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If $f(x)=\dfrac{2}{x+1}$, then what is the value of $f^{-1}\left(\frac{1}{5}\right)$?

 Dec 27, 2019
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\( $f(x)=\dfrac{2}{x+1}$ \)

 

 

\( $f^{-1}\left(\frac{1}{5}\right)$ \)

 

Note that  we  need  to  solve this

 

(1/5)  =   2 / ( x + 1)          and we can write

 

5   =  ( x + 1) / 2             multily both sides by 2

 

10  = x +  1                   subtract 1 from both sides

 

=  x  =  f-1(1/5)

 

So  the point (9, 1/5)  is on f(x)     and the point  (1/5 , 9)  is on f-1 (x)  

 

cool cool cool

 Dec 27, 2019

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