(a) ok
(b) ok
(c) g" = (x^2-4x+16)/(x-2)^2
Since g" is never negative, g is never concave down
(d)correct, but since you know g(3)=4,
y-4 = -9(x-3)
(e) since the graph is always concave up, any tangent lines must lie below the graph. Doodle around some, and you will see why this is so.
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.
a.Find all values of x where the graph of g has a critical value.
b.For each critical value, state whether the graph of g has a local maximum, local minimum or neither. You must justify your answers with a complete sentence.
c.On what intervals is the graph of g concave down? Justify your answer.
d.Write an equation for the tangent line to the graph of g at the point where x = 3.
e.Does this tangent line lie above or below the graph at this point? Justify your answer.
My answers so far
a.x=-4,4
b.Both points are a local minimum.
c.-infinity<x<2
2>x>infinity
d.y-g(3)=g’(3)(x-3)
e.?
Could someone check my answers and show me how to do e?
5 answers
I BELEAVCE IT IS CORRECT
i know this is old, but for anyone else, x=2 is also a critical point, but it is neither a maxima or minima
Um sigh bc steve did d incorrectly :// it's supposed to be times -7.....
Critical value of 16x-2x^2