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

Vedclass · Accepted Answer

decreasing for $x  0$.

Vedclass · Answer

(C) Given the function $f(x) = 1 - e^{-\frac{x^2}{2}}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(1 - e^{-\frac{x^2}{2}}) = 0 - e^{-\frac{x^2}{2}} \cdot \frac{d}{dx}(-\frac{x^2}{2}) = -e^{-\frac{x^2}{2}} \cdot (-x) = x e^{-\frac{x^2}{2}}$.Since $e^{-\frac{x^2}{2}} > 0$ for all $x \in R$,the sign of $f'(x)$ depends only on the sign of $x$.Case $1$: If $x > 0$,then $f'(x) > 0$,which means the function $f(x)$ is increasing.Case $2$: If $x Therefore,the function is decreasing for $x  0$.

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

Similar Questions

Let $y = x^2 e^{-x}$. The interval in which $y$ increases with respect to $x$ is:

If $x \in (0, 1)$,then what type of function is $f(x) = x^{100} + \sin x - 1$?

Show that the function given by $f(x) = \sin x$ is decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

Prove that the logarithmic function $f(x) = \log x$ is strictly increasing on $(0, \infty).$

What is the nature of the function $f(x) = \sin x$ at $x = 2\pi / 3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module