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?

(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.

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.....

Your answers for parts (a) through (d) are correct. Now let's move on to part (e).

To determine whether the tangent line lies above or below the graph at the point where x = 3, we need to compare the slope of the tangent line (given by g'(3)) with the slope of the graph at that point.

Recall that the derivative of a function gives the slope of the tangent line at any point on the graph. So, if g'(3) > 0, the tangent line will have a positive slope, indicating that it lies above the graph. Conversely, if g'(3) < 0, the tangent line will have a negative slope, indicating that it lies below the graph.

Using the equation for the derivative, g'(x) = (x^2 – 16) / (x – 2), we can substitute x = 3 to find g'(3):
g'(3) = (3^2 – 16) / (3 – 2) = (-7) / 1 = -7

Since g'(3) < 0, the tangent line to the graph at x = 3 will have a negative slope, indicating that it lies below the graph at that point.

Therefore, the answer to part (e) is: The tangent line to the graph of g at the point where x = 3 lies below the graph at this point.

Critical value of 16x-2x^2

I BELEAVCE IT IS CORRECT