Question

What is the equation of the tangent line of `f(x) =((x+3)(x-1))/e^x` at `x=-3`?

Answer 1

slope of the tangent is derivative of F(x).
then the equation of tangent can be found using y = m*x +c , where m is slope of tangent

Explanation:

First to find the slope of the tangent, find derivative of f(x):

f '(x) = `d/dx` (x+3)(x-1) * `e^-x`
= `d/dx` [ `(x^2 + 2x -3) * e^-x`]
we will differentiate using u.v form and chain rule.
u = `x^2 + 2x - 3`
`du` = `2x + 2`
v = `e^-x`
`dv = - e^-x`

f '(x) = `x^2 + 2x -3*(-e^-x) + e^-x*(2x+2)`

Now we want the particular slope at x = -3. So lets substitute x as -3 in f '(x)
f '(-3) = ( 9 -9)* `-e^3` + `e^3` * -4
= 0 + -4`e^3` = -4`e^3`
so equation of tangent at x = -3 is
y = mx + c
=> y = (-4`e^3`) x + c