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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?

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.

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2 Answers

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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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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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