If I have $a(x)=2x$ and $b(x)=x+1$. I am trying to find $ (a \circ b)^{-1}(x+3)$. However I have 2 different ways:
- Let $c(x)=a(b(x+3))=2x+8$. We have $c^{-1}(x)=\frac{x-8}{2}$. The answer is $\frac{x-8}{2}$ here
- Let $d(x)=a(b(x))=2x+2$. We have $d^{-1}(x)=\frac{x-2}{2}$. So $g(x)=d^{-1}(x+3)=\frac{x+1}{2}$. The answer is $\frac{x+1}{2}$ here
I see that the first one is correct, but I don't understand why is the second way wrong. Is it that $(a(b(x+3))^{-1}$ different from $(a \circ b)^{-1}(x+3)$?
Let $f(x) = x+3$. Then your defined functions are $$c(x) = d(x+3) = d(f(x))$$
Taking the inverse, $$c^{-1}(x) = f^{-1}(d^{-1}(x)) = d^{-1}(x) - 3$$
You need to apply the inverse of $d(x+3)$ as a nested function, but you took the inverse of $d(x)$ by itself, and then plugged in $x+3$ as the argument. Your calculation needs to include an inverse to the $+3$. Therefore the answer you are looking for in the end is not $$d^{-1}(x+3) = \tfrac{x+1}{2}$$ but rather $$d^{-1}(x) - 3 = \tfrac{x-2}{2} - 3 = \tfrac{x-8}{2}$$