The function f(x) = e^{-1/x} is strictly incr | vedclass

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(D) Given the function $f(x) = e^{-1/x}$.To find the intervals where the function is strictly increasing,we calculate the derivative $f'(x)$:$f'(x) =

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$x$ is a non-zero real number

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(D) Given the function $f(x) = e^{-1/x}$.To find the intervals where the function is strictly increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(e^{-1/x}) = e^{-1/x} \cdot \frac{d}{dx}(-x^{-1}) = e^{-1/x} \cdot (x^{-2}) = \frac{1}{x^2 e^{1/x}}$.For the function to be strictly increasing,we must have $f'(x) > 0$.Since $x^2 > 0$ for all $x 
eq 0$ and $e^{1/x} > 0$ for all $x 
eq 0$,the derivative $f'(x) = \frac{1}{x^2 e^{1/x}}$ is always positive for all $x$ in its domain.The domain of the function $f(x) = e^{-1/x}$ is all real numbers except $x = 0$.Therefore,the function is strictly increasing for all $x \in \mathbb{R} \setminus \{0\}$.

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

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