. I used the IVT to show that there is a solution in the domain $(\pi/2, 3\pi/2)$, so in particular there is a positive solution. $\endgroup$ – Alex G. May 19 '16 at 12:22

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