We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Math Help 2!!!

0
906
5

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

### 5+0 Answers

#1
+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   Jan 15, 2018
edited by CPhill  Jan 15, 2018
#2
+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   Jan 15, 2018
#3
+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   Jan 15, 2018
edited by CPhill  Jan 16, 2018
#4
+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   Jan 15, 2018
#5
+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   Jan 16, 2018