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

(C) Given $f(x) = x e^{x(1-x)}$.Applying the product rule and chain rule,we find the derivative:$f'(x) = 1 \cdot e^{x(1-x)} + x \cdot e^{x(1-x)} \cdot

Vedclass · Accepted Answer

increasing on $[-1/2, 1]$

Vedclass · Answer

(C) Given $f(x) = x e^{x(1-x)}$.Applying the product rule and chain rule,we find the derivative:$f'(x) = 1 \cdot e^{x(1-x)} + x \cdot e^{x(1-x)} \cdot (1-2x)$$f'(x) = e^{x(1-x)} [1 + x - 2x^2]$$f'(x) = -e^{x(1-x)} [2x^2 - x - 1]$Factoring the quadratic expression:$f'(x) = -e^{x(1-x)} (2x + 1)(x - 1)$$f'(x) = -2 e^{x(1-x)} (x + 1/2)(x - 1)$Since $e^{x(1-x)} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on the expression $-(x + 1/2)(x - 1)$.For $x \in [-1/2, 1]$,the product $(x + 1/2)(x - 1) \leq 0$.Therefore,$-(x + 1/2)(x - 1) \geq 0$.Thus,$f'(x) \geq 0$ for $x \in [-1/2, 1]$.Hence,$f(x)$ is increasing on $[-1/2, 1]$.

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

Similar Questions

In the interval $\left(\frac{1}{e}, e\right)$,a decreasing function among the following functions is

Let $\lambda^{}$ be the largest value of $\lambda$ for which the function $f_{\lambda}(x) = 4\lambda x^{3} - 36\lambda x^{2} + 36x + 48$ is increasing for all $x \in \mathbb{R}$. Then $f_{\lambda^{}}(1) + f_{\lambda^{*}}(-1)$ is equal to:

If the function $f(x)=x^3+2 p x^2+27 x+16$ is strictly increasing for all $x \in R$,then the range of $p$ is

Show that the function given by $f(x) = 7x - 3$ is strictly increasing on $R$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module