+267 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
+129899 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
+129899
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....!!!!
+129899 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
