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

Answer 1

Emily Ammerman

#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#

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Answer 2

Adam Adkins

The equation of the line tangent to f(x)=2x^3-2x^2 at x=3 is y=34x-54.

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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.