If f(x) = x e^{x(1-x)}, x in R,then f(x) is | vedclass

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

Question

(A) Given $f(x) = x e^{x(1-x)}$.Applying the product rule,$f'(x) = x \cdot e^{x(1-x)} \cdot \frac{d}{dx}(x-x^2) + e^{x(1-x)} \cdot 1$.$f'(x) = x e^{x(

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $f(x) = x e^{x(1-x)}$.Applying the product rule,$f'(x) = x \cdot e^{x(1-x)} \cdot \frac{d}{dx}(x-x^2) + e^{x(1-x)} \cdot 1$.$f'(x) = x e^{x(1-x)}(1-2x) + e^{x(1-x)}$.$f'(x) = e^{x(1-x)} [x(1-2x) + 1] = e^{x(1-x)} (x - 2x^2 + 1)$.$f'(x) = e^{x(1-x)} (-2x^2 + x + 1) = -e^{x(1-x)} (2x^2 - x - 1)$.$f'(x) = -e^{x(1-x)} (2x+1)(x-1) = e^{x(1-x)} (2x+1)(1-x)$.For $f(x)$ to be increasing,$f'(x) \geq 0$.Since $e^{x(1-x)} > 0$ for all $x \in R$,we need $(2x+1)(1-x) \geq 0$.This inequality holds when $x \in \left[-\frac{1}{2}, 1\right]$.Thus,$f(x)$ is increasing in $\left[-\frac{1}{2}, 1\right]$.

If $f(x) = x e^{x(1-x)}, x \in R$,then $f(x)$ is

Similar Questions

$y = x^2 (x - 3)^2$ increases for all values of $x$ lying in the interval :

Find the intervals in which the function $f(x) = -2x^{3} - 9x^{2} - 12x + 1$ is strictly increasing or strictly decreasing.

$f(x) = 10 - 6x - 2x^2$ is strictly increasing in the . . . . . . interval.

The function $f(x) = \tan x - x$ is:

Let $n$ be a natural number and let $a$ be a real number. The number of zeroes of $x^{2n+1} - (2n+1)x + a = 0$ in the interval $[-1, 1]$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module