Analyzing the Differentiability of \(f(x) = e^{|x|}\) at \(x=0\)

The question asks about the differentiability of the function \(f(x) = e^{|x|}\) at the point \(x=0\). To determine if a function is differentiable at a point, we need to check if the left-hand derivative (LHD) and the right-hand derivative (RHD) at that point exist and are equal.

Understanding the Function \(f(x) = e^{|x|}\)

The absolute value function \(|x|\) is defined as:

• \(|x| = x\) if \(x \ge 0\)
• \(|x| = -x\) if \(x < 0\)

Therefore, we can rewrite the function \(f(x) = e^{|x|}\) as a piecewise function:

\[ f(x) = \begin{cases} e^x & \text{if } x \ge 0 \\ e^{-x} & \text{if } x < 0 \end{cases} \]

First, let's check the value of the function at \(x=0\):

\[f(0) = e^{|0|} = e^0 = 1\]

Checking Differentiability at \(x=0\)

A function \(f(x)\) is differentiable at a point \(a\) if the limit

\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\] exists. For the limit to exist, the left-hand limit (LHL) and the right-hand limit (RHL) must exist and be equal.

In this case, we need to check differentiability at \(a=0\), so we evaluate:

\[f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 1}{h}\]

Calculating the Left-Hand Derivative (LHD) at \(x=0\)

The LHD is the limit as \(h\) approaches 0 from the left side (\(h < 0\)):

\[f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - 1}{h}\]

Since \(h < 0\), we use the definition \(f(x) = e^{-x}\) for \(x < 0\). So, \(f(h) = e^{-h}\).

\[f'(0^-) = \lim_{h \to 0^-} \frac{e^{-h} - 1}{h}\]

Let \(k = -h\). As \(h \to 0^-\), \(k \to 0^+\). The limit becomes:

\[f'(0^-) = \lim_{k \to 0^+} \frac{e^{k} - 1}{-k} = - \lim_{k \to 0^+} \frac{e^k - 1}{k}\]

We know the standard limit \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\). Using this, we get:

\[f'(0^-) = - (1) = -1\]

Calculating the Right-Hand Derivative (RHD) at \(x=0\)

The RHD is the limit as \(h\) approaches 0 from the right side (\(h > 0\)):

\[f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - 1}{h}\]

Since \(h > 0\), we use the definition \(f(x) = e^{x}\) for \(x \ge 0\). So, \(f(h) = e^{h}\).

\[f'(0^+) = \lim_{h \to 0^+} \frac{e^{h} - 1}{h}\]

This is the standard limit \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\). Using this, we get:

\[f'(0^+) = 1\]

Comparing LHD and RHD

For the derivative to exist at \(x=0\), the LHD must be equal to the RHD.

• Left-Hand Derivative at \(x=0\) is \(-1\).
• Right-Hand Derivative at \(x=0\) is \(1\).

Since \(f'(0^-) = -1\) and \(f'(0^+) = 1\), we have \(f'(0^-) \ne f'(0^+)\).

Because the left-hand derivative and the right-hand derivative at \(x=0\) are not equal, the derivative of \(f(x) = e^{|x|}\) at \(x=0\) does not exist.

Conclusion on Differentiability at \(x=0\)

Based on the calculations, the derivative \(f'(0)\) does not exist.

Let's examine the given options:

• Option 1: \(f'(0) = 1\). This is incorrect, as the LHD is -1 and RHD is 1.
• Option 2: \(f'(0) = -1\). This is incorrect, as the LHD is -1 and RHD is 1.
• Option 3: \(f'(0) = 0\). This is incorrect.
• Option 4: \(f'(0)\) does not exist. This matches our conclusion.

Revision Table: Key Concepts for Differentiability

Additional Information: Continuity and Differentiability

It is important to note the relationship between continuity and differentiability.

• If a function is differentiable at a point, then it must be continuous at that point.
• However, the converse is not true: a function can be continuous at a point but not differentiable at that point.

In this case, the function \(f(x) = e^{|x|}\) is continuous at \(x=0\).

• \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^{-x} = e^0 = 1\)
• \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{x} = e^0 = 1\)
• \(f(0) = e^{|0|} = e^0 = 1\)

Since \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)\), the function is continuous at \(x=0\). However, as shown by the different LHD and RHD, the function is not differentiable at \(x=0\).

Points where the derivative does not exist, but the function is continuous, often correspond to 'sharp corners' or 'cusps' on the graph of the function. The graph of \(f(x) = e^{|x|}\) has a sharp point at \(x=0\).