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

1 Answer
Mar 15, 2016

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

Explanation:

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