Is there a way to deduce graph $\frac{2x+1}{x(x+1)}$ from the graph $\frac{1}{x}$ & $\frac{1}{x+1}$?
My Work:
let, $\,\,\,y_1=\frac1x$
$\qquad y_2=\frac{1}{x+1}$
So $y_1+y_2$ would yield the required graph, since there are two different asymptotes, how to determine whether graph value is really 0 at the center of two asymptotes? also how to draw the shape within the two asymptotes. Thanks a lot.
The graph does not have to be zero at the center of the two asymptotes. For example, see the case of $\frac{2}{x}+\frac{1}{x+1}$. It is in the particular case of your problem. Just solve $2x+1=0$. Make sure the solution does not coincide with the vertical asymptotes.
As for how to graph the function, the limit at $-1$ when approaching from the positive side is $+\infty$, the limit at $0$ approached from the negative side is $-\infty$. If you calculate the first derivative, you can see that is always negative in the $(-1,0)$ interval. Second derivative changes sign at $-1/2$.