Is a differentiable function always continuous?

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 ?

Solution 1:

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.

Solution 2:

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.