# 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.

1. 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(x-2) + 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(x-2) - 5

when x = 3
g ' (x) = 9 - 16/1 = -7

tangent equation:
y-4 = -7(x-3)
or y = -7x + 25

see graph:

http://www.wolframalpha.com/input/?i=plot+y+%3D+%281%2F3%29x%5E3+-+16ln%28x-2%29+-+5+%2C+y+%3D+-7x%2B25

2. Steve

On the other hand, if you meant

g'(x) = (x^2–16)/(x−2)
= x+2 - 12/(x-2)

then

g(x) = x^2/2 + 2x - 12log(x-2) + c
g(3) = 9/2 + 6 - 12log(1) + c = 4, so c=-13/2

g'(3) = -7

and the tangent line is

y-4 = -7(x-3)

and the graph is at

http://www.wolframalpha.com/input/?i=plot+y%3Dx^2%2F2+%2B+2x+-+12log%28x-2%29+-+13%2F2%2C+y+%3D+-7%28x-3%29%2B4

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