+0  
 
0
122
5
avatar+33 

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

Jdaye  Jan 15, 2018
Sort: 

5+0 Answers

 #1
avatar+85864 
+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

CPhill  Jan 15, 2018
edited by CPhill  Jan 15, 2018
 #2
avatar+85864 
+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

CPhill  Jan 15, 2018
 #3
avatar+85864 
+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

CPhill  Jan 15, 2018
edited by CPhill  Jan 16, 2018
 #4
avatar+85864 
+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

CPhill  Jan 15, 2018
 #5
avatar+85864 
+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

CPhill  Jan 16, 2018

36 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details