# Example of how to use the Intermediate Value Theorem

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 $f(c)=\lambda$.

Question:

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.

Attempt:

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?

• 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 $\frac{\pi}{2}$. – almagest May 19 '16 at 11:33
• how do you know to choose $\pi$ and $\pi/2$? – UniStuffz May 19 '16 at 11:55
• you start by picking values which are easy to calculate :) – almagest May 19 '16 at 12:35

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.
• @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. – Alex G. May 19 '16 at 12:22