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Let $f: [a, b] \to \mathbb{R}^n$ where a and b are real numbers.

Prove that the curve $f$ is differentiable if and only if there is an open interval $I$ that contains the closed interval $[a, b]$, and a differentiable curve $g: I \to \mathbb{R}^n$ such that its restriction to the interval $[a, b]$ is the curve $f$.

Showing from 'down' to 'up' should be trivial, because if we have a differentiable $g$ on an interval $I$, then it should specifically be differentiable on the interval $[a, b]$, and if we let $f$ be the restriction of the function $g$ on that interval then $f$ must be differentiable.

Going from 'up' to 'down' is what is bothering me.

So now I construct a $g$ on $(a-1,b+1)$ as

$g(x)=f(x)$ for $x \in [a,b]$

g(x)=f(a)+f'(a)(x-a) for $x \in (a-1,a)$

g(x)=f(b)+f'(b)(x-b) for $x \in (b,b+1)$

We see that $g$ is differentiable on $(a-1, b+1)$ except for maybe on $a, b$. So I examine the left and right sided limits: the left sided limit \lim _ {t \to a-} \frac{g(a) - g(t)}{t} = \lim_{t \to a-} \frac{f(a) + f'(a)(t-a) - f(a) - f'(a)(a-a)}{t} = \lim_{t \to a-} \frac{f'(a)t - f'(a)a}{t}

And the right sided limit $\lim_{t \to a+} \frac{f(t) - f(a)}{t}$

I've been trying to bridge these two limits for over an hour now.

Am I approaching this problem incorrectly or am I missing something in the two one sided limits?

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    I'm not sure, but don't you need $a-t$ instead of $t$ in the denominator of the DQ of $g$, and $t-a$ in the denominator of the DQ of $f$?2012-03-05

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You should get this. $ \lim_{t\to a^-}\frac{g(t) - g(a)}{t-a} = \lim_{t\to a^-}\frac{f(a) + f^\prime(a)(t-a) - f(a) + 0}{t-a} = f^\prime(a). $ On the other end you should get $f^\prime(b)$.

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    Absolutely correct. I was working with a flawed definition of derivatives, thank you.2012-03-05