The question is
prove that $\dfrac {e^\pi}{x-\pi} + \dfrac {\pi^e}{x-e} + \dfrac {e^e+\pi^\pi}{x-e -\pi} = 0 $ . has one real root in $(e,\pi)$ and other $(\pi,\pi+e)$.
I tried this question hard but I couldn't find any easy way to solve the question , since simplifying these fraction doesn't really help , I also tried to manipulate the equation but I couldn't succeed .Any help , hint , solution would be really appreciated. Thanks..
On
According to Wolfy, there are roots at about 2.9 and 4.3.
Looking at the equation informally, if $x$ is just above $e$ it is positive, if $x$ us just below $\pi$ it us negative, if $x$ is just above $\pi$ it is positive, and for large $x$ it is negative.
The numerators do not matter except for their signs.
Let $f(x)=\dfrac {e^\pi}{x-\pi} + \dfrac {\pi^e}{x-e} + \dfrac {e^e+\pi^\pi}{x-e -\pi}.$
Thus, $$\lim_{x\rightarrow e^+}f(x)=+\infty$$ and $$\lim_{x\rightarrow \pi^-}f(x)=-\infty.$$ But $f$ is continuous on $(e,\pi)$, which says that there is a root of the equation on $(e,\pi)$.
By the same way we can obtain that there is a root on $(\pi,e+\pi).$
But, $f(x)=0$ is a quadratic equation...
Another way.
Rewrite our equation in the form $g(x)=0$, where $$g(x)=e^{\pi}(x-e)(x-e-\pi)+\pi^e(x-\pi)(x-e-\pi)+(e^{\pi}+\pi^{\pi})(x-e)(x-\pi).$$
Now, check $g(e),$ $g(\pi)$ and $g(e+\pi).$
Easy to see that $g(e)>0$, $g(\pi)<0$ and $g(e+\pi)>0$.