The function f defined by f(x) = (x + 2) e^{- | vedclass

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

Question

(C) To determine the intervals of increase and decrease,we find the derivative $f'(x)$.Given $f(x) = (x + 2) e^{-x}$.Using the product rule,$f'(x) = (

Vedclass · Accepted Answer

decreasing in $(-1, \infty)$ and increasing in $(-\infty, -1)$

Vedclass · Answer

(C) To determine the intervals of increase and decrease,we find the derivative $f'(x)$.Given $f(x) = (x + 2) e^{-x}$.Using the product rule,$f'(x) = (1) e^{-x} + (x + 2) (-e^{-x})$.$f'(x) = e^{-x} (1 - x - 2) = e^{-x} (-x - 1) = -(x + 1) e^{-x}$.For the function to be increasing,$f'(x) > 0$,which implies $-(x + 1) e^{-x} > 0$. Since $e^{-x} > 0$ for all $x$,we must have $-(x + 1) > 0$,or $x + 1 Thus,the function is increasing in $(-\infty, -1)$.For the function to be decreasing,$f'(x)  0$,or $x > -1$.Thus,the function is decreasing in $(-1, \infty)$.Therefore,the correct option is $C$.

The function $f$ defined by $f(x) = (x + 2) e^{-x}$ is

Similar Questions

If $f(x) = \sin x - \cos x,$ the function is decreasing in the interval $0 \le x \le 2\pi$ for which of the following?

In which interval does the function $f(x) = 2x^2 - \log |x|$ $(x \neq 0)$ monotonically increase?

Let $g(x) = f(x) + f(1 - x)$ and $f''(x) < 0$ for $0 \le x \le 1$. Which of the following is true regarding the monotonicity of $g(x)$?

The roots of $(x-41)^{49}+(x-49)^{41}+(x-2009)^{2009}=0$ are

If $R$ is the least value of $a$ such that the function $f(x) = x^{2} + ax + 1$ is increasing on $[1, 2]$ and $S$ is the greatest value of $a$ such that the function $f(x) = x^{2} + ax + 1$ is decreasing on $[1, 2]$,then the value of $|R - S|$ is ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module