Does translating a function change its domain?

Question

In Spivak's Calculus chapter 3, there is a part which essentially states:

$\textrm{if} ~~~ r(x)=x^2\ \textrm{such that} \ -17\leq x\leq \frac{\pi}{3}\\ \textrm{then} ~~~ r(x+1)=x^2+2x+1=r(x)+2x+1\ \textrm{such that} \ -17\leq x\leq \frac{\pi}{3}-1$

Why is the domain of $r(x+1)$ reduced by 1?

And would the domain of $r(x+a)$ be reduced by $a$?

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I have tried searching Google for this, which returns little other than programming related topics. I have also attempted sketching the graphs of both curves which has (wrongly?) convinced me that the domain should be $-18\leq x \leq \frac{\pi}{3}-1$.

Answer 1

Spivak writes:

You should have little difficulties checking the following assertions about the functions defined above.

An assertion is an equality that is supposed to be true. In this case.

$$ r(x+1) = r(x) + 2x + 1 \quad \text{if } -17 \le x \le \frac{\pi}{3} - 1 $$

So he reduced the domain to ensure that $r$ does not run out of the domain when used with $r(x+1)$. E.g. $x = \frac{\pi}{3}$ would turn the LHS to $r(\frac{\pi}{3}+1)$ which is not in the prior defined domain of $r$.

Valid would be the intersection of the domains for $r(x+1)$, which is $[-18,\frac{\pi}{3}-1]$ and for $r(x)$ which is $[-17,\frac{\pi}{3}]$, meaning $[-17,\frac{\pi}{3}-1]$.

Answer 2

We have some $r:\:x\in[-17;\pi/3]\mapsto r(x)$.

From this Spivak notices $r$ has some property from which it follows the equality $$r(x+1)=r(x)+2x+1\label{eq1}\tag{$\ast$}$$ is satisfied for, let us for the time being say, some $x$.

Let us now deduce for which $x$ does it hold.

Obviously, we must have $x+1\in[-17;\pi/3]$ because we have $r(x+1)$ in the LHS, but we also must have $x\in[-17;\pi/3]$ because we also have $r(x)$ in the RHS. Therefore the interval on which $\eqref{eq1}$ holds is given by $$[-18;\pi/3-1]\cap[-17;\pi/3]=[-17;\pi/3-1],$$ which is what had to be demonstrated.