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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

9  =  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)