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So the question asks me to find the equation of the tangent line passing through the point (1,-4) given this function. Isn't it undefined at x=1, so where do they get the value of -4 in the first place?

original function

I took the derivative of the function, as i'm assuming you'd need it to calculate the slope at that point, which is shown below.

derivative function

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  • $\begingroup$ The tangent passes through $(1,-4)$, but that is not the point where it touches the curve. In that sense $-4$ is more or less random; they want you to look at all possible tangents and figure out which one(s) happen(s) to go through the point you're given. $\endgroup$ – Arthur Oct 16 '14 at 0:21
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The tangent to the curve is given by (as you say):

$$\frac{4}{(x-1)^3}+1$$

So you can construct a line

$$y'=\left[\frac{4}{(x-1)^3}+1\right]x'+c$$

For what values of $x$ and $c$ does this line pass through $(x,x+1-\frac{2}{(x-1)^3})$ and (1,-4)? That is, solve the equations:

$$-4=\left[\frac{4}{(x-1)^3}+1\right]1+c$$

$$x+1-\frac{2}{(x-1)^3}=\left[\frac{4}{(x-1)^3}+1\right]x+c$$

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  • $\begingroup$ Alright, so I was trying to solve the equations given, and I expanded it all to have terms of x and c. I was unsure if you ignore the +c in the terms or not, could you just elaborate on that part a bit more? $\endgroup$ – Maddark Oct 16 '14 at 12:42
  • $\begingroup$ These are simultaneous equations so first step eliminate c by subtracting 1 from 2. Solve for x (you can use $x\ne1$). Then substitute for x to get c. $\endgroup$ – Dale M Oct 16 '14 at 19:09
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It's not true that the point of tangency has to be when x = 1. The condition just means that there is a point of tangency (p, f(p)) at which the slope of the curve is as you calculated and which passes through (1, -4).

So now, with the gradient of the tangent, as well as the two points it passes through (1,-4) and (p, f(p) you can now form an equation of the line through y = mx + c, and solve for c.

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