SOLUTION: Is there any one that knows how to work this problem? Find the zeros of f. f(x)= -x^(2)e^(-x)+ 2xe^(-x)

Algebra ->  Functions -> SOLUTION: Is there any one that knows how to work this problem? Find the zeros of f. f(x)= -x^(2)e^(-x)+ 2xe^(-x)      Log On


   



Question 113886: Is there any one that knows how to work this problem?
Find the zeros of f.
f(x)= -x^(2)e^(-x)+ 2xe^(-x)

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

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%28x%29=+-x%5E%282%29e%5E%28-x%29%2B+2xe%5E%28-x%29

0+=+-x%5E%282%29e%5E%28-x%29%2B+2xe%5E%28-x%29

Get 0 on the right:

x%5E%282%29e%5E%28-x%29-+2xe%5E%28-x%29=0

Factor out %28xe%5E%28-x%29%29

%28xe%5E%28-x%29%29%28x-2%29=0

Now we use the zero product principle.
There are three factors: x,e%5E%28-x%29, and x-2

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

Setting e%5E%28-x%29 = 0 gives no solution, since e%5E%28-x%29 is never 0.

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

So there are two zeros,  0 and 2.   

Edwin