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# 1. Let f(x) = x^2 - 2x. Find all real numbers x such that f(x) = f(f(x)).

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1. Let f(x) = x^2 - 2x. Find all real numbers x such that f(x) = f(f(x)).

2. The function f(x,y) accepts an ordered pair as input and gives another ordered pair as output. It is defined according to the following rules: If x > 4, f(x,y) = (x - 4,y). If \$x \le 4 but \$y > 4, f(x,y) = (x,y - 4). Otherwise, f(x,y) = (x + 5, y + 6). A robot starts by moving to the point (1,1). Every time it arrives at a point (x,y), it applies f to that point and then moves to f(x,y). If the robot runs forever, how many different points will it visit?

3. Let f(x) = px + q, where p and q are real numbers. Find p+q if \$f(f(f(x))) = 8x + 21.

Thanks so much! :D

edit: if you could post answers as you figure them out that would be greatly appreciated as Im kind of in a rush thanks :D

WhichWitchIsWhich  Nov 7, 2017
edited by Guest  Nov 7, 2017
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#1
+78762
+2

1. Let f(x) = x^2 - 2x. Find all real numbers x such that f(x) = f(f(x)).

f (f(x)  =  ( x^2 - 2x)^2  - 2(x^2 - 2x)

So

x^2 - 2x = ( x^2 - 2x)^2  - 2(x^2 - 2x)

We  can write

( x^2 - 2x)^2  - 2(x^2 - 2x) -  (x^2 - 2x)  = 0

x^4 - 4x^3 + 4x^2 - 2x^2+ 4x - x^2 + 2x  = 0

x^4 - 4x^3 + x^2 + 6x  = 0

x ( x^3 - 4x^2 + x  + 6 )

One solution  is  x  = 0

(x^3 -3x^2)  - (x^2 - x  - 6)  = 0

x^2 ( x - 3) - 1[ (x -3)( x + 2) ]  = 0

(x - 3) (x^2 -1(x + 2) )  = 0

(x - 3) (x^2 - x - 2)  = 0

(x-3) (x -2) (x+ 1)  = 0

Setting the  linear factors = 0 and solving for x, we have that

x = -1, 2  and 3

So....the solutions that make the original equation true are

x = -1, 0, 2 and 3

CPhill  Nov 7, 2017
#3
+272
0

Thanks so much :D

WhichWitchIsWhich  Nov 7, 2017
edited by Guest  Nov 7, 2017
#2
+78762
+1

2. The function f(x,y) accepts an ordered pair as input and gives another ordered pair as output. It is defined according to the following rules: If x > 4, f(x,y) = (x - 4,y). If \$x \le 4 but \$y > 4, f(x,y) = (x,y - 4). Otherwise, f(x,y) = (x + 5, y + 6). A robot starts by moving to the point (1,1). Every time it arrives at a point (x,y), it applies f to that point and then moves to f(x,y). If the robot runs forever, how many different points will it visit?

(1,1)      (5, 9)

(6, 7)     (1, 9)

(2, 7)     (1, 5)

(2, 3)     (1,1)

(7, 9)

(3, 9)

(3, 5)

(3, 1)

(8, 7)

(4, 7)

(4, 3)

(9, 9)

15 different points before the cycle begins anew....!!!!

CPhill  Nov 7, 2017
#4
+78762
+1

3. Let f(x) = px + q, where p and q are real numbers. Find p+q if \$f(f(f(x))) = 8x + 21.

f(f(x)) =  p (px + q)  + q    =    p^2x + pq + q

So

f ( f ( f(x)))  =  f  [   p^2x + pq + q ]  =

p [ p^2x + pq + q ] + q   =   8x + 21

[p^3x]  + [ p^2q + pq + q ] =  8x + 21

This must mean that

(p^3)x  = 8x   ⇒    p^3  = 8   ⇒  p  = 2

And it must also mean that

p^2q+ pq  + q  =  21           and using p = 2, we have that

2^2 q  + 2q +  q   = 21

4q  +  3q  = 21

7q  = 21

q  = 3

So....   p  +  q   =   2  +  3   =   5

CPhill  Nov 7, 2017

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