y = x^2 e^{-x} is increasing on . . . . . . | vedclass

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

Question

(D) To find the intervals where the function $y = f(x) = x^2 e^{-x}$ is increasing,we find its derivative $f'(x)$.Using the product rule,$f'(x) = \fra

Vedclass · Accepted Answer

$(0, 2)$

Vedclass · Answer

(D) To find the intervals where the function $y = f(x) = x^2 e^{-x}$ is increasing,we find its derivative $f'(x)$.Using the product rule,$f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x})$.$f'(x) = 2x e^{-x} - x^2 e^{-x} = x(2 - x) e^{-x}$.For the function to be increasing,we require $f'(x) > 0$.Since $e^{-x} > 0$ for all real $x$,the sign of $f'(x)$ depends on $x(2 - x)$.$x(2 - x) > 0$ implies $x(x - 2) This inequality holds when $x$ is between the roots $0$ and $2$.Thus,the function is increasing on the interval $(0, 2)$.

$y = x^2 e^{-x}$ is increasing on . . . . . . .

Similar Questions

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

The interval of the decreasing function $f(x) = x^3 - x^2 - x - 4$ is

Let $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$,for $x \in (-\infty, \infty)$. For which interval is $f(x)$ an increasing function?

The function $f(x) = \frac{x}{x^2-6x-16}$,where $x \in \mathbb{R} - \{-2, 8\}$,

For what values of $a$ is the function $f(x) = x^{2} + ax + 1$ increasing on the interval $[1, 2]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module