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