Let $f$ be a function given by $f(x)=\frac{1-2x}{1+x}$ and $g(x)$ and $f^{-1}(x+1)$ are symmetrical about $y=x$ then find $g(2)$
In first approach, I found inverse of $f(x)$ and then $f^{-1}(x+1)$ and got $f^{-1}(x+1)=\frac{-x}{x+3}$ and then used the fact that point $(g(2),2)$ will lie on $f^{-1}(x+1)=\frac{-x}{x+3}$ and got the answer.
But what I want to ask is that using the statement that $g(x)$ and $f^{-1}(x+1)$ are symmetrical about $y=x$, can't we say that $g(x)=f(x+1)$?
Yes, we can. That's the property of inverse function. And vice versa. If two functions are symmetrical on $y=x$ they are inverse of each other.
I was a little two quick. Actually $$f^{-1}(x+1) = f^{-1}(h(x))$$ where $h(x)=x+1$. So $f^{-1}(x+1)$ is composition of two functions. Now, how do we calculate the inverse function of composition? It is this rule: $$ (p\circ q)^{-1} = q^{-1}\circ p^{-1}$$ So in our case, since $h^{-1}(x) = x-1$ we have $$g(x) =f(x)-1$$