Examine the derivability of the following functions at the specified points :
`|x|" at "x = 0`

Answer

To examine the derivability of the function \( f(x) = |x| \) at the point \( x = 0 \), we will follow these steps:

### Step 1: Define the function
The function \( f(x) = |x| \) can be expressed as:
\[ f(x) = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \]

### Step 2: Check for continuity at \( x = 0 \)
To check if \( f(x) \) is continuous at \( x = 0 \), we need to verify that:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

Calculating the left-hand limit:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0 \]

Calculating the right-hand limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0 \]

Now, evaluate \( f(0) \):
\[ f(0) = |0| = 0 \]

Since both limits are equal and equal to \( f(0) \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0 \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 3: Check for differentiability at \( x = 0 \)
To check for differentiability, we need to find the left-hand derivative and the right-hand derivative at \( x = 0 \).

**Left-hand derivative:**
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h)}{h} = \lim_{h \to 0^-} \frac{-h}{h} = \lim_{h \to 0^-} -1 = -1 \]

**Right-hand derivative:**
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h)}{h} = \lim_{h \to 0^+} \frac{h}{h} = \lim_{h \to 0^+} 1 = 1 \]

### Step 4: Compare the left-hand and right-hand derivatives
Since the left-hand derivative \( f'(0^-) = -1 \) and the right-hand derivative \( f'(0^+) = 1 \) are not equal, we conclude that \( f(x) \) is not differentiable at \( x = 0 \).

### Conclusion
The function \( f(x) = |x| \) is continuous at \( x = 0 \) but not differentiable at that point.

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