What is the equation of the line tangent to #f(x)=(x-1)^3# at #x=3#?

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Answer 1 · 2015-12-18T05:53:44

#y = 12x - 28#

#f(3) = 8#

#f'(x) = 3(x - 1)^2#

#f'(3)=12#

For a line with equation #y = mx + c#,

#m = 12#

#c = y - mx#

#= 8 - 12(3)#

#= -28#

Equation of line: #y = 12x - 28#