Definition of class of continuously differentiable functions on a closed interval This is a technical simple question.  The space $C^1([a,b])$ is defined as

$$C^1([a,b])=\{f:[a,b]\to R: \exists\; f' \textrm{ and is continuous on } [a,b]  \}.$$

I think there is some explaining to do here, and is the fact that $f'$ cannot be defined at $a$ or $b$ because those points do not belong to the interior of the domain. The question is:

In the above definition, $f'(a)$ and $f'(b)$ are understood in the sense of right and left derivatives respectively? If not, in which sense?
A derivative can in fact be defined on the edge of a domain. For example, let $f(x): [0, 1] \to [0,1]$ be defined by $x \mapsto x$. Then $f'(0) = \lim\limits_{x \to 0}\frac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0}\frac{f(x)}{x}$. This limit exists and is equal to 1. Proof: let $\epsilon &gt; 0$ be arbritrary. Choose $\delta = \epsilon$. Let $x$ satisfy $0 &lt; |x - 0| &lt; \delta$ and $x \in [0, 1]$, which is $x \in ]0, \delta[$. Then $|f(x)/x - 1| = |x/x - 1| = 0 &lt; \epsilon$.
A derivative can in fact be defined on the edge of a domain. For example, let $f(x): [0, 1] \to [0,1]$ be defined by $x \mapsto x$. Then $f'(0) = \lim\limits_{x \to 0}\frac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0}\frac{f(x)}{x}$. This limit exists and is equal to 1. Proof: let $\epsilon &gt; 0$ be arbritrary. Choose $\delta = \epsilon$. Let $x$ satisfy $0 &lt; |x - 0| &lt; \delta$ and $x \in [0, 1]$, which is $x \in ]0, \delta[$. Then $|f(x)/x - 1| = |x/x - 1| = 0 &lt; \epsilon$.