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1.) 

Given that \[ f(x) = \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 - x}}}, \] compute $(f(f( - 2)))^{ - 2}$. Express your answer as a common fraction.

 

2.)

For positive integers $a$, $b$, and $c$, what is the value of the product $abc$? \[ \dfrac {1}{a + \dfrac {1}{b + \dfrac {1}{c}}} = \dfrac38 \]\

 

3.)

What is the value of the following expression? \[ \sqrt {6 + \sqrt {6 + \sqrt {6 + \cdots}}} \]

 

4.)

Fully simplify $\sqrt {14 + 8\sqrt {3}}$.

 

5.)

Suppose $f(x)=x+1$. For what value of $x$ is $\overbrace{f(f(\cdots(f}^\text{2015 total \textit{f}s}(x))\cdots)=0$?

 Jan 15, 2018
 #1
avatar+98107 
+1

1) 

 

 \( \[ f(x) = \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 - x}}}, \] \)

\($(f(f( - 2)))^{ - 2}$\)

 

 

 

f ( -2)  =  -2

And   f ( f(-2)) =  f (-2)   =  -2

 

So

 

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

 

 

cool cool cool

 Jan 15, 2018
edited by CPhill  Jan 15, 2018
 #2
avatar+98107 
+1

3)

 

Let   x  =  √[ 6 + √ [6 + √6 +.... ] ]     square both sides

 

x^2  =   6 +    √[ 6 + √ [6 + √6 +.... ] ] 

 

x^2  =  6  + x       rearrange as

 

 

x^2  - x  -  6  = 0       factor

 

(x - 3)  ( x + 2)  =  0

 

Setting  each factor to  0  and solving for x  gives that   x  = 3   or x  = -2

 

Reject the second

 

So.....x  =   3  =  the   value

 

 

cool cool cool

 Jan 15, 2018
 #3
avatar+98107 
+1

\( $\sqrt {14 + 8\sqrt {3}}$\)

 

In the form   √   [ a  + b√c ]      we  can  find a simplification  if

 

(b/2)^2  + ( √c ) ^2    =  a        

 

And  the simpification becomes  √ [   ( (b/2) + √c )^2  ]

 

(4)^2  + 3  =   19      isn't  true.....but notice that  we can  express this as

 

√ [ 2  ( 7  + 4√3)  ]   =  √2 * √ [ 7 + 4√3 ]

 

Then  considering the second radical....

 

(4/2)^2  + (√3)^2   =  2^2 + 3   =  7  is true

 

So  we can write

 

√2  * √ [   ( (4/2)  + √3 ) ^2 ]  =

 

√ 2  *  √ [ ( 2 + √3)^2 ]  =

 

√2 [ 2 + √3]  =

 

2√2  + √6  =

 

√8  +  √6

 

cool cool cool

 Jan 15, 2018
edited by CPhill  Jan 16, 2018
 #4
avatar+98107 
+1

5 )

 

\($f(x)=x+1$\)

 

\($\overbrace{f(f(\cdots(f}^\text{2015 total \textit{f}s}(x))\cdots)=0$?\)

 

 

Note  that  if  we had    x +  1.......then...

f ( f ( -2) )   =  f ( -1)  =  0

And

f ( f ( f (-3) ) )  =  f (f (-2)  )  =  f (-1)   =  0

 

So....this implies that  x  =  -2015

 

 

cool cool cool

 Jan 15, 2018
 #5
avatar+98107 
+1

2)   

 

The   simplifies to

 

[ bc + 1 ]  /  [ abc + a + c ]  =  3 / 8      which implies that

 

[ bc + 1 ]  =  3       and

 

[ a (bc + 1)  + c  ]  =  8

 

If  a,b,c  do not have to hold unique values....one solution  is  

 

a  = 2   b  = 1   and c  = 2

 

So.....  abc  =   2 * 1 * 2   =    4

 

 

cool cool cool

 Jan 16, 2018

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