The function f(x) = frac{x}{log_x e} is incre | vedclass

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

Question

(D) Given the function $f(x) = \frac{x}{\log_x e}$.Using the property $\log_x e = \frac{1}{\ln x}$,we can rewrite the function as:$f(x) = x \cdot \ln

Vedclass · Accepted Answer

$(\frac{1}{e}, \infty)$

Vedclass · Answer

(D) Given the function $f(x) = \frac{x}{\log_x e}$.Using the property $\log_x e = \frac{1}{\ln x}$,we can rewrite the function as:$f(x) = x \cdot \ln x$.To determine the interval where the function is increasing,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.For the function to be increasing,we require $f'(x) > 0$:$\ln x + 1 > 0$$\ln x > -1$$x > e^{-1}$$x > \frac{1}{e}$.Since the domain of the function is $x \in \mathbb{R}^+ - \{1\}$,the function is increasing on the interval $(\frac{1}{e}, 1) \cup (1, \infty)$. However,looking at the provided options,the most appropriate interval representing the increasing behavior is $(\frac{1}{e}, \infty)$ excluding the point $x=1$ where the function is undefined. Thus,the correct option is $D$.

The function $f(x) = \frac{x}{\log_x e}$ is increasing on the interval . . . . . . ,where $x \in \mathbb{R}^+ - \{1\}$.

Similar Questions

The interval in which $y = x^2 e^{-x}$ is decreasing is . . . . . . .

For what values of $x$ is the function $f(x) = x^9 + 3x^7 + 64$ strictly increasing?

If the function $y = \sin x(1 + \cos x)$ is defined in the interval $[-\pi, \pi]$,then $y$ is strictly increasing in the interval

The complete set of values of $m$ for which the function $f(x) = e^{\sin x} + 2m\sin x + 1$ is increasing for all $x \in \left( 0, \frac{\pi}{2} \right)$ is:

The function $f(x) = (1/2)^x$ on $R$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module