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

Answer 1

#y = x+8#

Since we have two functions that have been multiplied together, we must differentiate in this case using the product rule in order to find the gradient:

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

Consequently, (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)#

We now enter -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#

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Answer 2

Ezekiel Alexander

The equation of the line tangent to f(x)=(x^2-x)e^(x+2) at x=-2 is y = -4e^0(x+2) + 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.