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

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

If $f(x) = x + \log \left( \frac{x-1}{x+1} \right)$ is a well-defined real-valued function, then $f$ is

Find the intervals in which the function $f$ given by $f(x) = x^{2} - 4x + 6$ is:
$(a)$ increasing
$(b)$ decreasing

The maximum interval in which the slopes of the tangents drawn to the curve $y=x^4+5x^3+9x^2+6x+2$ increase is

If $f(x) = \int_{x^2}^{x^2 + 1} e^{-t^2} dt$,then $f(x)$ increases in

Vedclass Test Series

Exam Paper Generator

Online Exam Module