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

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

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $f(x) = (x + 2) e^{-x}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = (x + 2) \frac{d}{dx}(e^{-x}) + e^{-x} \frac{d}{dx}(x + 2)$$f'(x) = (x + 2)(-e^{-x}) + e^{-x}(1)$$f'(x) = e^{-x} [-(x + 2) + 1] = e^{-x}(-x - 1) = -e^{-x}(x + 1)$.Since $e^{-x} > 0$ for all $x \in \mathbb{R}$,the sign of $f'(x)$ depends on $-(x + 1)$.For $f(x)$ to be increasing,$f'(x) > 0 \Rightarrow -(x + 1) > 0 \Rightarrow x + 1 Thus,$f(x)$ is increasing in $(-\infty, -1)$.For $f(x)$ to be decreasing,$f'(x)  0 \Rightarrow x > -1$.Thus,$f(x)$ is decreasing in $(-1, \infty)$.Therefore,the correct option is $C$.

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

Similar Questions

Let $f(x) = \cos \left(\frac{\pi}{x}\right), x \neq 0$. Assuming $k$ is an integer, which of the following is true?

The range in which $y = -x^{2} + 6x - 3$ is increasing is

The interval in which $y = \ln(\ln(x)), x > 1$ is decreasing is

Let $\phi(x) = f(x) + f(2a - x)$, $x \in [0, 2a]$ and $f^{\prime \prime}(x) > 0$ for all $x \in [0, a]$. Then $\phi(x)$ is

Prove that the function given by $f(x) = x^{3} - 3x^{2} + 3x - 100$ is increasing in $R$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module