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 #3
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+5
Nov 11, 2014
 #1
avatar+130561 
+5

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 > 4f(x,y) = (x - 4,y). If x \le 4 but y > 4f(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 fto that point and then moves to f(x,y). If the robot runs forever, how many different points will it visit?  ...Sorry for the long question

(1,1) → (6, 7) → (2, 7) → (2, 3) → (7, 9) → ( 3, 9) → (3, 5) → (3, 1) → (8, 7) → (4, 7) → (4, 3) → (9, 9) →

(5, 9) → (1, 9) → (1, 5) → (1, 1)

15 points...and the cycle starts over......

 

Nov 11, 2014
 #1
avatar+130561 
+10

Let p(x) = 2x^3 - 113 and let q be the inverse of p. Find q(137)

Let's find the inverse.....let  p(x) = y

y = 2x3 - 113       add 113 to both sides

y + 113  = 2x3     divide both sides by 2

( y + 113 ) / 2   = x3    take the cube root of both sides

3√[( y + 113) / 2]  = x       exchange x and y

3√[( x + 113) / 2]  = y       for "y"  write q(x)

q(x) = 3√[( x + 113) / 2]     so q(137) = 3√[( 137 + 113) / 2] = 3√125  = 5

 

Nov 11, 2014
 #1
avatar+130561 
+7

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

f(f(f(x)))   = (f o f o f)(x)

We first want to put f into f...this gives

p(px + q) + q = p2x + pq + q

Now....we want to put this result into f, again...so we have

p[ p2x + pq + q] + q =  p3x + p2q + pq + q     and setting this to 8x + 21, we have

p3x + p2q + pq + q  = 8x + 21   equating coefficients .....   p3 = 8   so p = 2  ....so we have

8x + 4q + 2q + q  = 8x + 21     subtract 8x from both sides

4q + 2q + q = 21     simplify

7q = 21      divide both sides by 7

q = 3

So .... p + q =  2 + 3   = 5

 

Nov 11, 2014

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