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

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

Question

(D) Let $f(x) = \frac{1}{1 + x^2}$.To find the interval where the function is decreasing,we find its derivative $f'(x)$.Using the chain rule,$f'(x) =

Vedclass · Accepted Answer

$(0, \infty )$

Vedclass · Answer

(D) Let $f(x) = \frac{1}{1 + x^2}$.To find the interval where the function is decreasing,we find its derivative $f'(x)$.Using the chain rule,$f'(x) = -\frac{1}{(1 + x^2)^2} \cdot (2x) = -\frac{2x}{(1 + x^2)^2}$.$A$ function is decreasing when $f'(x) So,$-\frac{2x}{(1 + x^2)^2} Since $(1 + x^2)^2$ is always positive for all real $x$,the sign of $f'(x)$ depends only on $-2x$.Therefore,$-2x  0$.Thus,the function is decreasing in the interval $(0, \infty )$.

The function $f(x) = \frac{1}{1 + x^2}$ is decreasing in the interval

Similar Questions

For what values of $K$ is the function $f(x) = Kx^3 + 9x^2 + 9x + 3$ an increasing function on $R$?

Let $\phi (x) = (f(x))^3 - 3(f(x))^2 + 4f(x) + 5x + 3 \sin x + 4 \cos x$ for all $x \in R$,then -

The function $f(x) = (x - 2)|x - 3|$ is monotonically increasing in:

For what values of $x$ does the function $f(x) = x^2 e^{-x}$ strictly increase?

On the interval $(1, 3)$,the function $f(x) = 3x + \frac{2}{x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module