Find the interval in which the function f(x) | vedclass

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

Question

(D) Given $f(x) = \log x - \frac{2x}{x+2}$.The domain of the function is $x > 0$.Differentiating with respect to $x$:$f'(x) = \frac{1}{x} - \frac{(x+2

Vedclass · Accepted Answer

$x \in(0, \infty)$

Vedclass · Answer

(D) Given $f(x) = \log x - \frac{2x}{x+2}$.The domain of the function is $x > 0$.Differentiating with respect to $x$:$f'(x) = \frac{1}{x} - \frac{(x+2)(2) - 2x(1)}{(x+2)^2}$$f'(x) = \frac{1}{x} - \frac{4}{(x+2)^2}$$f'(x) = \frac{(x+2)^2 - 4x}{x(x+2)^2} = \frac{x^2 + 4x + 4 - 4x}{x(x+2)^2} = \frac{x^2 + 4}{x(x+2)^2}$Since $x^2 + 4 > 0$ and $(x+2)^2 > 0$ for all $x$ in the domain,$f'(x) > 0$ for all $x > 0$.Therefore,the function is strictly increasing for $x \in (0, \infty)$.

Find the interval in which the function $f(x) = \log x - \frac{2x}{x+2}$ is strictly increasing.

Similar Questions

The interval for which the given function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is decreasing,is

The interval in which the function $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is an increasing function,is

For every value of $x$,the function $f(x) = e^x$ is:

Show that the function $f$ given by $f(x) = x^{3} - 3x^{2} + 4x$,$x \in R$ is strictly increasing on $R$.

The function $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in

Vedclass Test Series

Exam Paper Generator

Online Exam Module