Prove That a Specific Equation has a Solution Over a Given Closed Interval

Question

how would you prove that the equation $e^x=\frac{9}{1+9\cos(x)}$ has a solution over the interval of $[0, \frac{\pi}{2}]$? Isn't it enough to just state, that $g(x)=e^x$ and $f(x) = \frac{9}{1+9\cos(x)}$ are continuous functions over $\mathbb{R}$ and therefore also over the interval and that since, $e^x=\frac{9}{1+9\cos(0)}<e^0$ and $e^x=\frac{9}{1+9\cos(\frac{\pi}{2})}>e^{\frac{\pi}{2}}$ are true, there has to be an intersection in the interval and so the equation has a solution?

Answer 1

Careful - $f(x)$ is not continuous over $\mathbb{R}$. In fact, it is not defined at (the infinitely many) points $x$ such that $\cos(x)=-1/9$. However, since $\cos$ is positive in $[0,\pi/2]$ we know $f$ is continuous there. Consider the function $h(x)=g(x)-f(x)$. Being the difference of continuous functions on $[0,\pi/2]$, it is also continuous on this interval. Furthermore, $$h(0)=g(x)-f(0)=1-9/10=1/10>0$$ and $$h(\pi/2)=e^{\pi/2}-9<0$$ Therefore, the Intermediate Value Theorem implies that there is some $c\in[0,\pi/2]$ with $h(c)=0$. For this value, $g(c)=f(c)$ so we have the solution.