An example of a function which is continuous every where but fails to be differentiable exactily at two point is `"……."`

Explanation

To find a function that is continuous everywhere but fails to be differentiable at exactly two points, we can consider the following steps:

### Step 1: Define the Function
Let's define the function \( f(x) = |x| + |x - 1| \).

### Step 2: Check Continuity
To check if \( f(x) \) is continuous everywhere, we need to analyze the function at all points, particularly at the points where the absolute value expressions change, which are \( x = 0 \) and \( x = 1 \).

- For \( x < 0 \):
\[ f(x) = -x + (1 - x) = -2x + 1 \]

- For \( 0 \leq x < 1 \):
\[ f(x) = x + (1 - x) = 1 \]

- For \( x \geq 1 \):
\[ f(x) = x + (x - 1) = 2x - 1 \]

Now, we check the limits at the points of interest:
- At \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = 1, \quad \lim_{x \to 0^+} f(x) = 1 \quad \Rightarrow \quad f(0) = 1 \]

- At \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = 1, \quad \lim_{x \to 1^+} f(x) = 1 \quad \Rightarrow \quad f(1) = 1 \]

Since \( f(x) \) is defined and the limits match at both points, \( f(x) \) is continuous everywhere.

### Step 3: Check Differentiability
Now we check differentiability at the points \( x = 0 \) and \( x = 1 \).

- At \( x = 0 \):
- Left-hand derivative:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-2h + 1 - 1}{h} = \lim_{h \to 0^-} -2 = -2 \]
- Right-hand derivative:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{1 - 1}{h} = 0 \]
Since \( f'(0^-) \neq f'(0^+) \), \( f(x) \) is not differentiable at \( x = 0 \).

- At \( x = 1 \):
- Left-hand derivative:
\[ f'(1^-) = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{1 - 1}{h} = 0 \]
- Right-hand derivative:
\[ f'(1^+) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(2(1 + h) - 1) - 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2 \]
Since \( f'(1^-) \neq f'(1^+) \), \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion
Thus, the function \( f(x) = |x| + |x - 1| \) is continuous everywhere but not differentiable at exactly two points, \( x = 0 \) and \( x = 1 \).