Question: I am solving the following problem:

> 
> Let $f(x) =\sqrt{x}$. Find a formula for a function $g$ whose graph is obtained from $f$ from the given sequence of transformations: 
> 
> 
> * shift right $3$ units
> * horizontal shrink by a factor of $2$
> * shift up $1$ unit
> 
> 
> 

I think that$ g(x) = f(2(x-3)) + 1 = \sqrt{2x-6} + 1$, but in the answers it says $\sqrt{2x-3} + 1$, so i assume $g(x) = f(2x-3) +1$, but wouldn't that mean that the horizontal shrink was done first and afterwards the right horizontal shift?

Answer: 
**Hint:** Consider $\phi : \mathbb R \to (-1,1)$ defined by $$\phi(x) = \frac{x}{1 +|x|}$$ and notice that $f: (-1,1) \to (a,b)$ defined by $$f(x) = \frac{1}{2}\bigg((b-a)x + a+ b\bigg)$$
 is a bijection.
**HINT:** Find an injection from $\Bbb R$ into the interval $(0,1)$. Then show there is a bijection between any two intervals $(a,b)$ and $(c,d)$.