Question

Function $f(x) = e^{-1/x}$ is strictly increasing for all $x$ where

• A. $x$ is only positive real number
• B. $x$ is only negative real number
• C. $x$ is a real number
• D. $x$ is a non-zero real number

Hint:

Whenever you differentiate a function of the type $e^{g(x)}$, the exponential part can be completely ignored when checking for a change of sign because $e^{\text{anything}}$ is always positive. The sign of the derivative depends entirely on $g'(x)$. Here, the derivative of $-\frac{1}{x}$ is $\frac{1}{x^2}$, which is obviously positive for any number except zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an exponential function $f(x) = e^{-1/x}$. We need to determine the complete domain or set of conditions on $x$ for which this function is strictly increasing.

Step 2: Key Formula or Approach:
A differentiable function is strictly increasing on an interval if its first derivative is strictly greater than zero ($f'(x) > 0$) for all points within that interval. We use the chain rule to differentiate the function: $$f'(x) = \frac{d}{dx}\left(e^{g(x)}\right) = e^{g(x)} \cdot g'(x)$$

Step 3: Detailed Explanation:
Given function: $$f(x) = e^{-1/x}$$ First, note that the function is undefined at $x = 0$ because division by zero is mathematically impossible. Therefore, $x$ cannot be zero ($x \neq 0$). Now, compute the first derivative with respect to $x$: $$f'(x) = e^{-1/x} \cdot \frac{d}{dx}\left(-\frac{1}{x}\right)$$ We know that the derivative of $-\frac{1}{x}$ (or $-x^{-1}$) is $-(-1 \cdot x^{-2}) = \frac{1}{x^2}$: $$f'(x) = e^{-1/x} \cdot \left(\frac{1}{x^2}\right) = \frac{e^{-1/x}}{x^2}$$ Let us analyze the signs of the components in this derivative expression for all valid real numbers: 1. The exponential function term $e^{-1/x}$ is always strictly positive ($e^k > 0$) for any real value of $k$.
2. The denominator term $x^2$ is a squared term, meaning it is strictly positive ($x^2 > 0$) for all real numbers except $x = 0$.
Since the numerator is always positive and the denominator is always positive (for $x \neq 0$), their quotient must be strictly positive: $$f'(x) = \frac{\text{positive}}{\text{positive}} > 0 \quad \text{for all } x \in \mathbb{R} \setminus \{0\}$$ This confirms that the function is strictly increasing everywhere except at $x=0$, meaning for all non-zero real numbers.

Step 4: Final Answer:
The function is strictly increasing for all non-zero real numbers, which corresponds to option (D).