Chapter 5 Section 5.1 First Derivative and Graphs Objectives: The student will be able to identify increasing and decreasing functions and local extrema The student will be able to apply the first derivative test The student will be able to apply the theory to applications in economics Increasing and Decreasing functions Theorem 1: For the interval (a,b) f (x) f(x) Graph of f + Increases Rises - Decreases Falls If a function rises from left to right on an interval then that function increases on that interval (x-axis) If a function falls from left to right on an interval then that function decreases on that interval (x-axis) If the graph of a function is horizontal on an interval then we say that function is constant on that interval. Ex: 1. 2. Test for Increasing and decreasing: f(x) is increasing where ever f (x) is positive, f (x) > 0 f(x) is decreasing where ever f (x) is negative, f (x) < 0 f(x) is constant where ever f (x) is zero, f (x) = 0 When a function changes from increasing to decreasing this creates a high point on the graph hill When a function changes from decreasing to increasing this creates a low point on the graph valley 1 P a g e
Such hills or valleys can occur in two ways 1. If the hill or valley is smooth and rounded, then the graph has a horizontal tangent line at the high or low point (zero derivative) 2. If the hill or valley is sharp and peaked, then the graph represents a function that is not differentiable at the high or low point. (derivative undefined or DNE). From Ex. #2 find the following: f (a) = f (b) = f (c) = f (d) = Ex: Find the intervals where f(x) = x 2 + 6x + 7 rising and falling. 2 P a g e
A partition number for the sign chart is a place where the derivative could change sign. Assuming that f is continuous wherever it is defined, this can only happen where f itself is not defined, where f is not defined or where f is zero Definition: The values of x in the domain of f where f (x) = 0 or does not exist are called the critical values (CV) of f. Let x be defined at x=c. If f (c) = 0 or if f (c) is undefined, then c is called a critical value 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. 3 P a g e
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When a function changes from increasing to decreasing, this creates a high point on the graph. High points on a graph are called relative maxima. When a function changes from decreasing to increasing, this creates a low point on the graph. Low points on a graph are called relative minima. High points (relative maxima) and low points (relative minima) only occur at x-values for which f (x) = 0 or f (x) is undefined. These are x-values are called critical values. **Caution** High points (relative maxima) and low points (relative minima) can only occur at critical values (CV), but the existence of a CV does not guarantee a relative maxima or minima there. Theorem: If f is continuous on a op[en interval (a,b), and c is a number in (a,b), and f(c) is a local extrema, then either f (c) = 0 or f (c) = DNE. That is, c is a critical point. Ex: 1. Find the intervals where ( ) rising and falling. Identify the relative extrema (max and min). Find the CV and discontinuities. 5 P a g e
2. Find the intervals where ( ) rising and falling. Identify the relative extrema (max and min). Find the CV and discontinuities. 3. Find the intervals where ( ) rising and falling. Identify the relative extrema (max and min). Find the CV and discontinuities. 6 P a g e
4. Find the intervals where ( ) rising and falling. Identify the relative extrema (max and min). Find the CV and discontinuities. First Derivative Test for relative extrema: Let c be a critical value of f. Construct a sign chart for f (x) close to and on either side of c. a. The function has a relative maxima at (c, f(c)) if the derivative is positive just left of x=c and negative just right of x=c. Ex. f(x) = -x 2 +2 b. The function has a relative minima at (c, f(c)) if the derivative is negative just left of x=c and positive just right of x=c. Ex. f(x) = 2x 2-4x+4 c. If the derivative does not change sign just left of x=c and just right of x=c, then (c, f(c)) is not a relative extrema. Ex. f(x) = (x-2) 1/3 f(x) left of c f(x) right of c f(c) Decreasing Increasing Local minima at c Increasing Decreasing Local maxima at c Decreasing Decreasing Not an extrema Increasing Increasing Not an extrema Polynomial Functions Theorem 3. If f (x) = a n x n + a n-1 x n-1 + + a 1 x + a 0, a n 0, is an n th -degree polynomial, then f has at most n x-intercepts and at most (n 1) local extrema. In addition to providing information for hand-sketching graphs, the derivative is also an important tool for analyzing graphs and discussing the interplay between a function and its rate of change. 7 P a g e
First Derivative Test Graphing Calculators Local extrema are easy to recognize on a graphing calculator. Method 1. Graph the derivative and use built-in root approximations routines to find the critical values of the first derivative. Use the zeros command under 2nd calc. Method 2. Graph the function and use built-in routines that approximate local maxima and minima. Use the MAX or MIN subroutine. Example: Preform Method 1 and 2 for f (x) = x 3 12x + 2. 8 P a g e
Section 5.2 - Second Derivative and Graphs Objectives: The student will be able to use concavity as a graphing tool. The student will be able to find inflection points. The student will be able to analyze graphs and do curve sketching. The student will be able to find the point of diminishing returns. Concavity The term concave upward (or simply concave up) is used to describe a portion of a graph that opens upward. Concave down(ward) is used to describe a portion of a graph that opens downward. Definition of Concavity A graph is concave up on the interval (a,b) if any secant connecting two points on the graph in that interval lies above the graph. It is concave down on (a,b) if all secants lie below the graph. 9 P a g e
Ex. Graph: A B C D The graph is concave up at (-, A)U(B,C)U(D, ) The graph is concave down at (A,B)U(C,D) Concavity Tests: The graph of a function f is concave upward on the interval (a,b) if f (x) is increasing on (a, b), and is concave downward on the interval (a, b) if f (x) is decreasing on (a, b). For y = f (x), the second derivative of f, provided it exists, is the derivative of the first derivative: The graph of a function f is concave upward on the interval (a, b) if f (x) is positive on (a, b), and is concave downward on the interval (a, b) if f (x) is negative on (a,b). Concavity: f(x) is concave up whenever f (x) is positive f(x) is concave down whenever f (x) is negative Procedure for deciding where a function is concave up or down 1. find potential points of inflection. That is find x-values where f (x) = 0 or f (x) = undefined (Note: not every PI will really be a point of inflection) 2. Use these values to split the x-axis into intervals 3. Choose a test value in each interval 4. Use the test value to decide if f (x) = + or for that interval 5. Decide on concavity 10 P a g e
Second Derivative Test Concavity Second derivative test is also a test for local min or max. This test can be applied on the critical values of the type where f (x) = 0, not to f (x) = undefined. Let c be a critical value of the function f(x), such that f (x) = 0, then. 1. if f (x) < 0, then f has a local max at x=c Concave down 2. if f (x) > 0, then f has a local min at x=c Concave up 3. if f (x) = 0, then the 2 nd derivative test is inconclusive. Ex. Find the intervals where the graph of f(x) = x 3 + 24x 2 + 15x 12 is concave up or concave down. 11 P a g e
Ex. Ex. Find the intervals where the graph of f(x) = 8x 2 2x 4 is concave up or concave down. Points of Inflection: Are those points where f(x) changes concavity. Points of inflection only occur at those x-values that make f (x) = 0 or f (x) = undefined An inflection point is a point on the graph where the concavity changes from upward to downward or downward to upward. This means that if f (x) exists in a neighborhood of an inflection point, then it must change sign at that point. Theorem 1. If y = f (x) is continuous on (a, b) and has an inflection point at x = c, then either f (c) = 0 or f (c) does not exist. The theorem means that an inflection point can occur only at critical value of f. However, not every critical value produces an inflection point. A critical value c for f produces an inflection point for the graph of f only if f changes sign at c, and c is in the domain of f. 12 P a g e
Summary Assume that f satisfies one of the conditions in the table below, for all x in some interval (a,b). Then the other condition(s) to the right of it also hold. Ex. Find the inflection points of f (x) = x 3 + 24x 2 + 15x 12 Solution: In example 1, we saw that f (x) was negative to the left of 8 and positive to the right of 8. At x = 8, f (x) = 0. This is an inflection point because f changes from concave down to concave up at this point. Inflection points can be difficult to recognize on a graphing calculator, but they are easily located using root approximation routines. For instance, the above example when f is graphed shows an inflection point somewhere between 6 and 10. Graphing the second derivative and using the zeros command on the calc menu shows the inflection point at 8 quite easily, because inflection points occur where the second derivative is zero. 13 P a g e
Curve Sketching Graphing calculators and computers produce the graph of a function by plotting many points. Although quite accurate, important points on a plot may be difficult to identify. Using information gained from the function and its dervative, we can sketch by hand a very good representation of the graph of f (x). This process is called curve sketching and is summarized on the example. Graphing Strategy Step 1. Analyze f (x). Find the domain and the intercepts. The x intercepts are the solutions to f (x) = 0, and the y intercept is f (0). Step 2. Analyze f (x). Find the partition points and critical values of f (x). Construct a sign chart for f (x), determine the intervals where f is increasing and decreasing, and find local maxima and minima. Step 3. Analyze f (x). Find the partition numbers of f (x). Construct a sign chart for f (x), determine the intervals where the graph of f is concave upward and concave downward, and find inflection points. Step 4. Sketch the graph of f. Locate intercepts, local maxima and minima, and inflection points. Sketch in what you know from steps 1-3. Plot additional points as needed and complete the sketch. Ex. Sketch the graph of y = x 3 /3 x 2 3x 14 P a g e
Analyzing Graphs - Applications A company estimates that it will sell N(x) units of a product after spending $x thousand on advertising, as given by N(x) = 2x 3 + 90x 2 750x + 2000 for 5 x 25 (a) When is the rate of change of sales, N (x), increasing? Decreasing? 15 P a g e
Point of Diminishing Returns If a company decides to increase spending on advertising, they would expect sales to increase. At first, sales will increase at an increasing rate and then increase at a decreasing rate. The value of x where the rate of change of sales changes from increasing to decreasing is called the point of diminishing returns. This is also the point where the rate of change has a maximum value. Money spent after this point may increase sales, but at a lower rate. The next example illustrates this concept. 16 P a g e
Section 5.5 - Absolute Maxima and Minima Objectives: The student will be able to identify absolute maxima and minima. The student will be able to use the second derivative test to classify extrema. Definition: f (c) is an absolute maximum of f if f (c) > f (x) for all x in the domain of f. f (c) is an absolute minimum of f if f (c) < f (x) for all x in the domain of f. If f is continuous on a closed interval [a,b] then f has both an absolute maxima and minima on the interval. Ex. Picture given in class Theorem 1. (Extreme Value Theorem) A function f that is continuous on a closed interval [a, b] has both an absolute maximum value and an absolute minimum value on that interval. Finding Absolute Maximum and Minimum Values Theorem 2. Absolute extrema (if they exist) must always occur at critical values of the derivative, or at end points. a. Check to make sure f is continuous over [a, b]. b. Find the critical values in the interval (a, b). c. Evaluate f at the end points a and b and at the critical values found in step b. d. The absolute maximum on [a, b] is the largest of the values found in step c. e. The absolute minimum on [a, b] is the smallest of the values found in step c. To find an absolute extrema on a closed interval [a,b] use the following steps: 1. Find the critical values of f in [a,b] Remember critical values (CV) are the values of x in the domain of f where f (x) = 0 or does not exist are called the critical values (CV) of f. 2. Evaluate f at each CV in [a,b] and at the ean points a and b. 3. Compare the values in step2 The largest value = absolute maxima of f in [a,b] The smallest value = absolute minima of f in [a,b] 17 P a g e
Ex. Find the absolute extrema of ( ) on the interval [-1, 2] Ex. Find the absolute extrema of ( ) on the interval [-1, 7] 18 P a g e
Second Derivative Test Theorem 3. Let f be continuous on interval (a,b) with only one critical value c in (a,b) If f (c) = 0 and f (c) > 0, then f (c) is the absolute minimum of f on I. If f (c) = 0 and f (c) < 0, then f (c) is the absolute maximum of f on I. Second Derivative and Extrema Ex: Find the local maximum and minimum values of ( ) on [ 1, 7]. Absolute Extrema in an Open Interval: Read text 19 P a g e
Section 5.6 - Optimization Objectives: The student will be able to calculate: Area and perimeter Revenue and profit Inventory control. Many practical problems require determining the maximum or minimum values. For example, business people wish to maximize profit and minimize cost. Builders wish to maximize the strength of their structures. The government has to be concerned about maximizing tax revenue, and the retailer wants to minimize inventory cost. Such problems are called optimization problems. They require the determination of absolute maxima or absolute minima. Steps: 1. Identify all given quantities and quantities to be determined. Draw a figure if possible. 2. Write a preliminary equation for the quantity that is to be maximized (or minimized) 3. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation. 4. Find critical values and locate absolute maxima and minima. (using second derivative test will often be faster that using the first derivative test) Ex: Using 120 feet of fencing, a farmer wishes to contain a cow in a rectangular plot of land that has one side along the bank of a river. (See the figure). If no fencing is needed along the river, what should be the dimensions of the rectangular field to provide the cox with the maximum grazing area? What is the maximum grazing area? 20 P a g e
Ex. Design a box with a square base and maximum volume using 216 square inches of cardboard. The box will have a top and a bottom. (See the figure). Your answer will give the length of the sides of the square base and the height of the box. 21 P a g e
Find the dimensions of a rectangular area of 225 square meters that has the least perimeter. 22 P a g e
Optimization Strategies 1. Introduce variables, look for relationships among these variables, and construct a math model of the form: Maximize (minimize) f (x) on the interval I. 2. Find the critical values of f (x). 3. Find the maximum (minimum) value of f (x) on the interval I. 4. Use the solution to the mathematical model to answer all the questions asked in the problem. Ex. A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are: C(x) = 60,000 + 60x p(x) = 200 x/50 for 0 x 6,000 a. Find the production level that will maximize the revenue, the maximum revenue, and the price that the company needs to charge at that level. b. Find the production level that will maximize the profit, the maximum profit, and the price that the company needs to charge at that level. 23 P a g e
Ex: A 300 room hotel in Las Vegas is filled to capacity every night at $80 per room. For each additional $1 increase in rent, 3 fewer rooms are rented. If each rented room cost $10 to service per day, how much should the management charge for each room to maximize gross profit? What is the maximum gross profit? 24 P a g e
Inventory Control A pharmacy has a uniform annual demand for 200 bottles of a certain antibiotic. It costs $10 per year for a storage place for one bottle, and $40 to place an order. How many times during the year should the pharmacy order the antibiotic in order to minimize total cost? Example: If you use 4 orders of 50 bottles each, you need 50 storage places. If you use 10 orders of 20 bottles each, you only need 20 storage places, but it costs more to order. Further Hint: Use $5 per storage place instead of $10. The reasoning is that some bottles will get used right away, and some need to be stored the full time. 25 P a g e