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

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

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the function $f(x) = (x+2)e^{-x}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$ using the product rule:$f'(x) = \frac{d}{dx}(x+2) \cdot e^{-x} + (x+2) \cdot \frac{d}{dx}(e^{-x})$$f'(x) = 1 \cdot e^{-x} + (x+2) \cdot (-e^{-x})$$f'(x) = e^{-x}(1 - (x+2))$$f'(x) = e^{-x}(1 - x - 2)$$f'(x) = -e^{-x}(x+1)$Now,we analyze the sign of $f'(x)$:Since $e^{-x} > 0$ for all real $x$,the sign of $f'(x)$ depends on $-(x+1)$.$1$. If $x  0$. Thus,$f'(x) > 0$,and the function is monotonically increasing in $(-\infty, -1)$.$2$. If $x > -1$,then $(x+1) > 0$,so $-(x+1) Therefore,the function is increasing in $(-\infty, -1)$ and decreasing in $(-1, \infty)$.

The function $f(x)$ is defined by $f(x)=(x+2) e^{-x}$. Which of the following statements is true?

Similar Questions

The function $y = 6 - 9x - x^2$ is strictly increasing on the interval . . . . . . .

The function $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$ is

The set of real values of $x$ for which $f(x) = \frac{x}{\log x}$ is increasing,is

The interval in which the function $f(x) = \operatorname{Tan}^{-1}(\sin x + \cos x)$ is an increasing function,is

$A$ function $y = f(x)$ is given by $x = \frac{1}{1 + t^2}$ and $y = \frac{1}{t(1 + t^2)}$ for all $t > 0$. Then $f$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module