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

If $f(x) = x e^{x(1-x)}, x \in R$,then $f(x)$ is

Let $f(x) = \sin x$ and $g(x) = x$.
Statement-$1$: For $x \in (0, \infty)$,$f(x) \leq g(x)$.
Statement-$2$: For $x \in (0, \infty)$,$f(x) \leq 1$ but as $x \rightarrow \infty$,$g(x) \rightarrow \infty$.

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

The roots of $(x-41)^{49}+(x-49)^{41}+(x-2009)^{2009}=0$ are

The function $f(x) = x^2$ is increasing in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module