The function f(x)=log (1+x)-frac{2 x}{2+x} is | vedclass

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

Question

(C) Given,$f(x)=\log (1+x)-\frac{2 x}{2+x}$.For the function to be defined,we must have $1+x > 0$,i.e.,$x > -1$.Differentiating the function with resp

Vedclass · Accepted Answer

$(-1, \infty)$

Vedclass · Answer

(C) Given,$f(x)=\log (1+x)-\frac{2 x}{2+x}$.For the function to be defined,we must have $1+x > 0$,i.e.,$x > -1$.Differentiating the function with respect to $x$,we get:$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{(2+x)(1) - x(1)}{(2+x)^2} \right]$$f^{\prime}(x) = \frac{1}{1+x} - 2 \left[ \frac{2+x-x}{(2+x)^2} \right] = \frac{1}{1+x} - \frac{4}{(2+x)^2}$$f^{\prime}(x) = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4+x^2+4x-4-4x}{(1+x)(2+x)^2}$$f^{\prime}(x) = \frac{x^2}{(1+x)(2+x)^2}$.Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for all $x > -1$,the sign of $f^{\prime}(x)$ depends on the term $\frac{1}{1+x}$.For $f(x)$ to be increasing,we require $f^{\prime}(x) > 0$.Since $x > -1$,$1+x > 0$,which implies $f^{\prime}(x) > 0$ for all $x \in (-1, \infty)$.Thus,the function is increasing on $(-1, \infty)$.

The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

Similar Questions

Statement $-1:$ The function $f(x) = x^2(e^x + e^{-x})$ is increasing for all $x > 0.$
Statement $-2:$ The functions $g(x) = x^2e^x$ and $h(x) = x^2e^{-x}$ are increasing for all $x > 0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b).$

For what values of $x$ is the function $f(x) = \tan^{-1}(\sin x + \cos x)$ strictly increasing?

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

If $f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x)$,then $f(x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module