# Thread: Continuity if g(x) = f(x-c)

1. ## Continuity if g(x) = f(x-c)

Suppose $f\rightarrow \Re, c \in \Re" alt="f\rightarrow \Re, c \in \Re" /> and g(x) = f(x-c)

1) What's the Domain of g?

I think it's $\Re$, am I right?

2) Suppose that f is continuous at $a \in D \Leftrightarrow$ g is continuous at c + a

So far I have this:
( $\Rightarrow$) Assume f is continuous. Then:
$\forall \epsilon >0 \exists \delta >0$ such that $x \in D, 0 < |x-a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon$.
Let g(x) = f(x-c).

We must prove that $\forall \epsilon'$ > 0 $\exists \delta'$ > 0 such that $x \in D, 0 < |x-c-a| < \delta' \Rightarrow |f(x-c) - f(a-c)| < \epsilon'$

I see the obvious outcome here because we'll have |x - (c + a)| < $\delta'$, but I'm confused on how to prove the rest of it so that it solves the first part of the iff statement.

The way I've structured it, I'm pretty sure the ( $\Leftarrow$) half with come naturally if I can figure out why this other part works.

Can anyone steer me towards the right wording?

2. $
\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {\left( {x + c} \right) - \left( {a + c} \right)} \right| < \delta \,$
$\Rightarrow \left| {f(x + c) - f(a + c)} \right|\, < \varepsilon \Rightarrow \,\,\left| {g(x) - g(a)} \right| < \varepsilon$