+0  
 
0
231
1
avatar

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.

 Apr 10, 2019
 #1
avatar+25259 
+2

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.

\(\begin{array}{|rcll|} \hline f(-2) &=& \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 - (-2) }}} \\\\ &=& \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 +2 }}} \\\\ &=& \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{3 }}} \\\\ &=& \dfrac {1}{1 - \dfrac {1}{ \dfrac {2}{3 }}} \\\\ &=& \dfrac {1}{1 - \dfrac {3}{ 2 }} \\\\ &=& \dfrac {1}{\dfrac {2}{ 2 } - \dfrac {3}{ 2 }} \\\\ &=& \dfrac {1}{-\dfrac {1}{ 2 } } \\\\ &=& \dfrac {2}{-1} \\\\ \mathbf{f(-2)} &\mathbf{=}& \mathbf{-2} \\ \hline \end{array}\)

 

I assume:

\(\begin{array}{|rcll|} \hline && \mathbf{\left(f(f( - 2)) \right)^{ - 2}} \\\\ &=&\mathbf{ \dfrac{1}{ \left(f(f( - 2)) \right)^2} } \quad | \quad f(-2) = -2 \\\\ &=& \dfrac{1}{ \left(f(-2) \right)^2} \quad | \quad f(-2) = -2 \\\\ &=& \dfrac{1}{ \left(-2 \right)^2} \\ \\ &\mathbf{=}& \mathbf{\dfrac{1}{4} } \\ \hline \end{array} \)

 

laugh

 Apr 11, 2019

14 Online Users

avatar
avatar
avatar