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Applications of the First Derivative By Dr. Julia Arnold using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text I recommend that you view the power point as a slide presentation first. After viewing then print the slides as notes according to the instructions below. Under view, click on slide show. To print the slides as notes go to the file command and select print. On the window that comes up toward the bottom it may say print what and next to that is a drop down menu. Select handouts and the number you want per page. Increasing and Decreasing Functions A function is increasing on an interval (a,b) if for any two numbers x1 and x2 in (a,b) f(x1) < f(x2) whenever x1 < x2 . Increasing Function A function is decreasing on an interval (a,b) if for any two numbers x1 and x2 in (a,b) f(x1) > f(x2) whenever x1 < x 2 . Decreasing Function How the first derivative can tell us where a function is increasing and where a function is decreasing. Decreasing Function First Derivative is negative. The one feature shared by all of these tangent lines is that their slopes (first derivative of f) are negative. Increasing function, First Derivative is positive. If the function is constant the first derivative will be 0. Example 1: Determine the intervals where the function is increasing and decreasing. f(x) x3 3x2 24x 32 Step 1: Find the derivative of f f(x) x 3x 24x 32 3 2 f(x) 3x2 6x 24 Step 2: Set the first derivative equal to 0 and solve for x. f(x) 3x2 6x 24 0 3x2 6x 24 0 3(x2 2x 8) 0 3(x 4)(x 2) x 4, x 2 We have 3 intervals -2 4 ,2 On this inverval f’ >0 2,4 On this inverval f’ <0 4, On this inverval f’ >0 Which is determined by the graph. We can also determine the same result by testing: ,2 Test point inside interval -3. 2 f(3) 3 3 6 3 24 3(9) 18 24 21 0 2,4 Test point inside interval 0. 4, Test point inside interval 5. 2 f(0) 3 0 60 24 24 0 2 f(5) 35 65 24 3(25) 30 24 21 0 Interpreting the results. ,2 Since f’(x) > 0 on this interval, f(x) is increasing on this interval. 2,4 Since f’(x) < 0 on this interval, f(x) is decreasing on this interval. 4, Since f’(x) > 0 on this interval, f(x) is increasing on this interval. 1 Example 2: Find the intervals where the function f(x) x x is increasing and where it is decreasing. First note that the domain does not contain 0. Why? Thus the intervals over which the function is defined are ,00, Next, find f’(x). 1 x 1 f(x ) 1 2 x f( x ) x Set f’(x) = 0 1 x2 0 x2 1 0 x 1x 1 x 1, x 1 0 1 Set up intervals using the domain and the zeros of f’(x) ,1 1,0 0,1 1, Now choose a test point contained inside the interval. Intervals ,1 1,0 0,1 1, Test Point -2 -.5 .5 2 Substitution into f’(x) 1 1 3 1 2 4 4 2 1 1 f(.5) 1 1 3 2 .25 .5 1 1 f(.5) 1 1 3 2 .25 .5 1 1 3 f(2) 1 2 1 4 4 2 f(2) 1 Interpretation of results: ,1 , since f’(x)>0 then f(x) is increasing. On 1,0 , since f’(x)<0 then f(x) is decreasing. On 0,1 , since f’(x)<0 then f(x) is decreasing. On 1, , since f’(x)>0 then f(x) is increasing. On Result + - + 0,1 , since On f’(x)<0 then f(x) is decreasing. -1 On ,1 , since f’(x)>0 then f(x) is increasing. On 1, , since f’(x)>0 then f(x) is increasing. 1 On 1,0 , since f’(x)<0 then f(x) is decreasing. Example 3: Find the intervals where the function f(x) x is increasing and where it is decreasing. Find f’(x). f(x) x 2 3 1 2 3 2 f (x ) x 3 3 3 x This fraction does not equal 0, and the domain of f’(x) does not contain 0, so to make the intervals, we will use the discontinuity at x = 0. ,0 0, 2 3 Choose a test point: Test Point ,0 0, Substitution into f’(x) -1 1 2 2 33 1 3 2 2 f(1) 3 3 1 3 f(1) Result + Interpretation of results: ,0 0, On this interval, f’(x)<0 so f(x) is decreasing. On this interval, f’(x)>0 so f(x) is increasing. f(x) is decreasing f(x) is increasing The Rule: 1.Find all values of x for which f’(x)=0 or f’ is discontinuous and identify the open intervals determined by these points. 2.Select a test point c in each interval found in step 1 and determine the sign of f’(c) in that interval. A. If f’(c) >0 in that interval, f is increasing on that interval. B. If f’(c) <0 in that interval, f is decreasing on that interval. Practice problem: Find the intervals where the following function is increasing or decreasing. f(x) x3 3x 4 The Rule: 1.Find all values of x for which f’(x)=0 or f’ is discontinuous and identify the open intervals determined by these points. Which of the following is the correct first step? f(x) x3 3x 4 f(x) x3 3x 4 f(x) 3x2 3 f(x) 3x2 3 0 3(x2 1) 0 3(x2 1) 0 x2 1 1 x 0 x2 1 1x 1,1 _ and _ 1, and _ 1, The _ open _ int ervals _ are _ ,1 The _ open _ int ervals _ are _ ,1 Practice problem: Find the intervals where the following function is increasing or decreasing. f(x) x3 3x 4 The Rule: 1.Find all values of x for which f’(x)=0 or f’ is discontinuous and identify the open intervals determined by these points. Which of the following is the correct first step? If you picked this one, you are on your way! f(x) x3 3x 4 f(x) 3x2 3 0 3(x2 1) 0 x2 1 1 x The _ open _ int ervals _ are _ ,1 1,1 _ and _ 1, 2.Select a test point c in each interval found in step 1 and determine the sign of f’(c) in that interval. A. If f’(c) >0 in that interval, f is increasing on that interval. B. If f’(c) <0 in that interval, f is decreasing on that interval. f(x) x3 3x 4 f(x) 3x2 3 1 x ,1 1,1 1, Now pick test points and determine the sign of f’ at that test point. Do this before you click the mouse again. Test Points -2 0 2 Value in f’ 3(4) - 3 = 9 3(0) - 3 = -3 3(4) -3 = 9 Sign of f’ + + Did you get the same result? f’(c) >0 means f’ is positive and f’(c) <0 means f’ is negative. Thus using A and B above what conclusion can you make about the function in the 3 intervals? Answer before you click the mouse. ,1 f’(-2) > 0 therefore f(x) is increasing on this interval 1,1 1, f’(0) < 0 therefore f(x) is decreasing on this interval f’(2) > 0 therefore f(x) is increasing on this interval Relative Extrema A function has a relative minimum at x = c if there exists an open interval (a,b) containing c such that f(x) f(c) for all x in (a,b). A function has a relative maximum at x = c if there exists an open interval (a,b) containing c such that f(x) f(c) for all x in (a,b). f(c1) rel.max c1 f(c2) rel.min c2 These are relative because they are maximums or minimums in a locale, not necessarily for the entire number line. To find the relative maximum and minimum, observe that the maximum and minimum represents places where the tangent line would be horizontal or have a slope of 0. Tangent lines c1 c2 Definition of Critical Point A critical point of a function f is any point x in the domain of f such that f’(x)=0 or f’(x) does not exist. Procedure for finding relative extrema The First Derivative Test 1. Determine the critical points of f. 2. Determine the sign of f’(x) to the left and right of each critical point. A. If f’(x) changes sign from positive to negative as we move across a critical point x = c, then f(c) is a relative maximum. B. If f’(x) changes sign from negative to positive as we move across a critical point x = c, then f(c) is a relative minimum. C. If f’(x) does not change sign as we move across a critical point x = c then f(c) is not a relative extremum. Find the relative maxima and relative minima of the function f(x) x3 3x2 24x 32 1. Determine the critical points of f. A critical point of a function f is any point x in the domain of f such that f’(x)=0 or f’(x) does not exist. f(x) x3 3x2 24x 32 f' (x) 3x2 6x 24 0 3(x2 2x 8) 0 3(x 4)(x 2) x 4, x 2 Since 4, and -2 are in the domain of f and cause f’ to be 0, they are critical points of f. 2. Determine the sign of f’(x) to the left and right of each critical point. + Test Points -2 max -3 + 4 min 0 5 f' (3) 3 3 6 3 24 27 18 24 21 2 f' (0) 30 60 24 24 2 =- f' (5) 35 65 24 75 30 24 21 2 =+ =+ f(x) x3 3x2 24x 32 x = -2 is a relative maximum because as you check the signs across the critical point from left to right you go from + to which means increasing then decreasing. Creates a max. The maximum is at (-2,60) f(2) 2 3 2 24 2 32 8 12 48 32 60 3 2 x = 4 is a relative minimum because as you check the signs across the critical point from left to right you go from - to + which means decreasing then increasing. Creates a min. The minimum is at (4,-48) f(4) 4 34 244 32 64 48 96 32 48 3 2 Remember Example 2: Find the intervals where the function 1 f(x) x is increasing and where it is decreasing. x We can use the work we did there for finding relative extrema + -1 -1 is a critical point and is a relative max 0 + 1 1 is a critical point and is a relative min (1,2) is a relative min -1 1 (-1,-2) is a relative max In Example 3 f(x) x f(x) x 2 3 2 3 1 2 2 f(x) x 3 3 3 3 x The derivative is undefined at 0 but 0 is in the domain of f, therefore 0 is a critical point. - + 0 From earlier work we see that the signs tell us 0 is a relative minimum. f(x) is decreasing f(x) is increasing (0,0) is a relative minimum Practice Problem: Find the relative max and relative min, if x any, for h(x ) x 1 Procedure for finding relative extrema The First Derivative Test 1. Determine the critical points of f. 2. Determine the sign of f’(x) to the left and right of each critical point. Which of the following is the correct first step? x x 1 x 11 x 1 1 h(x) x 12 x 12 1 0 _ has _ no _ solutions 2 x 1 h( x ) x 1 _ makes _ f' (x) _ discontinuous, x x 1 x 11 x 1 1 h(x) x 12 x 12 1 0 _ has _ no _ solutions 2 x 1 therefore x 1 _ makes _ f' (x) _ discontinuous. it _ is _ a _ cri tical_ po int . There _ are _ no _ critical_ po int . h(x) Practice Problem: Find the relative max and relative min, if x any, for h(x ) x 1 Procedure for finding relative extrema The First Derivative Test 1. Determine the critical points of f. 2. Determine the sign of f’(x) to the left and right of each critical point. Which of the following is the correct first step? This is correct because there are no zeroes and while x = -1 makes h’ discontinuous, x = -1 is not in the domain of h. Definition of Critical Point A critical point of a function f is any point x in the domain of f such that f’(x)=0 or f’(x) does not exist. x x 1 x 11 x 1 1 h(x) x 12 x 12 1 0 _ has _ no _ solutions 2 x 1 h(x) x 1 _ makes _ f' (x) _ discontinuous. There _ are _ no _ critical_ po int . Practice Problem: Find the relative max and relative min, if x any, for h(x ) x 1 Procedure for finding relative extrema The First Derivative Test 1. Determine the critical points of f. 2. Determine the sign of f’(x) to the left and right of each critical point. Which of the following is the correct next step? Since there are no critical points, there are no relative extrema. Although there are no critical points, you still set up the intervals using the undefined point x = -1 and create test points. ,1 1, Practice Problem: Find the relative max and relative min, if x any, for h(x ) x 1 Procedure for finding relative extrema The First Derivative Test 1. Determine the critical points of f. 2. Determine the sign of f’(x) to the left and right of each critical point. Which of the following is the correct next step? Since there are no critical points, there are no relative extrema. This is the correct conclusion concerning relative extrema. Practice Problem: Find the relative extrema for f(x) 1 4 x 3x2 4x 8 2 Which is the correct first step conclusion? 1 4 x 3x2 4x 8 2 f(x) 2x3 6x 4 f(x) 0 2x3 6x 4 0 x 3x 2 3 To solve we will need to use the rational zero theory from precalculus which is: if p _ is _ a _ factor _ of _ 2 p 1,2 q _ is _ a _ factor _ of _ 1 p 1 p 1,2 q continued Substitute x = 1 13 3(1) 2 0 The zeroes are 1,1 and -2 Thus, x = 1 is a zero Using synthetic division to reduce the equation 1 0 -3 2 1 +1 +1 -2 1 1 -2 0 x2 x 2 0 x 2x 1 0 x 2, x 1 continued There are no critical points. The critical points are 1 and -2. Practice Problem: Find the relative extrema for If you said: 1 4 x 3x2 4x 8 2 f(x) 2x3 6x 4 f(x) f(x) 1 4 x 3x2 4x 8 2 How do you determine whether the critical points are a relative max or a relative min? 0 2x3 6x 4 0 x3 3x 2 The zeroes are 1,1 and -2 The critical points are 1 and -2. A. Set up intervals on the number line and test points. If you go from + to across the number then the number is a rel max. If you go from - to + across the number then the number is a rel min. B. Substitute 1 and -2 into f(x), whichever is larger is the max and whichever is smaller is the min. Practice Problem: Find the relative extrema for If you said: 1 4 x 3x2 4x 8 2 f(x) 2x3 6x 4 f(x) 0 2x3 6x 4 0 x 3x 2 3 The zeroes are 1,1 and -2 The critical points are 1 and -2. f(x) 1 4 x 3x2 4x 8 2 How do you determine whether the critical points are a relative max or a relative min? A is the correct answer A. Set up intervals on the number line and test points. If you go from + to across the number then the number is a rel max. If you go from - to + across the number then the number is a rel min. Doing B may work some of the time. B. Substitute 1 and -2 into f(x), whichever is larger is the max and whichever is smaller is the min. Practice Problem: Find the relative extrema for If you said: 1 4 x 3x2 4x 8 2 f(x) 2x3 6x 4 f(x) 0 2x3 6x 4 The critical points are 1 and -2. 1 4 x 3x2 4x 8 2 A. Set up intervals on the number line and test points. If you go from + to across the number then the number is a rel max. If you go from - to + across the number then the number is a rel min. 0 x3 3x 2 The zeroes are 1,1 and -2 f(x) - + Rel Min ,2 2,1 1, Test -3 0 2 -2 + 1 Nothing f’(test) 2(-27)-6(-3)+4=-32 2(0)-6(0)+4=4 2(8)-6(2)+4=8 There is a relative minimum at x = -2. X = 1 is neither a relative max or a relative min. Practice Problem: Find the relative extrema for 1 4 x 3x2 4x 8 2 f(x) 2x3 6x 4 f(x) 0 2x 6x 4 3 0 x3 3x 2 The zeroes are 1,1 and -2 The critical points are 1 and -2. f(x) 1 4 x 3x2 4x 8 2 Doing B may work some of the time. B. Substitute 1 and -2 into f(x), whichever is larger is the max and whichever is smaller is the min. 1 4 1 312 4(1) 8 6.5 2 1 4 2 f(2) 2 3 2 4(2) 8 20 2 f(1) Which would lead you to conclude that 1 is a relative max and -2 is a relative min which would only be half right. Finally, let’s look at the graph and see what’s happening. 4y 3 2 1 4 3 2 Rel Min 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 1 2 Neither 3 4 x 5 Go to the HW