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?
Best Answer
from $\frac{c}{x+1} = -x+3$ and $-\frac{c}{(x+1)^2} = -1$ , you got $x+1=-x+3$