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serg [7]
3 years ago
12

The function f(x)= -1/4(x) +1 is getting closer to a particular number. Its initial values are 4 and -2, what happens to the val

ues of the function's after 10 iterations. What number are the function's values getting closer to each time?
Mathematics
1 answer:
Basile [38]3 years ago
6 0

Answer:

After 10 iterations, the value is approximately the value it gets closer each time, which is 0.8 (or 4/5)

Step-by-step explanation:

If the inicial value is 4, we have:

1: f(4) = -1/4(4) + 1 = 0

2: f(0) = -1/4(0) + 1 = 1

3: f(1) = -1/4(1) + 1 = 3/4

4: f(3/4) = -1/4(3/4) + 1 = 13/16

5: f(13/16) = -1/4(13/16) + 1 = 51/64

6: f(51/64) = -1/4(51/64) + 1 = 205/256

7: f(205/256) = -1/4(205/256) + 1 = 819/1024

8: f(819/1024) = -1/4(819/1024) + 1 = 3277/4096

9: f(3277/4096) = -1/4(4) + 1 = 13107/16384

10: f(13107/16384) = -1/4(13107/16384) + 1 = 52429/65536 = 0.8

If the inicial value is -2, we have:

1: f(-2) = -1/4(-2) + 1 = 3/2

2: f(3/2) = -1/4(3/2) + 1 = 5/8

3: f(5/8) = -1/4(5/8) + 1 = 27/32

4: f(27/32) = -1/4(27/32) + 1 = 101/128

5: f(101/128) = -1/4(101/128) + 1 = 411/512

6: f(411/512) = -1/4(411/512) + 1 = 1637/2048

7: f(1637/2048) = -1/4(1637/2048) + 1 = 6555/8192

8: f(6555/8192) = -1/4(6555/8192) + 1 =  26213/32768

9: f(26213/32768) = -1/4(26213/32768) + 1 = 104859/131072

10: f(104859/131072) = -1/4(104859/131072) + 1 = 419429/524288 = 0.8

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