Transforming points from function

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Currently, I am doing transformations on functions, with things like $y=f(x+2)$ representing a transformation of the original function 2 to the left, for example.

However, I am stuck on transforming inverse points. Given a point, say $(-3, 5)$ (made up) on $y=f(x)$, I want to transform it to the graph of $x+1=f(y-1)$.

I noticed the $x$ and $y$ have been switched, so I assume it's a inverse function, so my point is now at $(5, -3)$. I proceeded to isolate $x$, giving me $x=f(y-1)-1$. Then I decided I would need to translate the point 1 right, and 1 down, giving me $(6, -4)$.

However, the answer from the book (from the patterns I see, because I made up the point) is $(4, -2)$. From what I can tell, they left it at $x+1=f(y-1)$ and applied 1 left, and 1 up, the opposite of each to $x$ and $y$. However, I can't tell why.

Could anyone give a me a hint on what I have done wrong? Am I not allowed to move $-1$ over to the right side?

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I haven't looked at your geometric reasoning, but Algebraically, in order for $\color{Red}{x+1}=f(\color{Blue}{y-1})$ to correspond to the special values $\color{Red}5=f(\color{Blue}{-3})$, we need $\color{Red}{x+1}=\color{Red}5$ and $\color{Blue}{y-1}=\color{Blue}{-3}$, which when solved gives the textbook solution $x=4, y=-2$.


The part I don't follow in your argument is this:

Then I decided I would need to translate the point 1 right, and 1 down

I don't see why you would do that. Going from $x=\text{blah}$ to $x=\text{blah}-1$ involves translating 1 point to the left, not to the right. Similarly, replacing $y$ with $y-1$ in an equation will translate up, not down. So you should have gone $(-3,5)\mapsto (5,-3)\mapsto (4,-3)\mapsto (4,-2)$.