For what values of x does the function f(x) = | vedclass

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

Question

(A) To find the intervals where the function $f(x) = x^2 e^{-x}$ is strictly increasing,we calculate its derivative $f'(x)$.Using the product rule:$f'

Vedclass · Accepted Answer

$0 < x < 2$

Vedclass · Answer

(A) To find the intervals where the function $f(x) = x^2 e^{-x}$ is strictly increasing,we calculate 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})$$f'(x) = 2x e^{-x} - x^2 e^{-x}$$f'(x) = x e^{-x} (2 - x)$For the function to be strictly increasing,we must have $f'(x) > 0$.Since $e^{-x} > 0$ for all $x \in \mathbb{R}$,the inequality $x e^{-x} (2 - x) > 0$ simplifies to:$x(2 - x) > 0$Multiplying by $-1$ reverses the inequality sign:$x(x - 2) The roots of the quadratic expression are $x = 0$ and $x = 2$. The expression is negative between the roots.Therefore,the function is strictly increasing for $0 < x < 2$.

For what values of $x$ does the function $f(x) = x^2 e^{-x}$ strictly increase?

Similar Questions

The function $f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$ is strictly

If $f(x) = x$,$-1 \le x \le 1$,then the function $f(x)$ is

Let $h(x) = f(x) - (f(x))^2 + (f(x))^3$ for every real number $x$. Then

The function $f(x) = x^3 - 3x$ is....

The values of $a$ for which the function $f(x) = (a + 2)x^3 - 3ax^2 + 9ax - 1$ decreases monotonically for all real $x$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module