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

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Answer 1 · 2016-03-25T01:47:14

#y-36 = 42(x-3)# or #y=42x-90#

We know the #x# value of the original function, #f(x)# find the #y# value by plugging in 3:

#f(3) = 2(3)^3-2(3)^2 = 36# so our point is: #(3,36)#

Now find the derivative of #f(x)# using the power rule:

#f'(x) = 6x^2 - 4x#

The derivative gives us the slope at any given point, by plugging in 3 into #f'(x)# we get the slope of #f(x)# at #x=3#

#f'(3) = 6(3)^2 - 4(3) = 42# so #m=42, (3, 36)#

We now have enough information to create a line in point slope form: #y-y_1 = m(x-x_1)#

#y-36 = 42(x-3)#
If you want, you can solve for y:
#y=42x-90#