Can a function be continuous and non-differentiable on a given domain??

Jan 6, 2016

Yes.

Explanation:

One of the most striking examples of this is the Weierstrass function, discovered by Karl Weierstrass which he defined in his original paper as:

${\sum}_{n = 0}^{\infty} {a}^{n} \cos \left({b}^{n} \pi x\right)$

where $0 < a < 1$, $b$ is a positive odd integer and $a b > \frac{3 \pi + 2}{2}$

This is a very spiky function that is continuous everywhere on the Real line, but differentiable nowhere.

Jan 6, 2016

Yes, if it has a "bent" point. One example is $f \left(x\right) = | x |$ at ${x}_{0} = 0$

Explanation:

Continuous function practically means drawing it without taking your pencil off the paper. Mathematically, it means that for any ${x}_{0}$ the values of $f \left({x}_{0}\right)$ as they are approached with infinitely small $\mathrm{dx}$ from left and right must be equal:

${\lim}_{x \to {x}_{0}^{-}} \left(f \left(x\right)\right) = {\lim}_{x \to {x}_{0}^{+}} \left(f \left(x\right)\right)$

where the minus sign means approaching from left and plus sign means approaching from right.

Differentiable function practically means a function that steadily changes its slope (NOT at a constant rate). Therefore, a function that is non-differentiable at a given point practically means that it abruptly changes it's slope from the left of that point to the right.

Let's see 2 functions.

$f \left(x\right) = {x}^{2}$ at ${x}_{0} = 2$

Graph

graph{x^2 [-10, 10, -5.21, 5.21]}

Graph (zoomed)

graph{x^2 [0.282, 3.7, 3.073, 4.783]}

Since at ${x}_{0} = 2$ the graph can be formed without taking the pencil off the paper, the function is continuous at that point. Since it is not bent at that point, it's also differentiable.

$g \left(x\right) = | x |$ at ${x}_{0} = 0$

Graph

graph{absx [-10, 10, -5.21, 5.21]}

At ${x}_{0} = 0$ the function is continuous as it can be drawn without taking the pencil off the paper. However, since it bents at that point, the function is not differentiable.