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

1 answer:

8 0

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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Step by step explanation-

Given-

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Now,

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