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

The number of real solutions of $x^{7}+5x^{3}+3x+1=0$ is equal to............

On which of the following intervals is the function $f$ given by $f(x)=x^{100}+\sin x-1$ decreasing?

The length of the longest interval,in which the function $f(x) = 3\sin x - 4\sin^3 x$ is increasing,is

Find the intervals in which the function $f$ given by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is $(i)$ increasing $(ii)$ decreasing.

The complete set of values of $x$ for which $f(x) = 2 \log_e(x - 2) - x^2 + 4x + 1$ is increasing is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module