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: State the test.
A function is strictly increasing wherever its derivative $f'(x)$ is positive, so we differentiate $f(x) = e^{-1/x}$.
Step 2: Note the domain restriction.
The term $\dfrac{1}{x}$ is undefined at $x = 0$, so $x = 0$ is excluded from the start.
Step 3: Differentiate with the chain rule.
$f'(x) = e^{-1/x}\cdot \dfrac{d}{dx}\!\left(-\dfrac{1}{x}\right)$. Since $\dfrac{d}{dx}\!\left(-\dfrac{1}{x}\right) = \dfrac{1}{x^2}$, we get $f'(x) = \dfrac{e^{-1/x}}{x^2}$.
Step 4: Sign of the numerator.
$e^{-1/x}$ is an exponential, which is always positive for every allowed $x$.
Step 5: Sign of the denominator.
$x^2 > 0$ for every $x \neq 0$.
Step 6: Combine the signs.
A positive over a positive is positive, so $f'(x) > 0$ for all $x \neq 0$. Hence $f$ is strictly increasing for every non-zero real number, which is option 4 and matches the key.
\[ \boxed{x \in \mathbb{R}\setminus\{0\}} \]