What is the equation of the line tangent to # f(x)=(x^2-x)e^(x+2) # at # x=-2 #?

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Answer 1 · 2015-12-29T11:00:34

#y = x+8#

To find the gradient, we need to differentiate, here, using the product rule, as we have two functions multiplied together:

(uv)' = u'v + v'u
u = #x^2- x#
u' = #2x - 1#
v = #e^(x+2)#
v' = #e^(x+2)#

Hence (uv)' = u'v + v'u

= #(2x-1)e^(x+2)# + #(e^(x+2))(x^2-x)#
=#(e^(x+2))(2x-1+x^2-x)#

#f'(x)#=#(e^(x+2))(x^2+x-1)#

Now, we fill in -2 for x:

#f'(-2)# = #(e^((-2)+2))((-2)^2+(-2)-1)#
= #e^0(4-2-1)#

As #e^0 = 1#:
#f'(-2) = 1*1 = 1#

Now you know the gradient, you need to sub into #y-b = m(x-a)#; to get point b, you need to sub #x =-2# into the original equation.

#f(x) = ((-2)^2 - (-2))*(e^((-2)+2))#
= #((4+2)*1)#

=#6#

Now we have our gradient #m = 1# and the point #(-2,6)#, so the equation of the tangent is:

#y-b = m(x-a)#

#y-6 = 1(x-(-2))#

#y = x+2+6#

#y = x+8#