The function f(x) = x cdot e^{x(1-x)} is | vedclass

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

Question

(A) Given the function $f(x) = x \cdot e^{x-x^2}$.To find the intervals of increase or decrease,we find the derivative $f'(x)$.Using the product rule

Vedclass · Accepted Answer

increasing in $\left[-\frac{1}{2}, 1\right]$

Vedclass · Answer

(A) Given the function $f(x) = x \cdot e^{x-x^2}$.To find the intervals of increase or decrease,we find the derivative $f'(x)$.Using the product rule and chain rule: $f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$.$f'(x) = e^{x-x^2} [1 + x(1-2x)] = e^{x-x^2} (1 + x - 2x^2)$.Factor the quadratic expression: $1 + x - 2x^2 = -(2x^2 - x - 1) = -(2x+1)(x-1) = (2x+1)(1-x)$.So,$f'(x) = e^{x-x^2} (2x+1)(1-x)$.For the function to be increasing,$f'(x) \geq 0$.Since $e^{x-x^2} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on $(2x+1)(1-x)$.$(2x+1)(1-x) \geq 0$ implies $-\frac{1}{2} \leq x \leq 1$.Thus,the function is increasing in the interval $\left[-\frac{1}{2}, 1\right]$.

The function $f(x) = x \cdot e^{x(1-x)}$ is

Similar Questions

In the interval $(-3,3)$,the function $f(x) = \frac{x}{3} + \frac{3}{x}, x \neq 0$ is :

The function $f(x) = 2{x^3} + 18{x^2} - 96x + 45$ is an increasing function when:

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

If $f(x)=k x^3-9 x^2+9 x+3$ $(k>0)$ is increasing for all $x$,then

In the open interval $\left(0, \frac{\pi}{2}\right)$,which of the following is true for the expression $\cos x + x \sin x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module