2.  Recall MARGINAL Costs, Revenue, Profit & Sales are ALL first derivatives of C(x), R(x), P(x), S(x)  For our purposes, marginal functions represent the approximation of producing the next “x” item. ◦ Therefore:  marginal cost = the cost of producing (x + 1) item

3.  Thus far with derivatives we have inferred their meanings as it might relate to costs, revenues or profits.  Now, let’s consider how a graph of these inferences can be made.

5. ( ⅔, -1.48) (-7, 0)(1, 0)

6. (-7, 0)(1, 0)

10. (-2, 8) ( ⅔, -1.48)

11. (-2, 8) ( ⅔, -1.48)

12. (-2, 8) ( ⅔, -1.48)

13. (-2, 8) ( ⅔, -1.48)

15. Definition. The values of x in the domain of f where f ‘(x) = 0 or does not exist are called the critical values of f. Insight: All critical values are also partition numbers, but there may be partition numbers that are not critical values (where f itself is not defined). If f is a polynomial, critical values and partition numbers are both the same, namely the solutions of f ‘(x) = 0.

18. Using f ’ (x) = (3x 2 + 4x – 4), substitute values in the interval to determine how the GRAPH of f(x) behaves ++++++++++0--------------0++++++++++ Use -4 -2Use -1 ⅔ Use 1 Since our critical points are -2 and ⅔, we construct a number line these two values highlighted.

19. Using f ′ (x) = (3x 2 + 4x – 4), substitute values in the interval to determine how the GRAPH of f(x) behaves ++++++++++0-------------0++++++++++ Use (-4)-2Use (-1) ⅔ Use 1

20. Critical values will ALWAYS occur in the domain of the function under consideration Any value that causes a 0 in the denominator is considered a Critical Value. A continuous function can INCREASE or Decrease on an interval containing a value of x for which f ′ (x) does NOT EXIST. To be a critical value, the number under consideration MUST be in the domain of the function. Values where a function is INCREASING or Decreasing must be written in an Open Interval.

21. When the graph of a continuous function changes from rising to falling, a high point or local maximum occurs. When the graph of a continuous function changes from falling to rising, a low point or local minimum occurs. Theorem. If f is continuous on the interval (a, b), c is a number in (a, b), and f (c) is a local extremum, then either f ‘(c) = 0 or f ‘(c) does not exist. That is, c is a critical point.

22.  Let c be a critical value of f, ◦ f(c) is a relative minimum if there exists within an open interval (a, b) value c such that f(c) ≤ f(x)  What does this really mean????  In an interval test the values of f ′ (x) are as follows Since the values go from negative to positive, the graph will appear to decrease before “c” and increase after “c” making a shape. (hence, the minimum) This is f ' (x) --------------------0+++++++++++++++ Values to the left of “c”cValues to the right of “c”

23.  Let c be a critical value of f, ◦ f(c) is a relative MAXIMUM if there exists within an open interval (a, b) value c such that f(c) ≥ f(x)  What does this really mean????  In an interval test the values of f '(x) are as follows Since the values go from positive to negative, the graph will appear to increase before “c” and decrease after “c” making a shape. (hence, the MAXIMUM) This is f ' (x) ++++++++++++++++++0------------------------- Values to the left of “c”cValues to the right of “c”

24.  Let c be a critical value of f, ◦ f will NOT a relative minimum nor MAXIMUM if the sign does not change around f '(c).  What does this really mean????  In an interval test the values of f '(x) are as follows This is f ' (x) ++++++++++++++++++0 Values to the left of “c”cValues to the right of “c” This is f ' (x) -------------------------0 Values to the left of “c”cValues to the right of “c”

25.  Determine the CRITICAL POINT(S) & intervals of increase/decrease for the following:

26.  Step 1: Find derivative.

27.  Determine the CRITICAL POINT(S) & intervals of increase/decrease for the following:  Step 2: Set derivative = 0, find critical points.

28.  Step 3  Create an interval test table for the critical points. This is f ’ (x) +++++++++0++++++0------------- 012

29.  Step 4  Create intervals using critical values.  NOTICE: at 0 no extrema occurs +++++++++0++++++0------------- 0= ND12 = MAX

30.  Step 5  Make determinations: +++++++++0++++++0------------- 0= ND12 = MAX

31.  Graph:

32. Notice the tangents in the intervals we made in STEP 3 go in the direction of the signs:

33.  According to some studies, oil companies are investing less in exploration since they continue to find less oil than they pump out of the ground. The function A(t) = 0.0265t 3 – 0.453t 2 + 1.796t + 7.47 (0 ≤ t ≤ 10), represents in billions of dollars, the amount invested by large oil companies in exploration for new reserves t years after 1995.  Find the relative extrema and graph the function.

34.  A(t) = 0.0265t 3 – 0.453t 2 + 1.796t + 7.47  A ’ (t) = 0.0795t 2 – 0.906t + 1.796

35.  Find the relative extrema and graph the function.  Create an interval table  A ’ (t) = 0.0795t 2 – 0.906t + 1.796  t = 2.5553 or t = 8.841 A ’ (t) = 0.0795t 2 – 0.906t + 1.796 (-∞, 2.5553)(2.5553, 8.841)(8.841, ∞) +++++++++0--------0++++++++ 2.555358.8419

36.  Find the relative extrema and graph the function.  Create an interval table  A(t) = 0.0265t 3 – 0.453t 2 + 1.796t + 7.47  Relative MAX at x = 2.553  Relative min at x = 8.841

37.  The Graph: A(t) = 0.0265t 3 – 0.453t 2 + 1.796t + 7.47

38. Write a description of the graph of the marginal revenue function, y = R ’ (x), including a discussion of any x-intercepts.

39. Analyzing the graph, we can see that from (0, 12500) the marginal revenue is positive. At x=12500, the marginal revenue is 0, and on the interval (12500, 25000) the marginal revenue is negative.

40. Write a description of the graph of the marginal profit function, y = P ’ (x), including a discussion of any x-intercepts.

41. Analyzing the graph, we can see that from (0, 1000) the Marginal Profit is positive. At x=1000, the marginal revenue is 0, and on the interval (1000, 1500) the Marginal Profit is negative.

42.  First Derivative Test: determines where a function is increases or decreases  Critical Values occur where the First Derivative = 0.  Sign charts help to see the change in signs of the First Derivative which allows for locating the local maximum/minimum points.  Local maximum is where f ′(x) changes from positive to negative  Local minimum occurs when f ′(x) changes from negative to positive  If f ′(x) is discontinuous at “c,” then f ′(x) DNE (Does Not Exist)

44.  2 nd Derivative Test stated: ◦ For y = f(x), the second derivative test, if it exists, is ◦ If the 1 st Derivative Test determines intervals of increasing/decreasing & critical points;  Then the 2 nd Derivative Test determines CONCAVITY.

45.  Understanding:  When the 1 st Derivative is increasing and the 2 nd Derivative is positive, a graph of f will be concave UP.  However, when the 1 st Derivative is decreasing and the 2 nd Derivative is negative, a graph of f will be concave Down.

46.  Like with the 1 st Derivative Test, the 2 nd Derivative Test has critical points called:  Points of inflection  A point of inflection is that point “c” on an open interval (a, b) where f ″(x)= 0 or f ″(x)= Does Not Exist

48. CRITICAL Points =

49. Inflection Point =

50. f ′(x)=(3x – 2)(x + 2)++++0 ---------------0++++ -5 -20 ⅔ 2 f ″(x) = 6x + 4--------------- 0+++++++++++++++++++++ -⅔  The graph increases from (-∞, -2) and (2, ∞) and decreases from (-2, ⅔).  The graph is concave Up from (-⅔, ∞) and concave down (-∞, -⅔)

51. Critical point x = -2 is a MAXIMUM value Critical Point x= ⅔ is a local Minimum value The inflection point x = - ⅔, is where the graph turns from concave Down to concave UP x = - ⅔

52. On what intervals is this graph concave Up? On what intervals is this graph concave down? Where is f ″(x) < 0? Where is f ″(x) > 0? Where is f '(x) increasing ? Where is f '(x) decreasing ? Where are the extrema?

53. On what intervals is this graph concave Up? (a, b) (b, c) (d, e) On what intervals is this graph concave down? (c, d) Where is f ″(x) < 0? (c, d) Where is f ″(x) > 0? (a, b) (b, c) (d, e) Where is f '(x) increasing ? (a, b) (d, e) Where is f '(x) decreasing ? (b, c) (c, d) Where are the extrema? x = b x = d

55.  With FIRST DERIVATIVE TEST: Interval Tables provide information about a function’s intervals of Increase & Decrease  With Second DERIVATIVE TEST: Interval Tables provide information about a function’s CONCAVITY x-301245 f(x)-40210 ++ND +++++0--------------0++++++++ 01 24 + +ND --------------0+++++++++++++++++++

56.  In the EXAMPLE below: ◦ the Intervals of increase are (-∞, 0) and (4, ∞) ◦ Decrease Interval (1, 4) ◦ The Function is CONCAVED Up on (-∞, 0) & (2, ∞) ◦ Concaved down on (-∞, 2) x-301245 f(x)-40210 ++ND +++++0--------------0++++++++ 01 24 + +ND --------------0+++++++++++++++++++

64. This graph below is f ’ (x). Create a table based on this graph determine intervals of increase and decrease, any inflection points, intervals of concavity, then graph f(x).

65. x f ’ (x) f(x) -∞ < x < -2 Negative & increasingDecreasing & concave UP x = -2 Local maxInflection Point -2 < x < 0 Negative & decreasingDecreasing & concave down x = 0 Local minInflection Point 0 < x < 2 Negative & increasingDecreasing & concave UP x = 2 Local maxInflection Point -2 < x < ∞ Negative & decreasingDecreasing & concave down Negative means the value of f ’ (x) is negative Positive means the value of f ’ (x) is positive.