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# HELP!!

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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
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Given that $f(x) = \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 - x}}},$ compute $(f(f( - 2)))^{ - 2}$.

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

I assume:

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

Apr 11, 2019