Find c for which tangent line meets $c/(x+1)$ Here is the problem: 
Determine $c$ so that the straight line joining $(0, 3)$ and
$(5, -2)$ is tangent to the curve $y = c/(x + 1)$. 
So, what I know: the line connecting the points is $y=-x+3$.
I need to find a value for c such that the derivative of $c/(x+1)$ can be $-x+3$.
I took the derivative of $c/(x+1)$, and found a line via point-slope form, so 
$$y-3=-\frac{cx}{(x+1)^2}  \quad \Rightarrow \quad y=-\frac{cx}{(x+1)^2} + 3 $$
At this point the tangent line is $y=-x+3$.
So, $-x = (-cx/(x+1)^2)$. From there, I can conclude $c=x^2+2x+1$. How do I conclude, without looking at a graph, what value of $x$ to choose? I know from the answer that at $x=1$, $c=4$, and the tangent line meets the curve at that point $x=1$. But how would I determine this? 

Hint:
Here are two curves whose slope are supposed to be equal:
$$
y=-x + 3
$$
and
$$
\frac{c}{x+1}=y
$$
equating the slopes of both of curves,
we have
$$
-1=\frac{-c}{(x+1)^2}
$$
this gives you
$$
(x+1)^2=c\implies x=-1\pm \sqrt c
$$
note that from here we need $c$ should be non-negative. i.e. $c\geq 0$
Now at $x=-1\pm \sqrt c$ both curves touch each other. So,
$$
-x + 3=\frac{c}{(x+1)}
$$
at $x=-1\pm \sqrt c$.
Solve this equation replacing $x=-1\pm \sqrt c$. and you will have value of $c$ you want. 
