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What is the equation of the line tangent to #f(x)=x^3-6x # at #x=2#?

Answer 1

The equation of the tangent at #f(2)# is #y=6x-16#

#f(x) = x^3-6x# #f'(x) = 3x^2 - 6#

The equation of any tangent to #f(x)# in slope and intercept form is: #y = mx+c# Where: #m# = slope of #f(x)# and #c# = the intercept on the y axis

Slope of the tangent to #f(x)# at #x=2# is: #f'(2)= 3*2^2-6# #=6 -> m=6#

Since the tangent touches #f(x)# at #f(2)# #f(2) = m*2 +c#

Replacing for #f(2)# and #m=6#: #2^3-6*2 = 6*2 +c#

#c = 8-12-12# #c=-16#

Hence the equation of the tangent at #x=2# is:

#y = 6x-16#

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Answer 2

Emma Aldridge

The equation of the line tangent to f(x)=x^3-6x at x=2 is y = -8x + 16.

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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.