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

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(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}

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

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(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 $\left[-\frac{1}{2}, 1\right]$.

$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

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

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