What is the equation of the tangent line of #f(x)=x^2e^(x-2)-xe^x# at #x=4#?

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Answer 1 · 2016-03-15T07:01:35

#y=(24e^2 - 5e^4)x+16e^4 - 80e^2#

#f(x)=x^2e^(x-2)-xe^x#

#f'(x)=2xe^(x-2)+x^2e^(x-2)-e^x-xe^x#

#f(4) = 16e^2 - 4e^4#

#f'(4) = 8e^2 + 16e^2 - e^4 - 4e^4 = 24e^2 - 5e^4#

tangent line eq: #y=f'(x_0)x+n#

#16e^2 - 4e^4 = 4(24e^2 - 5e^4) + n#

#n = 16e^2 - 4e^4 - 96e^2 + 20e^4 = 16e^4 - 80e^2#

#y=(24e^2 - 5e^4)x+16e^4 - 80e^2#