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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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) =

Vedclass · Accepted Answer

$x$ is a non-zero real number

Vedclass · Answer

(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

Similar Questions

For every value of $x$,the function $f(x) = \frac{1}{5^x}$ is:

Prove that the function $f$ given by $f(x) = \log(\sin x)$ is increasing on $\left(0, \frac{\pi}{2}\right)$ and decreasing on $\left(\frac{\pi}{2}, \pi\right)$.

If $a < 0$,then the function $f(x) = e^{ax} + e^{-ax}$ is monotonically decreasing for all values of $x$ where ...

$f(x) = \cos x - 1 + \frac{x^2}{2!}, x \in R$. Then $f(x)$ is

The function $f(x) = \frac{x - 2}{x + 1}, (x \neq -1)$ is increasing on the interval:

Vedclass Test Series

Exam Paper Generator

Online Exam Module