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

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Mathematics

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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Steve · Answer 1

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