Key Concepts and Formulas

• Continuity of a function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c} f(x) = f(c)$. For a function defined piecewise, continuity at the boundary points requires checking the left-hand limit, right-hand limit, and the function value at that point.
• Differentiability of a function: A function $f(x)$ is differentiable at a point $x=c$ if the limit of the difference quotient exists, i.e., $\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$ exists. For piecewise functions, differentiability at boundary points requires checking if the left-hand derivative and the right-hand derivative are equal.
• Properties of Logarithms and Exponentials:
  • $|\log_e x| = -\log_e x$ for $0 < x < 1$
  • $|\log_e x| = \log_e x$ for $x \geq 1$
  • $e^{-\log_e x} = e^{\log_e (x^{-1})} = x^{-1} = \frac{1}{x}$
  • $e^{\log_e x} = x$

Step-by-Step Solution

Step 1: Analyze the function and rewrite it in a piecewise form. The given function is $f(x) = e^{-|\log_e x|}$ for $x \in (0, \infty)$. We need to consider the definition of the absolute value of $\log_e x$.

• If $0 < x < 1$, then $\log_e x < 0$, so $|\log_e x| = -\log_e x$.
• If $x \geq 1$, then $\log_e x \geq 0$, so $|\log_e x| = \log_e x$.

Substituting these into the function definition: For $0 < x < 1$: $f(x) = e^{-(-\log_e x)} = e^{\log_e x} = x$.

For $x \geq 1$: $f(x) = e^{-(\log_e x)} = e^{\log_e (x^{-1})} = x^{-1} = \frac{1}{x}$.

So, the piecewise definition of $f(x)$ is: $f(x) = \begin{cases} x & \text{if } 0 < x < 1 \\ \frac{1}{x} & \text{if } x \geq 1 \end{cases}$

Step 2: Determine points of discontinuity (find $m$). A function is discontinuous at points where it is not defined, or where the limit does not exist, or where the limit does not equal the function value. The domain of $f(x)$ is $(0, \infty)$. The function is defined by elementary functions ($x$ and $\frac{1}{x}$) over their respective intervals.

• For $0 < x < 1$, $f(x) = x$, which is a polynomial and is continuous everywhere.
• For $x > 1$, $f(x) = \frac{1}{x}$, which is a rational function and is continuous everywhere except where the denominator is zero. Since $x > 1$, the denominator is never zero.

We only need to check the continuity at the point where the definition of the function changes, which is $x=1$. We need to check if $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.

• Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$.
• Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = \frac{1}{1} = 1$.
• Function value at $x=1$: $f(1) = \frac{1}{1} = 1$ (using the second case of the piecewise definition, $x \geq 1$).

Since the left-hand limit, the right-hand limit, and the function value at $x=1$ are all equal to 1, the function $f(x)$ is continuous at $x=1$. Therefore, there are no points in the domain $(0, \infty)$ where $f(x)$ is not continuous. So, $m=0$.

Step 3: Determine points of non-differentiability (find $n$). A function is not differentiable at points where the left-hand derivative and the right-hand derivative are not equal, or where the function is not continuous. We already established that $f(x)$ is continuous everywhere in its domain. So, we only need to check for differentiability at the point where the definition changes, $x=1$.

We need to find the left-hand derivative ($f'_-(1)$) and the right-hand derivative ($f'_+(1)$) at $x=1$.

• For $0 < x < 1$, $f(x) = x$. The derivative is $f'(x) = \frac{d}{dx}(x) = 1$. So, the left-hand derivative at $x=1$ is $f'_-(1) = \lim_{x \to 1^-} f'(x) = 1$. Alternatively, using the definition of the derivative: $f'_-(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$. For $h < 0$ and small, $1+h$ is in the interval $(0, 1)$. $f(1+h) = 1+h$. $f(1) = 1$. $f'_-(1) = \lim_{h \to 0^-} \frac{(1+h) - 1}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1$.

• For $x > 1$, $f(x) = \frac{1}{x}$. The derivative is $f'(x) = \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}$. So, the right-hand derivative at $x=1$ is $f'_+(1) = \lim_{x \to 1^+} f'(x) = -\frac{1}{1^2} = -1$. Alternatively, using the definition of the derivative: $f'_+(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$. For $h > 0$ and small, $1+h$ is in the interval $(1, \infty)$. $f(1+h) = \frac{1}{1+h}$. $f(1) = 1$. $f'_+(1) = \lim_{h \to 0^+} \frac{\frac{1}{1+h} - 1}{h} = \lim_{h \to 0^+} \frac{\frac{1 - (1+h)}{1+h}}{h} = \lim_{h \to 0^+} \frac{-h}{h(1+h)} = \lim_{h \to 0^+} \frac{-1}{1+h} = -1$.

Since $f'_-(1) = 1$ and $f'_+(1) = -1$, the left-hand derivative and the right-hand derivative are not equal at $x=1$. Therefore, the function $f(x)$ is not differentiable at $x=1$. There is exactly one point ($x=1$) where $f(x)$ is not differentiable. So, $n=1$.

Step 4: Calculate $m+n$. We found $m=0$ and $n=1$. Therefore, $m+n = 0 + 1 = 1$.

Common Mistakes & Tips

• Incorrectly simplifying the absolute value: Ensure the conditions for $|\log_e x|$ are correctly applied based on whether $x$ is less than 1 or greater than or equal to 1.
• Forgetting to check continuity/differentiability at boundary points: For piecewise functions, the points where the definition changes are critical for checking continuity and differentiability.
• Confusing continuity and differentiability: A function can be continuous at a point but not differentiable there (e.g., corners or cusps). Differentiability implies continuity, but continuity does not imply differentiability.

Summary

The function $f(x) = e^{-|\log_e x|}$ was rewritten as a piecewise function: $f(x) = x$ for $0 < x < 1$ and $f(x) = \frac{1}{x}$ for $x \geq 1$. We analyzed the continuity by checking the limits at the boundary point $x=1$. Since the left-hand limit, right-hand limit, and function value all matched, the function is continuous everywhere in its domain, so $m=0$. We then analyzed differentiability by comparing the left-hand and right-hand derivatives at $x=1$. The left-hand derivative was 1, and the right-hand derivative was -1. Since they are not equal, the function is not differentiable at $x=1$, so $n=1$. The sum $m+n$ is $0+1=1$.

The final answer is $\boxed{1}$.