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If $f(x)=\frac{x^5-1}3$, find $f^{-1}(-31/96)$.

 Mar 2, 2020
 #1
avatar+25485 
+2

If

\(f(x)=\dfrac{x^5-1}3\),

find \(f^{-1} \left( \dfrac{-31}{96} \right)\).

 

\(\begin{array}{|rcll|} \hline \mathbf{f\Big( f^{-1} \left( x \right) \Big)} &=& \mathbf{x} \quad | \quad x = -\dfrac{31}{96} \\\\ f\Bigg( f^{-1} \left( -\dfrac{31}{96} \right) \Bigg) &=& -\dfrac{31}{96} \\ \hline \\ && \boxed{f(x)=\dfrac{x^5-1}3 \\ x=f^{-1}\left( -\dfrac{31}{96} \right)} \\ \dfrac{\left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 -1 }{3} &=& -\dfrac{31}{96} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 -1 &=& -\dfrac{31}{32} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 &=&1 -\dfrac{31}{32} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 &=& \dfrac{1}{32} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 &=& \dfrac{1}{2^5} \\\\ f^{-1}\left( -\dfrac{31}{96} \right) &=& \dfrac{1}{\sqrt[5]{2^5}} \\\\ \mathbf{ f^{-1}\left( -\dfrac{31}{96} \right) } &=& \mathbf{ \dfrac{1}{2} } \\ \hline \end{array}\)

 

laugh

 Mar 2, 2020
 #2
avatar+26006 
+2

y = x^5-1   /3          find the inverse function

 

x =  y^5-1   /3

(3x +1)1/5 =y         this is f-1

 

(3(-31/96)+1 )1/5 =y

y = 1/2

 Mar 2, 2020

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