+152 If $f(x)=\dfrac{2}{x+1}$, then what is the value of $f^{-1}\left(\frac{1}{5}\right)$?
+129852 \( $f(x)=\dfrac{2}{x+1}$ \)
\( $f^{-1}\left(\frac{1}{5}\right)$ \)
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)
