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Journal 1/27/16 I’m a good person. Shouldn’t that be enough for me to get into heaven? To learn about max and min numbers on closed intervals pp 329: 1,

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Presentation on theme: "Journal 1/27/16 I’m a good person. Shouldn’t that be enough for me to get into heaven? To learn about max and min numbers on closed intervals pp 329: 1,"— Presentation transcript:

1 Journal 1/27/16 I’m a good person. Shouldn’t that be enough for me to get into heaven? To learn about max and min numbers on closed intervals pp 329: 1, 2, 3, 4, 5 Objective Tonight’s Homework

2 Homework Help Let’s spend the first 10 minutes of class going over any problems with which you need help.

3 Notes on the Closed Interval Theorem Let’s imagine we take a snippet of a function: The critical number theorem says that as long as this function is continuous in this span, we can say something about where to find maxima or minima. - At an endpoint - There is none because the function is flat - Somewhere in between on the span

4 Notes on the Closed Interval Theorem Example: A 10-inch-long string is to be cut into two pieces. One of the pieces will be bent to form a square, and the other piece will be formed into a circle. Find where the string should be cut to maximize the combined area of both.

5 Notes on the Closed Interval Theorem Example: A 10-inch-long string is to be cut into two pieces. One of the pieces will be bent to form a square, and the other piece will be formed into a circle. Find where the string should be cut to maximize the combined area of both. x 10 – x We need to get the area of both shapes in terms of just x.

6 Notes on the Closed Interval Theorem x 10 – x We need to get the area of both shapes in terms of just x. Circle x = 2 π r r = x/2 π Area = π (x/2 π ) 2

7 Notes on the Closed Interval Theorem x 10 – x We need to get the area of both shapes in terms of just x. CircleSquare x = 2 π r4s = 10 – x r = x/2 π s = (10 – x)/4 Area = π (x/2 π ) 2 Area = ((10 – x)/4) 2

8 Notes on the Closed Interval Theorem x 10 – x We need to get the area of both shapes in terms of just x. CircleSquare x = 2 π r4s = 10 – x r = x/2 π s = (10 – x)/4 Area = π (x/2 π ) 2 Area = ((10 – x)/4) 2 Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2

9 Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0.

10 Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95

11 Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95 f’(x) = ((1/2 π ) + (1/8))x – 5/4 0 = ((1/2 π ) + (1/8))x – 5/4 x = 4.399

12 Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95 f’(x) = ((1/2 π ) + (1/8))x – 5/4 0 = ((1/2 π ) + (1/8))x – 5/4 x = 4.399(plug this back in to f(x)) x = 4.399Area = 3.5

13 Notes on the Closed Interval Theorem Total Area = π (x/2 π ) 2 + ((10 – x)/4) 2 We can use the closed interval theorem to find the max area. This theorem tells us that it will either be an endpoint (x=0, or x=10), or will happen when f’(x) = 0. x = 0Area = 100/16 = 6.25 x = 10Area = 100/4 π = 7.95  max area f’(x) = ((1/2 π ) + (1/8))x – 5/4 0 = ((1/2 π ) + (1/8))x – 5/4 x = 4.399(plug this back in to f(x)) x = 4.399Area = 3.5

14 Group Practice Look at the example problems on pages 327 and 328. Make sure the examples make sense. Work through them with a friend. Then look at the homework tonight and see if there are any problems you think will be hard. Now is the time to ask a friend or the teacher for help! pp 329: 1, 2, 3, 4, 5

15 Exit Question How do we define a "closed interval"? a) A segment on which our function is continuous b) Something like a circle or square where it connects c) An integral d) We don’t. “Closed interval” doesn’t e) All of the above f) None of the above


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