Is a differentiable function always continuous?
Matthew Barrera 
Continuous Functions are not Always Differentiable. But can we safely say that if a function f(x) is differentiable within range $(a,b)$ then it is continuous in the interval $[a,b]$ . If so , what is the logic behind it ?
$\endgroup$ 84 Answers
$\begingroup$I will assume that $a<b$.
Consider the function $g: [a,b]\to {\mathbb R}$ which equals $0$ at $a$, and equals $1$ on the interval $(a,b]$. This function is differentiable on $(a,b)$ but is not continuous on $[a,b]$. Thus, "we can safely say..." is plain wrong.
However, one can define derivatives of an arbitrary function $f: [a,b]\to {\mathbb R}$ at the points $a$ and $b$ as $1$-sided limits: $$ f'(a):= \lim_{x\to a+} \frac{f(x)-f(a)}{x-a}, $$ $$ f'(b):= \lim_{x\to b-} \frac{f(x)-f(b)}{x-b}. $$ If these limits exist (as real numbers), then this function is called differentiable at the points $a, b$. For the points of $(a,b)$ the derivative is defined as usual, of course. The function $f$ is said to be differentiable on $[a,b]$ if its derivative exists at every point of $[a,b]$.
Now, the theorem is that a function differentiable on $[a,b]$ is also continuous on $[a,b]$. As for the proof, you can avoid $\epsilon$-$\delta$ definitions and just use limit theorems. For instance, to check continuity at $a$, use: $$ \lim_{x\to a+} (f(x)-f(a))= \lim_{x\to a+} (x-a) \lim_{x\to a+} \frac{f(x)-f(a)}{x-a} = 0\cdot f'(a)=0. $$ Hence, $$ \lim_{x\to a+} f(x)=f(a), $$ hence, $f$ is continuous at $a$. For other points the proof is the same.
$\endgroup$ 2 $\begingroup$The logic behind it lies in a little pool of definitions which, if you meditate on them, will start to make math come alive in a new way. Any good math textbook will take you there; my favorite is Spivak's Calculus.
Until then, intuitively, a function is continuous if its graph has no breaks, and differentiable if its graph has no corners and no breaks. So differentiability is stronger.
$\endgroup$ 0 $\begingroup$A function is only differentiable on an open set, then it has no sense to say that your function is differentiable en $a$ or on $b$. But if $\lim_{x\to a^+}f'(x)$ and $\lim_{x\to b^-}f'(x)$ exists, then your function is $\mathcal C^1([a,b])$ and so yes your function is continuous on $[a,b]$. But this is stronger than just to check the continuity of $f$ on $a$ and on $b$.
$\endgroup$ 5 $\begingroup$Let a point M(y,f(y)) that f is differentiable
For x $\ne$ y , f(x) - f(y) = $\frac{f(x) - f(y)}{x - y}$ * (x - y)
$\lim\limits_{x \to y}( {f(x) - f(y)} )$ = $\lim\limits_{x \to y} (\frac{f(x) - f(y)}{x - y} * (x - y))$ = f'(y) * 0 = 0
So $\lim\limits_{x \to y} {f(x) - f(y)}$ = 0. and $\lim\limits_{x \to y+} f(x)$ = f(y)
Hence f is continuous at y.
Now I will also prove that if a function is differentiable in [a,b] then f' is continuous on (a,b).
Let M(y,f(y)) be a point inside (a,b). In order for f to be differentiable at M,
$\lim\limits_{x \to y+} \frac{f(x) - f(y)}{x - y}$ =$\lim\limits_{x \to y-} \frac{f(x) - f(y)}{x - y}$ = $\lim\limits_{x \to y} \frac{f(x) - f(y)}{x - y}$ = f'(y).
Therefore, $\lim\limits_{x \to y+}{f'(x)}$ = $\lim\limits_{x \to y-}{f'(x)}$ = $\lim\limits_{x \to y}{f'(x)}$ = f'(y).
Hence, since M(y,f(y)) was chosen randomly f' is continuous in the (a,b) interval.
$\endgroup$
