If x 0,then frac{x}{1+x} - log(1+x) | vedclass

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

Question

(A) Let $f(x) = \frac{x}{1+x} - \log(1+x)$ for $x > 0$. First,we find the derivative $f'(x)$: $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} - \frac{1}{1+x}

Vedclass · Accepted Answer

is less than zero

Vedclass · Answer

(A) Let $f(x) = \frac{x}{1+x} - \log(1+x)$ for $x > 0$. First,we find the derivative $f'(x)$: $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} - \frac{1}{1+x}$ $f'(x) = \frac{1}{(1+x)^2} - \frac{1}{1+x} = \frac{1 - (1+x)}{(1+x)^2} = \frac{-x}{(1+x)^2}$. Since $x > 0$,we have $f'(x)  0$. This implies that $f(x)$ is a strictly decreasing function for $x > 0$. As $x 	o 0^+$,$f(x) 	o \frac{0}{1} - \log(1) = 0$. Since the function starts at $0$ and is strictly decreasing for $x > 0$,it follows that $f(x)  0$.

If $x > 0$,then $\frac{x}{1+x} - \log(1+x)$

Similar Questions

Which of the following functions are strictly decreasing on $\left(0, \frac{\pi}{2}\right)?$

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

Function $f(x) = \frac{|x - 1|}{x^2}$ is monotonically decreasing in

Let $f(x) = \begin{cases} x e^{3x}, & x \le 0 \\ 2x^3 + x, & x > 0 \end{cases}$. Find the complete set of values of $x$ for which $f'(x)$ is an increasing function.

Define $f(x) = \frac{1}{2}[|\sin x| + \sin x]$, $0 < x \leq 2\pi$. Then, $f$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module