If f(x) = frac{log x}{x} (x 0),then it is i | vedclass

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

Question

(A) Given the function $f(x) = \frac{\log x}{x}$.To determine where the function is increasing,we find its derivative $f'(x)$ using the quotient rule:

Vedclass · Accepted Answer

$(0, e)$

Vedclass · Answer

(A) Given the function $f(x) = \frac{\log x}{x}$.To determine where the function is increasing,we find its derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.For the function to be increasing,we require $f'(x) > 0$.Since $x^2 > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 - \log x > 0$.This simplifies to $1 > \log x$,which means $\log x Taking the exponential of both sides,we get $x Given the domain $x > 0$,the function is increasing in the interval $(0, e)$.

If $f(x) = \frac{\log x}{x}$ $(x > 0)$,then it is increasing in

Similar Questions

The range of values of $x$ for which $f(x)=x^3+6x^2-36x+7$ is increasing is:

The function $f(x) = 1 - e^{-x^2/2}$ is

The function $f(x) = \tan x - x$ is:

The function $f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$ is strictly

Let $(2, 3)$ be the largest open interval in which the function $f(x) = 2 \log_e(x-2) - x^2 + ax + 1$ is strictly increasing and $(b, c)$ be the largest open interval in which the function $g(x) = (x-1)^3(x+2-a)^2$ is strictly decreasing. Then $100(a+b-c)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module