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§4.1 Graphing Using the First and Second Derivatives. The student will learn about relative extreme points and critical numbers, and the first derivative test, and concavity, inflections points, and the second derivative test. 1 Introduction We are going to use derivatives in two ways. First, in this section, we will look at how the derivative helps us to graph a function. Then, in the next section, we will look at how the derivative and graph help us to optimize a function. That means we will find the maximums and/or minimums of a function. Could be pretty useful stuff!! 2 Relative Extreme Points and Critical Numbers f has a relative maximum at c if f (c) ≥ f (x) for all x near c. f has a relative minimum at c if f (c) ≤ f (x) for all x near c. The relative maximums and relative minimums are called collectively relative extreme points. 3 Critical Values Definition. The values of x in the domain of f where f ’(x) is 0 or does not exist are called the critical values of f. f′ (x) = 0 or f′ (x) is undefined A critical value for x tells us that something important is happening. 4 First – A Basic Fact about Derivatives For the interval (a, b) f ' (x) f (x) Graph of f + Increasing Rising - Decreasing Falling 0 Flat Steady REMEMBER! The derivative describes the slope and when the slope is positive the curve is rising and when the slope is negative the curve is falling. 5 Graphing Strategies for y = f (x). Step 1: Analyze f (x): [Graphing calculator in blue.] • Find the domain of f. Remember the two step process, assume the domain is all real numbers and delete division by 0 and the even roots of negative numbers. (Check graph and table.) • Find the x intercepts. Remember that this occurs at y = 0. (Zeros or roots.) • Find the y intercept. Remember that this occurs at x = 0. (Value with x = 0.) • Find the vertical asymptotes. (Check graph and table.) This is algebra! 6 Graphing Strategies for y = f (x). Step 2: Analyze f ′ (x): [Graphing calculator in blue.] Today's Lesson • Find any critical values for f ′ (x). • Construct a sign chart for f ′ (x) to determine the intervals where f (x) is increasing and decreasing. (Use the graph of f ′ (x) OR f (x).) • Find local maxima and minima. (Use the graph of f (x).) 7 Graphing Strategies for y = f (x). OR USE YOUR GRAPHING CALCULATOR. • Make sure you know what you are looking at. 8 Example 1 Find the intervals where f (x) = x2 + 6x + 7 is rising and falling. The function will be rising when the derivative is positive. WHY? f ’ (x) = 2x + 6 so then 2x + 6 > 0 2x + 6 > 0 when 2x > - 6 or x > - 3 So the graph is rising when x > - 3. - 3 is a critical value. f (x) will be falling when f ’ (x) = 2x + 6 < 0 OR 2x + 6 < 0 when 2x < - 6 or x < - 3 So the graph is falling when x < - 3. Continued 9 Example 1 continued f (x) = x2 + 6x + 7 Sign charts of f ’ are helpful. f ’ (x) = 2x + 6 (- , -3) f ’ (x) f (x) (-3, ) - - - - - - 0 + + + + + + Decreasing -3 - 3 is a critical Increasing value. f (x) = x2 + 6x + 7 -3 10 Example 2 f (x) = 1 + x3 f ’(x) = 3x2 Critical value at x = 0. (- , 0) f ’ (x) f (x) (0, ) + + + + + 0 + + + + + + Increasing 0 Increasing f (x) = 1 + x3 11 0 Example 3 f (x) = (1 – x)1/3 f ’(x) = 1 2 3 1 x 3 Critical value at x = 1. f ’ (x) (- , 1) (1, ) - - - - - - ND - - - - - - f (x) Decreasing 1 Decreasing f (x) = (1 – x)1/3 1 12 Example 4 f (x) = 1/(1 – x). f ’(x) = 1/(1 – x)2 Critical value at x = 1. f ’ (x) (- , 1) (1, ) + + + + + ND + + + + + f (x) Increasing 1 Increasing f (x) = 1/(1 – x). 1 13 First Derivative Test Let c be a critical value of f (i.e. f (c) is defined and either f ’ (c) = 0 or f ’ (c) is not defined). Construct a sign chart for f ’ (x) close to and on either sign of c. f (x) left of c f (x) right of c f (c) Decreasing Increasing Local minimum at c y’ is negative Increasing y’ is positive Decreasing y’ is positive y’ is negative Decreasing Decreasing y’ is negative Increasing y’ is positive Local maximum at c Not an extremum y’ is negative Increasing y’ is positive Not an extremum 14 First Derivative Test Graphing Calculators Local extrema are easy to recognize on a graphing calculator. • Graph the derivative and use built-in root approximations routines to find the critical values. Use the zeros subroutine. • Graph the function and use built-in routines that approximate local maxima and minima. Use the MAX or min subroutine. 15 * * Example * * Example. f (x) = x3 – 12x + 2. f ‘ (x) = 3x2 – 12. Graph this derivative and look for critical values f (x) = x3 – 12x + 2. Graph this and look for maxima and minima. -10 ≤ x ≤ 10 and -15 ≤ y ≤ 20 -10 ≤ x ≤ 10 and -15 ≤ y ≤ 20 f ‘ (x) + + + + + 0 - - - 0 + + + + + increases decrease f (x) increases f (x) increases decrs increases Critical values at –2 and 2 Maximum at - 2 and minimum at 2. 16 Summary. • We can use the first derivative test to determine where relative maxima or minima occur. • Use the first derivative to determine where the critical values are located. • Then examine the function on either side of these critical values to find if there is a max or min at the critical value. 17 More Let’s continue this study of using derivatives to tell us about the original function by examining what the second derivative tells us. Concavity. The term concave upward is used to describe a portion of a graph that opens upward. Concave downward is used to describe a portion of a graph that opens downward. Concave down Concave up 19 More 20 Summary of Facts f is increasing f ‘ (x) > 0 f is decreasing f ‘ (x) < 0 f is constant f ‘ (x) = 0 f is concave up f ‘ (x) increasing f “ (x) > 0 f concave down f ‘ (x) decreasing f “ (x) < 0 Inflection point f ‘ (x) is constant f “ (x) = 0 21 Example 2 Find the inflection points of f (x) = x3 + 3x2 - 9x + 5 f ‘ (x) = 3x2 + 6x – 9 = 3 ( x 2 + 2x – 3) = 3 (x + 3)(x – 1) with critical values at x = - 3 and x = 1 f “ (x) = 6x + 6 with a critical value at x = - 1. Sign charts for f ‘ (x) and f “ (x) may be useful. Next slide please. 22 What It All Means f ‘ (x) = 3x2 + 6x – 9 = 3 (x + 3)(x – 1) with critical values at x = - 3 and x = 1 f “ (x) = 6x + 6 with a critical value at x = - 1. f “ (x) - - - - - - - 0 + + + + + + + + f ’ (x) + + + 0 - - - - - -3 f (x) increasing f (x) f (x) -1 1 decreasing Maximum 0 + + + + + increasing minimum concave down - inflect - concave up 23 What It All Means f (x) increasing f (x) f (x) -3 -1 decreasing Maximum 1 increasing minimum concave down - inflect - concave up 24 Note: It is difficult to find the inflection point on a calculator. You can graph y’ = 3x 2 + 6x - 9 and find where the max or min are located. You can graph y” = 6x + 6 and find the x-intercepts. 25 Second Derivative Test Let c be a critical value for f ' (x), f ' (c) f '' (c) Graph of f is f (c) 0 + Concave up Local minimum 0 - Concave down Local maximum 0 0 ? Test fails 26 Summary. • We can use the first derivative test to determine where relative maxima or minima occur. • Use the first derivative to determine where the critical values are located. • Then examine the function on either side of these critical values to find if there is a max or min at the critical value. 27 Summary. • We can use the second derivative to determine when a function is concave up or concave down. • When the second derivative is zero we get an inflection point in f (x). • The second derivative test may be used to determine if a point is a local maxima or minima. 28 ASSIGNMENT §4.1; Page 76; 1 - 9, odd. 29