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if $f(x)=\frac{x-1}{x+1}$ , x∈R-{-1}, then f-1(x) is equal to

$a. \frac{x-1}{x+1} b.\frac{x+1}{x-1} c.\frac{2}{1+x}$ d.Does Not exist

edited | 101 views

• $f(x)=\frac{x-1}{x+1}$ , here f(1) = 0
• By assigning $x = 0$ in the given options , none of them results $1$
• So, it should be none of these gievn options for inverse. And f(x) can never be equal to 1 and $f^{-1}(x)$ does not exist for 1

\begin{align*} &\Rightarrow\frac{y}{1} =\frac{x-1}{x+1} \\ &\Rightarrow \frac{y-1}{y+1} =\frac{(x-1)-(x+1)}{(x-1)+(x+1)} \\ &\Rightarrow\frac{y-1}{y+1} = \frac{-2}{2x} \\ &\text{x and y are dummy only} \\ &\Rightarrow f^{-1}(x) = \frac{1+x}{1-x} \\ \end{align*}

answered by Veteran (56.9k points) 36 193 500
edited by
sorry! 4th option is doesn't exist. what is your answer (D) or b

f-1(x) does not exist for x=1
so option D is correct..

Awesome! you're right f^-1(x) doesn't exist for 1.
this $\frac{x-1}{x+1}$ fraction  can never be equal to one !
@Debashish

You mean to say that if we are not getting image=1 from F(x), then there is no need to consider 1 as input for inverse function?