CPhill

OK

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OK

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Ah....thanks, hectictar.....I forgot about the word "tangent"    !!!!!

[-81 / 625]   /   [243 / 3125]    =

[-81 / 625 ]  *  [ 3125 / 243 ]          swap denominators

[-81 / 243 ]  *  [ 3125  / 625 ] 

[ -81 / ( 81 * 3 ) ]  *  [  (625 * 5)  / 625 ]

[  -1 / 3 ] *  [ 5 / 1 ]  =

-5 / 3

y = 2x^2  + kx +  6

y  = 4 - x

Set these equal

4 - x  = 2x^2 + kx + 6          rearrange as

2x^2 + (k + 1)x  +  2  =   0

If there is one solution, the discriminant must   = 0

So we have that

(k + 1)^2  - 4(2) (2)  = 0

(k + 1) ^2   - 16  =  0

( k + 1)^2   = 16        take the negative root

k + 1   =   -4

k  =   - 5

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

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....!!!!

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

That's pretty neat, PM  !!!!

Let  me try !!!!!

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