Question

What is the equation of the tangent line of `f(x) =x-4/e^x+x^2/e^x-1` at `x=4`?

Answer 1

`y = -8e^-4x + 3 + 44e^-4`

Explanation:

Given:

`y = f(x) = x - 4e^-x + x^2e^-x - 1`

Evaluate a `x = 4` so that we know a point on the tangent line

`y = f(4) = 4 - 4e^-4 + (4)^2e^-4 - 1`

`y = 3 + 12e^-4`

The point for the tangent line is `(4, 3 + 12e^-4)`

Compute the first derivative:

`f'(x) = 4e^-x + 2xe^-x - x^2e^-x`

Evaluate at `x = 4`:

`f'(4) = 4e^-4 + 2(4)e^-x - (4)^2e^-4 = -8e^-4`

The slope `m = -8e^-4`

Using the point-slope form of the equation of a line:

`y - (3 + 12e^-4) = -8e^-4(x - 4)`

`y - (3 + 12e^-4) = -8e^-4x + 32e^-4`

`y = -8e^-4x + 3 + 44e^-4`