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

The question asks us to consider two statements about the function \(f(x) = e^{-|x|}\) specifically at the point \(x = 0\). The statements are regarding its continuity and differentiability at this point.

First, let's understand the function \(f(x) = e^{-|x|}\). The absolute value function \(|x|\) is defined differently for non-negative and negative values of \(x\).

• If \(x \ge 0\), then \(|x| = x\). So, for \(x \ge 0\), \(f(x) = e^{-x}\).
• If \(x < 0\), then \(|x| = -x\). So, for \(x < 0\), \(f(x) = e^{-(-x)} = e^x\).

Thus, we can write the function \(f(x)\) piecewise as:

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

Checking Continuity of \(f(x)\) at \(x = 0\)

A function \(f(x)\) is continuous at a point \(x=a\) if the following three conditions are met:

1. The function \(f(a)\) is defined.
2. The limit of the function as \(x\) approaches \(a\) exists, i.e., \(\lim_{x \to a} f(x)\) exists. This requires the left-hand limit and the right-hand limit to exist and be equal: \(\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)\).
3. The limit of the function at \(a\) is equal to the function value at \(a\): \(\lim_{x \to a} f(x) = f(a)\).

Let's check these conditions for \(f(x) = e^{-|x|}\) at \(x = 0\) (\(a=0\)).

• Condition 1: Is \(f(0)\) defined?
  Using the piecewise definition for \(x \ge 0\), we have \(f(0) = e^{-0} = e^0 = 1\). So, \(f(0)\) is defined and \(f(0) = 1\).
• Condition 2: Does \(\lim_{x \to 0} f(x)\) exist?
  We need to check the left-hand limit and the right-hand limit at \(x=0\).
  • Right-hand limit: As \(x \to 0^+\), \(x\) is slightly greater than 0. We use the definition \(f(x) = e^{-x}\) for \(x \ge 0\). \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{-x} = e^{-0} = e^0 = 1\).
  • Left-hand limit: As \(x \to 0^-\), \(x\) is slightly less than 0. We use the definition \(f(x) = e^x\) for \(x < 0\). \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^x = e^0 = 1\).
  Since the left-hand limit (\(\lim_{x \to 0^-} f(x) = 1\)) equals the right-hand limit (\(\lim_{x \to 0^+} f(x) = 1\)), the limit \(\lim_{x \to 0} f(x)\) exists and is equal to 1.
• Condition 3: Is \(\lim_{x \to 0} f(x) = f(0)\)?
  We found that \(\lim_{x \to 0} f(x) = 1\) and \(f(0) = 1\). Since \(\lim_{x \to 0} f(x) = f(0)\), all conditions for continuity at \(x=0\) are met.

Therefore, statement 1, "The function is continuous at \(x = 0\)", is correct.

Checking Differentiability of \(f(x)\) at \(x = 0\)

A function \(f(x)\) is differentiable at a point \(x=a\) if the derivative \(f'(a)\) exists. The derivative at a point is defined as the limit of the difference quotient:

\(\qquad f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\)

For the derivative to exist at \(x=a\), the left-hand derivative and the right-hand derivative must exist and be equal:

• Right-hand derivative at \(a\): \(f'(a^+) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}\)
• Left-hand derivative at \(a\): \(f'(a^-) = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}\)

Let's check differentiability for \(f(x) = e^{-|x|}\) at \(x = 0\) (\(a=0\)). We know \(f(0)=1\).

• Right-hand derivative at \(x=0\):
  We need to evaluate \(f'(0^+) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h}\). As \(h \to 0^+\), \(h\) is positive, so \(0+h = h > 0\). We use \(f(x) = e^{-x}\) for \(x \ge 0\). \(f(0+h) = f(h) = e^{-h}\). \(f'(0^+) = \lim_{h \to 0^+} \frac{e^{-h} - 1}{h}\). This is a standard limit form \(\lim_{u \to 0} \frac{e^u - 1}{u} = 1\). Here, \(u = -h\). As \(h \to 0^+\), \(u \to 0^-\). \(\lim_{h \to 0^+} \frac{e^{-h} - 1}{h} = \lim_{h \to 0^+} -\frac{1 - e^{-h}}{h} = \lim_{h \to 0^+} - \left( \frac{e^{-h} - 1}{-h} \right)\). Let \(u = -h\). As \(h \to 0^+\), \(u \to 0^-\). \(\lim_{u \to 0^-} - \left( \frac{e^u - 1}{u} \right) = - (1) = -1\). Alternatively, using L'Hopital's rule, \(\lim_{h \to 0^+} \frac{\frac{d}{dh}(e^{-h} - 1)}{\frac{d}{dh}(h)} = \lim_{h \to 0^+} \frac{-e^{-h}}{1} = -e^0 = -1\). So, the right-hand derivative \(f'(0^+) = -1\).
• Left-hand derivative at \(x=0\):
  We need to evaluate \(f'(0^-) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h}\). As \(h \to 0^-\), \(h\) is negative, so \(0+h = h < 0\). We use \(f(x) = e^x\) for \(x < 0\). \(f(0+h) = f(h) = e^{h}\). \(f'(0^-) = \lim_{h \to 0^-} \frac{e^h - 1}{h}\). This is the standard limit \(\lim_{h \to 0} \frac{e^h - 1}{h} = 1\). As \(h \to 0^-\), the limit is 1. So, the left-hand derivative \(f'(0^-) = 1\).

For the function to be differentiable at \(x=0\), the left-hand derivative must equal the right-hand derivative. We found \(f'(0^+) = -1\) and \(f'(0^-) = 1\). Since \(-1 \ne 1\), the left-hand derivative is not equal to the right-hand derivative. Therefore, the function \(f(x) = e^{-|x|}\) is not differentiable at \(x = 0\). Statement 2 is incorrect.

Summary of Findings

Let's summarize our analysis of the statements for \(f(x) = e^{-|x|}\) at \(x=0\):

• Statement 1: The function is continuous at \(x = 0\). We found this to be Correct.
• Statement 2: The function is differentiable at \(x = 0\). We found this to be Incorrect.

Based on our analysis, only statement 1 is correct.

Revision Table: Continuity and Differentiability

Additional Information: Understanding the Graph of \(e^{-|x|}\)

The graph of \(y = e^{-|x|}\) has a distinctive shape. For \(x \ge 0\), the graph is \(y = e^{-x}\), which is an exponentially decreasing curve starting at \(y=1\) when \(x=0\). For \(x < 0\), the graph is \(y = e^x\), which is an exponentially increasing curve ending at \(y=1\) as \(x\) approaches 0 from the left.

At \(x=0\), the two parts of the graph meet at the point \((0, 1)\). Because the function is continuous at \(x=0\), the graph does not have a break or a gap at this point.

However, the slopes of the two parts of the graph are different as they approach \(x=0\). The slope of \(e^{-x}\) is \(-e^{-x}\), which approaches \(-1\) as \(x \to 0^+\). The slope of \(e^x\) is \(e^x\), which approaches \(1\) as \(x \to 0^-\).

Since the slopes from the left and right are different at \(x=0\), the graph has a sharp corner or a "cusp" at \((0, 1)\). This sharp corner indicates that the function is not smooth at this point, which is consistent with our finding that the function is not differentiable at \(x=0\). Differentiability at a point means the graph is smooth at that point, with a unique tangent line.