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Question: \(f(x) = xe^{x(1 - x)}\) then \(f(x)\) is...

f(x)=xex(1−x)f(x) = xe^{x(1 - x)} then f(x)f(x) is

A

Increasing on [−12,1]\left\lbrack \frac{- 1}{2},1 \right\rbrack

B

Decreasing on R

C

Increasing on R

D

Decreasing on [−12,1]\left\lbrack \frac{- 1}{2},1 \right\rbrack

Answer

Increasing on [−12,1]\left\lbrack \frac{- 1}{2},1 \right\rbrack

Explanation

Solution

f′(x)=ex(1−x)+x.ex(1−x).(1−2x)f^{'}(x) = e^{x(1 - x)} + x.e^{x(1 - x)}.(1 - 2x)

= ex(1−x){1+x(1−2x)}=ex(1−x).(−2x2+x+1)e^{x(1 - x)}\{ 1 + x(1 - 2x)\} = e^{x(1 - x)}.( - 2x^{2} + x + 1) Now by the sign-scheme for −2x2+x+1- 2x^{2} + x + 1

f′(x)≥0,f^{'}(x) \geq 0, if x∈[−12,1]x \in \left\lbrack - \frac{1}{2},1 \right\rbrack, because ex(1−x)e^{x(1 - x)} is always positive. So, f(x)f(x) is increasing on [−12,1]\left\lbrack - \frac{1}{2},1 \right\rbrack.