`-(x^2)e^(-x) + 2xe^(-x) = 0` Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility.

Textbook Question

Chapter 3, 3.4 - Problem 74 - Precalculus (3rd Edition, Ron Larson).
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kspcr111 | In Training Educator

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Given

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

=>  e^(-x) [2x - x^2]=0

=> (2x - x^2)/(e^(x) )=0

as (e^(x) ) cannot be zero so

2x- x^2 = 0

=> x(2-x) =0

=> x= 0 or x= 2

the graphs of the equations

-(x^2)e^(-x) + 2xe^(-x) = 0 and

y= -(x^2)e^(-x) + 2xe^(-x)  is as follows

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