Critical and Inflection Points

Transcription

1 Critical and Inflection Points 1 Finding and Classifying Critical Points A critical point is a point on the graph where the tangent slope is horizontal, (0) or vertical, ( ). or not defined like the minimum of y = x.

2 x 4 Some people call critical points stationary points or extrema but we will always call them critical points. First what s so special about critical points? Why are they critical? They are critical because they could be maximum or minimum points. If the tangent slope at a 2

3 point is horizontal, that is, y 0 =0, then the curve has one of four shapes near that critical point. You can see that the first two situations are a maximum and minimum respectively. The bottom two situations are 3

4 neither, they are just places on the graph with a horizontal tangent slope. If the tangent slope at the critical point is vertical, that is, y 0 =, then there are four possible shapes near the critical point. The top two are called 4

5 cusps. The first one has a critical point that is a maximum and has a vertical tangent slope. The second is a minimum and also has a vertical tangent slope. The bottom two are just critical points where the tangent slope is vertical. They are neither maximum nor minimum. Ok, let s do the firstpartof this activity, find the critical points. All we have to do is take the derivative of the function in question and see 5

6 what x 0 s make the tangent slope 0 or. That is, set the derivative equal to zero, or see where the derivative is undefined. By undefined we usually mean the denominator of a fraction is zero. Let s do some examples to flesh this out. Example Q? Find the critical points of the function y = 2x 3 9x 2 +12x +6. A. Take the derivative 6

7 of the function to get y 0 =6x 2 18x +12 Now, the derivative or tangent slope function is a polynomial so y 0 is never going to be. Set y 0 =0and solve. 0=6x 2 18x +12 =6(x 1)(x 2) Therefore at the points where x =1and x =2the tangent slope is horizontal. This gives us the x-values of the two critical points. What are the y-values? We get the y-values by plugging the x-values into 7

8 the original equation, y = 2x 3 9x 2 +12x +6. Thus the two critical points are (1, 11) and (2, 10). Now notice the graph of the function from Example and spot the critical points x Example Q? Find the critical 8

9 points of the function y = x3 2 2x 2 A. Take the derivative y 0 = x 1 2 x 1 2 and then factor out the x to the lowest power, in our case, x 1 2. i y 0 = x 1 2 hx h i x = x 1 2 We can see that if x =0,then the denominator of the above fraction would be zero, but x =0is a defined point on the 9

10 graph. Plug x =0into the original equation and see that the graph goes through the origin. We conclude that at the critical point where x =0we have a vertical tangent slope. To check for horizontal tangent slopes we set y 0 =0and solve. Note that for a fraction to be zero, the top part must be 10

11 zero. h i y 0 x = x 1 2 h i x = x 1 2 0=x =x 1 2 x =1 Therefore (1, 4 3 ) is a critical point where the tangent slope is horizontal. When we look at the graph of the function from Example we can see the critical points but we also notice that there 11

12 are no points where the x- values are negative. Why? x Because the domain of the function is [0, ). Look back to the beginning of the term to refresh your recollection of range and domains. Ok, now we can find critical points. The next step is to classify them. We need to know whether a critical point 12

13 is a local maximum local minimum or neither To do this we will draw a slope line. This is an important illustrative tool. After the slope line is drawn we will use the First Derivative Test to classify the critical points. There is a Second Derivative Test that we will learn later Example Q? Classify the critical 13

14 points of the function y = 2x 3 9x 2 +12x First Derivative Test After we draw the slope line we can use these patterns to classify the critical points /-\ indicates a maximum \-/indicates a minimum 14

15 \ / indicates a cusp that s a minimum / \ indicates a cusp that s a maximum /-/ indicates a critical point that is neither a maximum nor a minimum \- \ indicates a critical point that is neither a maximum 15

16 nor a minimum / / indicates a critical point that is neither a maximum nor a minimum \ \indicates a critical point that is neither a maximum nor a minimum 16

17 Steps for drawing a Slope Line 1. Find all the critical points of the function. 2. Draw a number line with the critical points on it and their slopes represented above, either as a horizontal or vertical line. 3. Pick test points ( x-values ) in between the critical points andplugthemintothefirst derivative to tell if the tangent slope is positive or negative in that interval. 17

18 4. Add those slopes to the number line to represent the tangent slopes in the intervals 5. Use the First derivative test to classify the critical points. This concept of a slope line can answer many questions concerning a particular function. Wehavealreadyusedit to classify critical point with the First Derivative Test. We can also use the slope line to find intervals of increase or decrease, that is, intervals 18

19 where the graph is increasing or decreasing respectively. Example Q? Find the intervals of increase and decrease of the function y =2x 3 9x 2 +12x +6 from Example. Another benefit of the slope line is that we get a rough idea of the shape of the graph. The slope line shown in Example implies the curve increases until x =1where it reaches a peak and drops down until x =2where it bottoms out 19

20 and increases continually after x =2. This implication is confirmed by the actual graph shown after Example. Local versus Global We use the terms local and global to describe maximum or minimum points. All maximum or minimum points are local maximums or minimums respectively. They are the highest (or lowest) spot on the graph in their immediate vicinity. The highest point on the 20

21 whole graph is the global maximum. Likewise, the lowest point on the whole graph is the global minimum. 2 Concavity and Points of Inflection The first derivative is the tangent slope formula, the second derivative is the concavity formula. Concavity is the bend in the curve. Positive concavity or concave up occurs when y 00 > 0 and negative concav- 21

22 ity or concave down occurs when y 00 < 0. There can be any combination of slope and concavity. Positive concavity with positive slope, y 00 > 0,y 0 > 0 Positive concavity with negative slope, y 00 > 0,y 0 < 0 Negative concavity with positive slope, y 00 < 0,y 0 > 0 22

23 Negative concavity with negative slope, y 00 < 0,y 0 < 0 A good way to remember the difference between positive concavity and negative concavity is this cheesy child s drawing. Anyway, just like when we drew the slope line, we draw a concavity line. The steps 23

24 to make it are similar to the slope line. Get the second derivative of the function and find the x-values that make y 00 zero or undefined and mark those points on your concavity line. These points are called possible points of inflection. Use test points on either sides of these possible points of inflection to determine the concavity of the intervals. Wait you say. What s an inflection point? Definition 1 An inflection point is 24

25 a point on the graph where the concavity changes from concave up to concave down or from concave down to concave up. So a possible point of inflection still has to prove itself to be a point of inflection by having different concavities on either side of it. Example Q? Find the intervals of positive concavity and the intervals of negative concavity 25

26 of the function y = x5 20 x x and any inflection points. A. We can see the concavity in the graph of the function from Example x Oh, I promised we would learn the Second Derivative Test, eh. Here it is, as 26

27 promised. A point on a curve with a horizontal tangent slope and negative concavity must be a maximum. Likewise a point with a horizontal tangent slope and positive concavity must be a minimum. This thinking forms the basis for the second derivative 27

28 test. First find the critical points, then test them with the: 2.1 Second Derivative Test To test a critical point x = a, y 0 (a) =0to classify it as a maximum or a minimum: If y 00 (a) < 0 then x = a is a maximum point If y 00 (a) > 0 then x = a is a minimum point That s it. This test is used frequently in economics. So 28

29 nowwehavetwowaysto classify critical points. My personal favorite is the First Derivative Test. It s much better. You only have to take one derivative. Notice the Second Derivative Test is only good for critical points where the slope is horizontal and there is no conclusion to be made when y 00 (a) =0. Then what? You have to do the First Derivative Test anyway. Also, you can get so much more information from the 29

30 slope line than you can with the Second Derivative Test. I ll show you an example of the inferior Second Derivative Test just for completeness. Example Q? Find and classify the critical points of the curve y = x 4 2x Homework Section 2.10 #10, 13, 14-18, 21, 22 Submit 10, 16, 22 Section 3.4 #25-29 Submit 26 30

31 Section 3.5 #49, 50 Section 3.6 #3-12 Submit 4, 6, 10 31