f(x) = (-x^2)(e^-x) + 2xe^-x

finding the zeros of this darn function, help please :)

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- Jul 28th 2010, 03:45 PMdanielh9103Finding the zeros of a function (algebriacally)
f(x) = (-

*x*^2)(*e*^-x) + 2x*e^-x*

finding the zeros of this darn function, help please :)

- Jul 28th 2010, 03:51 PMArchie Meade
That's the purpose of factoring, writing an expression as a product.

Fact: Zero multiplied by anything is zero because any amount of zeros is zero.

So, can we factor that ?

Yes, there is a common factor $\displaystyle x$ and another $\displaystyle e^{-x}$

$\displaystyle -x^2e^{-x}+2xe^{-x}=xe^{-x}(2-x)=(x)\left(e^{-x}\right)(2-x)$

Now you can see which values of x causes that to be zero.