karleyHart
Answered

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.

  1. Find all values of x where the graph of g has a critical value.

  2. 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.

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

  4. Write an equation for the tangent line to the graph of g at the point where x=3
  5. Does this tangent line lie above or below the graph at this point? Justify your answer.

Verified Answer :

nobillionaireNobley

1.) g has a critical value when g'(x) = 0
(x^2 - 16)/(x - 2) = 0
(x - 4)(x + 4) = 0
x = 4 or x = -4
Therefore, g has critical values at x = 4 and x = -4

2.) g''(x) = [(x - 2)(2x) - (x^2 - 16)]/(x - 2)^2 = [2x^2 - 4x - x^2 + 16]/(x - 2)^2 = (x^2 - 4x + 16)/(x - 2)^2
g''(4) = 4^2 - 4(4) + 16 = 16 - 16 + 16 = 16 => local minimum
g''(-4) = (-4)^2 - 4(-4) + 16 = 16 + 16 + 16 = 48 => local minimum

3.) The graph of g is not concave down at any intervals.

4.) Required equation is y - 4 = g'(3)(x - 3)
y - 4 = [((3)^2 - 16)/(3 - 2)](x - 3)
y - 4 = -7(x - 3)
y = -7x + 25

5.) the tangent line lies below the graph.

Answer:

Remember the critical values will be where g'(x)=0 or is undefined. Then use a sign chart to determine the local minimums or maximums. Use the second derivative to determine concavity.

Step-by-step explanation:

Analyze the given function to calculate where it equals zero and where it is undefined as these will all be critical values, not just the zeros.

Set up a sign chart to check the sign on each side of the three critical values. If it changes signs, it will be an extrema. Negative to positive is a minimum, but positive to negative is a maximum.

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