math, a.p. calc
posted by Darrius
Let g be a function that is defined for all x, x ≠ 2, such that g(3)=4 and the derivative of g is g′(x)=
x^2–16/x−2, with x≠2.
Write an equation for the tangent line to the graph of g at the point where x = 3.
Does this tangent line lie above or below the graph at this point? Justify your answer.

Reiny
I will assume you meant
if g ' (x) = x^2  16/(x  2)
or else why would you have x ≠ 2
g(x) = (1/3)x^3  16ln(x2) + c
but (3,4) lies on it, so
4 = (1/3)(27)  16ln1 + c
4 = 9  0 + c
c = 5
so g(x) = (1/3)x^3  16ln(x2)  5
when x = 3
g ' (x) = 9  16/1 = 7
tangent equation:
y4 = 7(x3)
or y = 7x + 25
see graph:
http://www.wolframalpha.com/input/?i=plot+y+%3D+%281%2F3%29x%5E3++16ln%28x2%29++5+%2C+y+%3D+7x%2B25 
Steve
On the other hand, if you meant
g'(x) = (x^2–16)/(x−2)
= x+2  12/(x2)
then
g(x) = x^2/2 + 2x  12log(x2) + c
g(3) = 9/2 + 6  12log(1) + c = 4, so c=13/2
g'(3) = 7
and the tangent line is
y4 = 7(x3)
and the graph is at
http://www.wolframalpha.com/input/?i=plot+y%3Dx^2%2F2+%2B+2x++12log%28x2%29++13%2F2%2C+y+%3D+7%28x3%29%2B4
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