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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.

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## Applications of the First Derivative

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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 < x2 . Decreasing Function**How the first derivative can tell us where a function is**increasing and where a function is decreasing. The one feature shared by all of these tangent lines is that their slopes (first derivative of f) are negative. Decreasing Function First Derivative is negative.**Example 1: Determine the intervals where the function is**increasing and decreasing. Step 1: Find the derivative of f Step 2: Set the first derivative equal to 0 and solve for x.**We have 3 intervals**On this inverval f’ >0 On this inverval f’ <0 On this inverval f’ >0 -2 4 Which is determined by the graph. We can also determine the same result by testing: Test point inside interval -3. Test point inside interval 0. Test point inside interval 5.**Interpreting the results.**Since f’(x) > 0 on this interval, f(x) is increasing on this interval. Since f’(x) < 0 on this interval, f(x) is decreasing on this interval. Since f’(x) > 0 on this interval, f(x) is increasing on this interval.**Example 2: Find the intervals where the function**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 Set up intervals using the domain and the zeros of f’(x) Set f’(x) = 0 Next, find f’(x).**Now choose a test point contained inside the interval.**Intervals Test Point Substitution into f’(x) Result + - - + -2 -.5 .5 2 Interpretation of results: On , since f’(x)>0 then f(x) is increasing. On , since f’(x)<0 then f(x) is decreasing. On , since f’(x)<0 then f(x) is decreasing. On , since f’(x)>0 then f(x) is increasing.**-1**1 On , since f’(x)<0 then f(x) is decreasing. On , since f’(x)>0 then f(x) is increasing. On , since f’(x)>0 then f(x) is increasing. On , since f’(x)<0 then f(x) is decreasing.**Example 3: Find the intervals where the function**is increasing and where it is decreasing. Find f’(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.**Choose a test point:**Substitution into f’(x) Result Test Point -1 1 - + Interpretation of results: 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. 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?**Practice problem: Find the intervals where the following**function is increasing or decreasing. 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!**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. Now pick test points and determine the sign of f’ at that test point. Do this before you click the mouse again. Sign of f’ + - + Value in f’ 3(4) - 3 = 9 3(0) - 3 = -3 3(4) -3 = 9 Test Points -2 0 2 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. f’(-2) > 0 therefore f(x) is increasing on this interval f’(0) < 0 therefore f(x) is decreasing on this interval f’(2) > 0 therefore f(x) is increasing on this interval**c1 c2**Relative Extrema A function has a relative minimum at x = c if there exists an open interval (a,b) containing c such that 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 for all x in (a,b). f(c1) rel.max These are relative because they are maximums or minimums in a locale, not necessarily for the entire number line. f(c2) rel.min**c1 c2**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**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**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. 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. + - + -2 4 max min Test Points -3 0 5 =+ =- =+**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. The maximum is at (-2,60) Creates a max. 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)**+**- - + -1 0 1 Remember Example 2: Find the intervals where the function is increasing and where it is decreasing. We can use the work we did there for finding relative extrema -1 is a critical point and is a relative max 1 is a critical point and is a relative min**-1**1 (1,2) is a relative min (-1,-2) is a relative max**In Example 3**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 any, for 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?**Practice Problem: Find the relative max and relative min,**if any, for 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.**Practice Problem: Find the relative max and relative min,**if any, for 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? Although there are no critical points, you still set up the intervals using the undefined point x = -1 and create test points. Since there are no critical points, there are no relative extrema.**Practice Problem: Find the relative max and relative min,**if any, for 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? This is the correct conclusion concerning relative extrema. Since there are no critical points, there are no relative extrema.**Practice Problem: Find the relative extrema for**Which is the correct first step conclusion? The zeroes are 1,1 and -2 Substitute x = 1 Thus, x = 1 is a zero Using synthetic division to reduce the equation There are no critical points. To solve we will need to use the rational zero theory from precalculus which is: if 1 0 -3 2 1 +1 +1 -2 1 1 -2 0 The critical points are 1 and -2. continued continued**Practice Problem: Find the relative extrema for**If you said: How do you determine whether the critical points are a relative max or a relative min? 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. The zeroes are 1,1 and -2 The critical points are 1 and -2. 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: 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. The zeroes are 1,1 and -2 The critical points are 1 and -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.**Practice Problem: Find the relative extrema for**If you said: 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. - + + The zeroes are 1,1 and -2 Rel Min -2 1 Nothing Test -3 0 2 f’(test) 2(-27)-6(-3)+4=-32 2(0)-6(0)+4=4 2(8)-6(2)+4=8 The critical points are 1 and -2. 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**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. The zeroes are 1,1 and -2 The critical points are 1 and -2. 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.**Neither**Rel Min

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