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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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Tell us the point you get stuck at exactly. You differentiate $f(x) = x^2 e^{-x}$, what do you find to be the derivative? Then, the function is increasing when $f'(x)>0$ and decreasing if $f'(x)<0$, can you solve these inequalities? –  Ragib Zaman Jul 21 '12 at 13:23
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Now, can our "helpful" users refrain from answering until Arpit returns to show what he can do? –  GEdgar Jul 21 '12 at 13:32
    
my mistake one part of derivative is always positive –  Arpit Bajpai Jul 21 '12 at 15:24
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3 Answers

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:

enter image description here

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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.

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