If f(x) = frac{1}{x + 1} - log(1 + x),where x | vedclass

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

Question

(B) Given $f(x) = \frac{1}{x + 1} - \log(1 + x)$ for $x > 0$.To determine the nature of the function,we find its derivative $f'(x)$:$f'(x) = \frac{d}{

Vedclass · Accepted Answer

Decreasing function

Vedclass · Answer

(B) Given $f(x) = \frac{1}{x + 1} - \log(1 + x)$ for $x > 0$.To determine the nature of the function,we find its derivative $f'(x)$:$f'(x) = \frac{d}{dx} \left( \frac{1}{x + 1} \right) - \frac{d}{dx} (\log(1 + x))$$f'(x) = -\frac{1}{(x + 1)^2} - \frac{1}{1 + x}$$f'(x) = -\left( \frac{1 + (x + 1)}{(x + 1)^2} \right) = -\frac{x + 2}{(x + 1)^2}$Since $x > 0$,we have $x + 2 > 0$ and $(x + 1)^2 > 0$.Therefore,$f'(x) = -\frac{x + 2}{(x + 1)^2}  0$.Since the derivative $f'(x) < 0$ for all $x$ in the domain,the function $f$ is a decreasing function.

If $f(x) = \frac{1}{x + 1} - \log(1 + x)$,where $x > 0$,then what type of function is $f$?

Similar Questions

Given $f(x) = \int_{-2}^{x} t \cdot g'(t) \, dt$ for $x \geq -2$,where $g$ is an increasing function,then:

The equation $x^3+x-1=0$ has

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 $(a, b).$

In the interval $(-3,3)$,the function $f(x) = \frac{x}{3} + \frac{3}{x}, x \neq 0$ is :

If $f(x) = x \cdot e^{x(1-x)}$,then $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module