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I am unable to find values of $x$ for which following function

$x^2 \cdot e^{-x}$

When this function is increasing or decreasing?

My normal approach is "differentiate" but I am stuck at some point...

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    my mistake one part of derivative is always positive2012-07-21

3 Answers 3

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Let $f(x)=x^2e^{-x}$. Differentiate. Using the Product Rule and Chain Rule, we find that $f'(x)=x^2(-e^{-x})+(2x)e^{-x}$, which simplifies to $e^{-x}(2x-x^2)$. But $e^{-x}$ is always positive, and therefore $\dots$.

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Not meant to be an answer, but an addendum to André's answer:

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Set the derivative $ (2x-x^2)\,e^{-x} > 0 $ to find the region where your function increases. This implies $ x(2-x) > 0 \Rightarrow x>0$ and $(x<2) \Rightarrow x\in(0,2). Try to find the region where your function decreases by setting the derivative of the function less than 0.