The function f(x) = 1 - e^{-x^2/2} is | vedclass

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

Question

(C) Given the function $f(x) = 1 - e^{-x^2/2}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx

Vedclass · Accepted Answer

Decreasing for $x  0$

Vedclass · Answer

(C) Given the function $f(x) = 1 - e^{-x^2/2}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(1 - e^{-x^2/2}) = 0 - e^{-x^2/2} \cdot \frac{d}{dx}(-x^2/2) = -e^{-x^2/2} \cdot (-x) = x e^{-x^2/2}$.Since $e^{-x^2/2}$ is always positive for all real $x$,the sign of $f'(x)$ depends solely on $x$.$1$. For $f(x)$ to be increasing,we require $f'(x) > 0$,which implies $x e^{-x^2/2} > 0$. Since $e^{-x^2/2} > 0$,this holds when $x > 0$.$2$. For $f(x)$ to be decreasing,we require $f'(x)  0$,this holds when $x Therefore,the function is decreasing for $x  0$.

The function $f(x) = 1 - e^{-x^2/2}$ is

Similar Questions

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

For what values of $K$ is the function $f(x) = x^3 + 6x^2 + (9 + 2K)x + 1$ strictly increasing for all $x \in \mathbb{R}$?

The function $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in the interval.....

For all real values of $x$,the increasing function $f(x)$ is

The function $f(x) = \frac{\log x}{x}$ is increasing in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module