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

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

Question

(D) Given,$f(x)=x e^{x(1-x)}$.To find the intervals of increase and decrease,we calculate the derivative $f^{\prime}(x)$:$f^{\prime}(x) = \frac{d}{dx}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given,$f(x)=x e^{x(1-x)}$.To find the intervals of increase and decrease,we calculate the derivative $f^{\prime}(x)$:$f^{\prime}(x) = \frac{d}{dx} [x e^{x-x^2}] = e^{x-x^2} + x e^{x-x^2} (1-2x)$$f^{\prime}(x) = e^{x-x^2} [1 + x(1-2x)] = e^{x-x^2} (1 + x - 2x^2)$$f^{\prime}(x) = -e^{x-x^2} (2x^2 - x - 1) = -e^{x-x^2} (2x^2 - 2x + x - 1)$$f^{\prime}(x) = -e^{x-x^2} [2x(x-1) + 1(x-1)] = -e^{x-x^2} (x-1)(2x+1)$.Since $e^{x-x^2} > 0$ for all $x \in R$,the sign of $f^{\prime}(x)$ depends on $-(x-1)(2x+1)$.Setting $f^{\prime}(x) = 0$,we get critical points $x = 1$ and $x = -\frac{1}{2}$.Analyzing the sign of $f^{\prime}(x)$ on the number line:For $x \in \left(-\infty, -\frac{1}{2}\right)$,$f^{\prime}(x) For $x \in \left[-\frac{1}{2}, 1\right]$,$f^{\prime}(x) \geq 0$ (increasing).For $x \in (1, \infty)$,$f^{\prime}(x) Thus,$f(x)$ is increasing in the interval $\left[-\frac{1}{2}, 1\right]$.

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

Similar Questions

Identify $acetaldoxime$.

If $f(x) = \begin{cases} \frac{\sin(1+[x])}{[x]}, & \text{for } [x] \neq 0 \\ 0, & \text{for } [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer not exceeding $x$,then $\lim_{x \rightarrow 0^{-}} f(x)$ is equal to

When a system is taken from state $i$ to state $f$ along the path $iaf$,$Q = 50 \, cal$ and $W = 20 \, cal$. Along the path $ibf$,if $Q = 36 \, cal$,find the work $W$ for the path $ibf$ in $cal$.

In a $\triangle ABC$,if $3a = b + c$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

Which domain includes well-known pathogenic organisms?

Vedclass Test Series

Exam Paper Generator

Online Exam Module