What is the critical point of this function? The problem reads :
$$f(x)=7\frac{e^{2x}}{x} + 4.$$
I am unsure of how to approach this problem to find the derivative. If someone could break down the steps that would be greatly appreciated. 
Also, the question asks for intervals of the increasing and decreasing parts of the function. How would I figure this out? I'm thinking I'd use a sign chart. But if you have any other useful methods, I am all ears, or rather eyes. 
Thanks!
 A: You'd need to use the quotient rule!
A: As John has mentioned, you would have to use the quotient rule, and also note that in applying the quotient rule, you will have to use the chain rule in differentiating e^2x. 
As for intervals where the function is increasing or decreasing, remember that the derivative of a function must be positive where it is increasing and decreasing otherwise. When you have successfully found the derivative of the function, some of the terms in the derivative will be positive for all real x, so you will only need to find the values for which one of the terms is positive or negative. Good luck.
A: Let $f(x)=7\dfrac{e^{2x}}{x}+4$ for $x\ne 0$.  Then, we have for $x\ne 0$
$$f'(x)=7e^{2x}\left(\dfrac{2}{x}-\dfrac{1}{x^2}\right) \tag 1$$
where we used the product rule $(gh)'=g'h+gh'$ with $g(x)=e^{2x}$, $h(x)=1/x$, $g'(x)=2e^{2x}$, and $h'(x)=-1/x^2$.
Now, we note from $(1)$ that $f'(x)=0$ when $x=1/2$.  
Similarly, from $(1)$, we see that when $x>1/2$, $f'>0$ and $f$ is increasing.   When $0<x<1/2$ or $x<0$, $f'<0$ and $f$ is decreasing.
