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

1 Answer
anantha n.
Dec 8, 2017

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/dxddx (x+3)(x-1) * e^-xe−x
= d/dxddx [ (x^2 + 2x -3) * e^-x(x2+2x−3)⋅e−x]
we will differentiate using u.v form and chain rule.
u = x^2 + 2x - 3x2+2x−3
dudu = 2x + 22x+2
v = e^-xe−x
dv = - e^-xdv=−e−x

f '(x) = x^2 + 2x -3*(-e^-x) + e^-x*(2x+2)x2+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−e3 + e^3e3 * -4
= 0 + -4e^3e3 = -4e^3e3
so equation of tangent at x = -3 is
y = mx + c
=> y = (-4e^3e3) x + c

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