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

$f(x) = x^3 - 27x + 5$ is an increasing function,when

The function $f(x) = \tan^{-1}(\sin x + \cos x), x > 0$ is an increasing function in the interval.....

Show that the function given by $f(x) = e^{2x}$ is strictly increasing on $R$.

Function $f(x) = \log_{10} \cos x$ is . . . . . . function in the interval $\left(0, \frac{\pi}{2}\right)$.

Let $I$ be any interval such that $I \cap [-1, 1] = \phi$. Prove that the function $f$ given by $f(x) = x + \frac{1}{x}$ is strictly increasing on $I$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module