If $f:\mathbb R \rightarrow \mathbb R$ is differentiable at $v$, why is $ x\rightarrow f(x-3)$ differentiable at $v+3$?

Question

So, we know that this unknown f is differentiable at v. Thus, it has to have a limit at v, so we can therefore say:

$\forall \varepsilon > 0$, $ \exists u> 0$ such that: $|x-v| < u\implies |f(x) - L|< \varepsilon $, where $L$ is the limit.

With this information, I have to prove that

$\forall \varepsilon' > 0$, $ \exists u' > 0$ such that: $|x-(v+3)| < u\implies |f(x-3) - M|< \varepsilon $, where $M$ is the limit.

This is what I have to show, right? I'm not really sure how to do it; I'm assuming it has to do with manipulating the first implication to come out with the second implication, but I'm not sure how to do this at all. Any help is appreciated. Thanks.

Answer 1

Let $$g(x)=f(x-3)$$

You want to show that g(x) is differentiable at $v+3$

Note that $$lim _{h\to 0 } \frac {g(v+3+h) -g(v+3)}{h} = lim _{h\to 0 }\frac {f(v+h) -f(v)}{h} = f'(v)$$

Thus g(x) is differentiable at $x=v+3$