 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

 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.

( ⅔, -1.48) (-7, 0)(1, 0)

(-7, 0)(1, 0)

(-2, 8) ( ⅔, -1.48)

(-2, 8) ( ⅔, -1.48)

(-2, 8) ( ⅔, -1.48)

(-2, 8) ( ⅔, -1.48)

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.

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

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

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.

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.

 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) Values to the left of “c”cValues to the right of “c”

 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) Values to the left of “c”cValues to the right of “c”

 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) Values to the left of “c”cValues to the right of “c” This is f ' (x) Values to the left of “c”cValues to the right of “c”

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

 Step 1: Find derivative.

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

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

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

 Step 5  Make determinations: = ND12 = MAX

 Graph:

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

 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) = t 3 – 0.453t t (0 ≤ t ≤ 10), represents in billions of dollars, the amount invested by large oil companies in exploration for new reserves t years after  Find the relative extrema and graph the function.

 A(t) = t 3 – 0.453t t  A ’ (t) = t 2 – 0.906t

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

 Find the relative extrema and graph the function.  Create an interval table  A(t) = t 3 – 0.453t t  Relative MAX at x =  Relative min at x = 8.841

 The Graph: A(t) = t 3 – 0.453t t

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

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.

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

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.

 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)

 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.

 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.

 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

CRITICAL Points =

Inflection Point =

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

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

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?

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

 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 f(x) ND ND

 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 f(x) ND ND

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).

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.