In which interval is the function f(x) = log | vedclass

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

Question

(B) To find the interval where the function $f(x) = \log x - \frac{2x}{2 + x}$ is increasing,we first find its derivative $f'(x)$.$f'(x) = \frac{d}{dx

Vedclass · Accepted Answer

$(0, \infty)$

Vedclass · Answer

(B) To find the interval where the function $f(x) = \log x - \frac{2x}{2 + x}$ is increasing,we first find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(\log x) - \frac{d}{dx}\left(\frac{2x}{2 + x}\right)$Using the quotient rule for the second term: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$$f'(x) = \frac{1}{x} - \frac{(2)(2 + x) - (2x)(1)}{(2 + x)^2}$$f'(x) = \frac{1}{x} - \frac{4 + 2x - 2x}{(2 + x)^2}$$f'(x) = \frac{1}{x} - \frac{4}{(2 + x)^2}$$f'(x) = \frac{(2 + x)^2 - 4x}{x(2 + x)^2}$$f'(x) = \frac{4 + x^2 + 4x - 4x}{x(2 + x)^2} = \frac{x^2 + 4}{x(2 + x)^2}$Since $x^2 + 4 > 0$ and $(2 + x)^2 > 0$ for all $x$ in the domain of $\log x$ (which is $x > 0$),the sign of $f'(x)$ depends only on $x$.For $f(x)$ to be increasing,we require $f'(x) > 0$.$\frac{x^2 + 4}{x(2 + x)^2} > 0 \implies x > 0$.Thus,the function is increasing in the interval $(0, \infty)$.

In which interval is the function $f(x) = \log x - \frac{2x}{2 + x}$ an increasing function?

Similar Questions

Let $g(x) = 2f(\frac{x}{2}) + f(2 - x)$ and $f''(x) < 0$ for all $x \in (0, 2)$. Then $g(x)$ increases in

If $f(x) = \sin x - \cos x - ax + b$ is a decreasing function for all $x \in R$,then:

Let $f(x) = x^3 + bx^2 + cx + d$ where $0 < b^2 < c$. Then $f(x)$:

For every value of $x \in [1, 3]$,the function $f(x) = \frac{1}{8^x}$ is

For what values of $K$ is the function $f(x) = x^3 + 6x^2 + (9 + 2K)x + 1$ strictly increasing for all $x \in \mathbb{R}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module