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?