Question

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

Answer 1

`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`