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? 

The main idea of these questions is "equating the slopes", i.e.,  $m_1=m_2$: $m_1$ is the slope of the given line which you already computed here as $m_1=-1$ and $m_2$ is the derivative of the curve function at the point of tangency, which is the slope of the tangent line.
Since the derivative of the curve is $m_2=y'=-\frac{c}{(x+1)^2}$, at the point of tangency $P=(x_0,y_0)$, we have $-\frac{c}{(x_0+1)^2}=-1$ and hence $c$ must be non-negative and 1) $x_0=\sqrt{c}-1$ or 2) $x_0=-\sqrt{c}-1$. Let's consider the two options separetely.
If $x_0=\sqrt{c}-1$, then from the equation of the curve, $y_0=\sqrt{c}$ and hence $P=(\sqrt{c}-1,\sqrt{c})$. But, $P$ also lies on the line connecting the given points which you found as $y=-x+3$. So, $\sqrt{c}=-(\sqrt{c}-1)+3\implies2\sqrt{c}=4\implies c=4.$
If $x_0=-\sqrt{c}-1$, then from the equation of the curve, $y_0=-\sqrt{c}$ and hence $P=(-\sqrt{c}-1,-\sqrt{c})$. So, $-\sqrt{c}=-(-\sqrt{c}-1)+3\implies\sqrt{c}=-2$. No solution, since $\sqrt{c}>0$ by definition. This convention is a little tricky and not practical. Since, you may argue that there is a solution again and it is $c=4$.
Anyway, $c=4$. And also $P=(1,2).$ Also, in my opinion, using subscripts like $x_0, y_0$ is not practical in tests.
