Let f:(-1, infty) rightarrow mathbb{R} be def | vedclass

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

Question

(A) For $x 
eq 0$,$f'(x) = \frac{d}{dx} \left( \frac{\ln(1+x)}{x} ight) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} = \frac{x - (1+x)\ln(1+x)}{x

Vedclass · Accepted Answer

decreases in $(-1, \infty)$

Vedclass · Answer

(A) For $x 
eq 0$,$f'(x) = \frac{d}{dx} \left( \frac{\ln(1+x)}{x} ight) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} = \frac{x - (1+x)\ln(1+x)}{x^2(1+x)}$.Let $h(x) = x - (1+x)\ln(1+x)$.Then $h'(x) = 1 - [\ln(1+x) + (1+x) \cdot \frac{1}{1+x}] = 1 - \ln(1+x) - 1 = -\ln(1+x)$.For $x \in (-1, 0)$,$1+x \in (0, 1)$,so $\ln(1+x)  0$.For $x \in (0, \infty)$,$1+x > 1$,so $\ln(1+x) > 0$,which implies $h'(x) Since $h(0) = 0 - (1)\ln(1) = 0$,$h(x)$ increases on $(-1, 0)$ and decreases on $(0, \infty)$.Thus,$h(x) Since $x^2(1+x) > 0$ for all $x \in (-1, \infty) \setminus \{0\}$,$f'(x) = \frac{h(x)}{x^2(1+x)} Therefore,the function $f$ is strictly decreasing on $(-1, \infty)$.

Let $f:(-1, \infty) \rightarrow \mathbb{R}$ be defined by $f(0)=1$ and $f(x)=\frac{1}{x} \ln(1+x), x \neq 0$. Then the function $f$

Similar Questions

The function $f(x) = (x(x - 2))^2$ is an increasing function on the set.....

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

If $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$ decreases for all values of $x$,then

Find the intervals in which the function $f$ given by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is $(i)$ increasing $(ii)$ decreasing.

Let $\phi(x) = f(x) + f(1-x)$ and $f^{\prime \prime}(x) < 0$ in $[0, 1]$, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module