Question 113886
<pre><b>
Is there any one that knows how to work this problem?
Find the zeros of f.

To find the zeros substitute 0 for f(x) and solve for x.

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

{{{0 = -x^(2)e^(-x)+ 2xe^(-x)}}}

Get 0 on the right:

{{{x^(2)e^(-x)- 2xe^(-x)=0}}}

Factor out {{{(xe^(-x))}}}

{{{(xe^(-x))(x-2)=0}}}

Now we use the zero product principle.
There are three factors: {{{x}}},{{{e^(-x)}}}, and {{{x-2}}}

Setting x = 0 gives one solution, namely x = 0

Setting {{{e^(-x)}}} = 0 gives no solution, since {{{e^(-x)}}} is never 0.

Setting x-2 = 0 gives a second solution, namely x = 2

So there are two zeros,  0 and 2.   

Edwin</pre>