This is a continuous from my last question

Question

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.

On what intervals is the graph of g concave down? Justify your answer.

It looks like its between 1 and 3 but I do not know the exact points

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.

Answer 1

well, you can find the exact interval: it's where g" < 0

Since g" = (x^2-4x+16)/(x-2)^2

g" < 0 where x^2-4x+16 < 0

Since that is never negative, there is no interval where g(x) is concave down.

g(x) = (x^2+4x-13)/2 - 12log(x-2)

See that this is so at

http://www.wolframalpha.com/input/?i=%28x%5E2%2B4x-13%29%2F2+-+12log%28x-2%29

As for the tangent line at (3,4), it has slope g'(3) = -7, so it is

g-4 = -7(x-3)

Since g is concave up everywhere, the tangent must lie below the curve.

Plot that with g(x), and you have

http://www.wolframalpha.com/input/?i=plot+g%3D%28x%5E2%2B4x-13%29%2F2+-+12log%28x-2%29%2C+g%3D-7x%2B25