give an example of function which is continuous but not differentiable? give 2

One example would be the modulus function, f(x)=|x|. It is continuous at x=0, but not differentiable at x=0.

Another example would be, $f\left(x\right)=\left\{\begin{array}{l}1-x, x\le 2\\ x-3, x>2\end{array}\right.$

It is defined at x=2, but not differentiable at x=2.

This is because, if you look at the definition of continuity, it says that f(x) is continuous at x=a if:-

• f(a) is defined, and
• the value that f(x) approaches when x approaches a from the negative side (i.e., x is very slightly less than a) is equal to the value that f(x) approaches when x approaches a from the positive side (i.e., x is very slightly greater than a), which is also equal to the value of f(x) at x=a.

Here, we notice that f(2)=1-2=-1, i.e., f(2) is defined.

Also, $\underset{x\to 2^{-}}{\lim }f\left(x\right)=1-2=-1$
And, $\underset{x\to 2^{+}}{\lim }f\left(x\right)=2-3=-1$

So, $\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }f\left(x\right)=f\left(2\right)$

Thus, f(x) is continuous at x=2.

But, f(x) is not differentiable at x=2. This is because, if you look at the definition of differentiability, a function f(x) is differentiable at x=a if $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ exists. But, if you look at our f(x) here, $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ does not exist at x=2, it cannot be defined at x=2. So, f(x) is not differentiable at x=2.

You can understand this more intuitively by visualizing the graph of f(x). Take f(x)=|x| for example. When x approaches 0 from left side, it has a slope which is different than the slope it has when x approaches 0 from right side. In other words, we have a "kink" at x=0. So, because of this kink's presence, we don't have one, unique slope of f(x)=|x| at x=0. Hence, |x| is not differentiable at x=0.

The graph of $f\left(x\right)=\left\{\begin{array}{l}1-x, x\le 2\\ x-3, x>2\end{array}\right.$is similar to the graph of f(x)=|x|; you just need to shift it to the right (positive x-axis) by 2 units, and shift it down (negative y-axis) by 1 unit. So, we have a kink at x=2. And thus, we have 1 slope towards the left of x=2, and another slope towards the right of x=2. Hence, f(x) is not differentiable at x=2.

You can make an ​infinite number of such functions. The only key is that, the function should be defined at all points (which takes care of the continuity), and there should be a "kink" on the curve, a sudden bump, like we have in the curves of these 2 functions (which will make it undifferentiable at such points).

Another example would be the Weirstrass Function. It has multiple kinks on its curve.

NOTE:- Kink means a steep bump, like a needle, it must have a pointed end. So, if a curve is not sharp or pointed, but rather smooth and rounded at some point, it is not called a kink, and it is differentiable at that point. In fact, if there is a smooth and rounded point at which f(x) changes slope, it is called a critical point of f(x). It means that, the slope of f(x) at that point is 0, but the slope to its left and right have different signs (one is positive, one is negative).