Question

Can anyone solve functions

Answer 1

10

`f(x) = (x-1)/(x) = 1 - (1)/(x)`

`f(f(x)) =1 - (1)/(1 - (1)/(x)) = 1 - (x)/(x-1) = (x-1-x)/(x-1)=(1)/(1-x)`

`f(f(f(x)) = 1 - (1)/((1)/(1 - x)) = x`

`f^4(x) = 1-(1)/(x) =f(x)`

`f^(3k)(x)=x`

`2017 = 3(672)+1`

`f^(2017)(x) = f(f^(3(672))(x))=f(x)=(x-1)/(x)`

11

f(x)=f(x+1)+f(x-1)

f(20)=15

f(15) - 5 = 15

f(15) = 20

f(16)=f(15)+f(17)=20 +f(17)

f(17)=f(16)+f(18)= 20+f(17)+f(18)

f(18)=-20

f(18)=f(17)+f(19)=-20

f(19)=f(20)+f(18)=15 + f(18)=15 + -20 = -5

f(19) = -5

-20 = f(17) + -5

f(17) = -15

f(16) = 20 + -15= 5

f(16) = 5

Check:

f(15)=20, f(16)=5, f(17)=-15, f(18)=-20, f(19)=-5, f(20)=15

Each has to be the sum of the neighbors.

5=20+-15, -15=5+-20, -20=-15+-5, -5=-20+15, all good

f(x)=f(x+1)+f(x-1)

f(x+1)=f(x)+f(x+2)

f(x)=f(x)+f(x+2)+f(x-1)

f(x+2)=-f(x-1)

f(x+3)=-f(x)

f(x)=g[x mod 6]

where g is the the table

g(0)=-20, g(1)=-5, g(2)=15, g(3)=20, g(4)=5, g(5)=-15

20152015 = 1 mod 6

f(20152015) = g(1) = -5

12

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

2f(0) = f(0)

f(0) = 0

f(1) = 1

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

2f(1) = (1/2) f(2)

f(2) = 4 f(1)

f(2) = 4

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

8 = (1/2) f(4)

f(4) = 16

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

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

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

f(3) = (1/2) 16 + (1/2) 4 - 1 = 8 + 2 - 1 = 9

f(3) = 9

13

P(x) = x^2 + 2007 x + 1

Let x=P(x)

x = x^2 + 2007 x + 1

0 = x^2 + 2006 x + 1

`x = -1003 \pm √(1003^2-1) = -1003 \pm 2 √(251502)`

Two real fixed points. P(x)=x

Given any x such that P(x)=x, then P(P(x))=x, and it's still x after composing P n times.

Answer 2

10) The only thing I can provide to help you is:

f°ⁿ(x) = f(f°ⁿ⁻¹(x))

Hopefully, that can help you to figure out the answer

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11) Answer: f(15) = 20, f(16) = 5, f(17) = -15, f(18) = -20, f(19) = -5

Step-by-step explanation:

f(x) = f(x + 1) + f(x - 1) f(20) = 15

f(15) - 5 = 15 ⇒ f(15) = 20

f(16) = f(16 + 1) + f(16 - 1)

f(16) = f(17) + f(15)

f(16) = f(17) + 20

f(17) = f(17 + 1) + f(17 - 1)

f(17) = f(18) + f(16)

f(17) = f(18) + f(17) + 20

0 = f(18) + 20

-20 = f(18)

f(19) = f(19 + 1) + f(19 - 1)

f(19) = f(20) + f(18)

f(19) = 15 + -20

f(19) = -5

f(18) = f(18 + 1) + f(18 - 1)

f(18) = f(19) + f(17)

-20 = -5 + f(17)

-15 = f(17)

f(16) = f(17) + 20

f(16) = -15 + 20

f(16) = 5

*****************************************************************************************

12a) Answer: 0

Step-by-step explanation:

f(m - n) + f(m + n) =
`(1)/(2)`f(2m) +
`(1)/(2)`f(2n)

• m - n = 0 ⇒ m = n

f(0) + f(2m) =
`(1)/(2)`f(2(m)) +
`(1)/(2)`f(2(m))

f(0) + f(2m) = f(2m)

f(0) = 0

*****************************************************************************************

12b) Answer: f(2) = 4, f(3) =
`(1)/(2)`f(4) + 1

Step-by-step explanation:

f(m - n) + f(m + n) =
`(1)/(2)`f(2m) +
`(1)/(2)`f(2n)

Let m = 1 and n = 0, then

f(1 - 0) + f(1 + 0) =
`(1)/(2)`f(2(1)) +
`(1)/(2)`f(2(0))

f(1) + f(1) =
`(1)/(2)`f(2) +
`(1)/(2)`f(0)

2f(1) =
`(1)/(2)`f(2) +
`(1)/(2)`(0)

2f(1) =
`(1)/(2)`f(2)

2[1] =
`(1)/(2)`f(2)

(2)2 = (2)
`(1)/(2)`f(2)

4 = f(2)

Let m = 2 and n = 1, then

f(2 - 1) + f(2 + 1) =
`(1)/(2)`f(2(2)) +
`(1)/(2)`f(2(1))

f(1) + f(3) =
`(1)/(2)`f(4) +
`(1)/(2)`f(2)

1 + f(3) =
`(1)/(2)`f(4) +
`(1)/(2)`(4)

1 + f(3) =
`(1)/(2)`f(4) + 2

f(3) =
`(1)/(2)`f(4) + 1