What is the equation of the line tangent to # f(x)=(e^x-x)^2 # at # x=3#?

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Answer 1 · 2016-10-23T14:01:18

#y=652.1732x-1664.6040#

First, we must differentiate the function,

#y=(e^x-x)^2#

#y'=2(e^x-x)(e^x-1)# (by the chain rule)

at the point 3

#y'= 2(e^3-3)(e^3-1) #

so our tangent line is equal to

#y=("gradient")x+c#

#y=(2(e^3-3)(e^3-1))x+c#

Now we have to solve for c using the point where x=3,

#(e^3-3)^2=(2(e^3-3)(e^3-1))(3)+c#

#c=-5e^6+18e^3-9#

so our tangent line is,

#y=(2(e^3-3)(e^3-1))x+ (-5e^6+18e^3-9)#

which can be approximated to (since e is irrational),

#y=652.1732x-1664.6040#