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5 November, 15:30

What is the instantaneous rate of change of f (x) = xe^x - (x 2) e^ (x-1) at x=0?

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  1. 5 November, 15:50
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    You have to derive for a multiplication in both terms:

    =e^x+xe^x - (e^x-1 + (x-2) e^x-1) now apply distributive property in the last term:

    =e^x+xe^x+e^x-1-xe^x-1 now replace each x by 0 (x=0)

    =1 + 0 + e^-1 + 0 = 1 + e^-1 = 1.3679
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