The function f(x) = frac{log x}{x} is increas | vedclass

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

Question

(B) Given the function $f(x) = \frac{\log x}{x}$.To find the interval where the function is increasing,we first find the derivative $f'(x)$ using the

Vedclass · Accepted Answer

$(0, e)$

Vedclass · Answer

(B) Given the function $f(x) = \frac{\log x}{x}$.To find the interval where the function is increasing,we first find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2}$$f'(x) = \frac{x \cdot (\frac{1}{x}) - \log x \cdot (1)}{x^2} = \frac{1 - \log x}{x^2}$.For the function to be increasing,we must have $f'(x) > 0$.Since $x^2 > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 - \log x > 0$.$1 > \log x \implies \log_e x Since the domain of $\log x$ is $x > 0$,the function is increasing in the interval $(0, e)$.

The function $f(x) = \frac{\log x}{x}$ is increasing in the interval

Similar Questions

$F(x) = \log |\sin x|$,where $x \in (0, \pi)$,is strictly increasing on

If $f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x}$ is strictly increasing for all real values of $x$,then

If $f(x)=x^3-10x^2+200x-10$,then

If $f(x) = \int_x^{x+1} e^{-t^2} dt$,then the interval in which $f(x)$ is decreasing is

If $f(x) = x^{3/2}(3x - 10)$,$x \geq 0$,then in which interval is $f(x)$ an increasing function?

Vedclass Test Series

Exam Paper Generator

Online Exam Module