For #f(x) =(x-1)^3#, what is the equation of the line tangent to #x =2 #?

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Answer 1 · 2015-11-20T19:57:45

#y =3*x -5#

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

Derive the function:
#f'(x)=3*(x-1)^2#

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

The derivative in x =2 is the slope of the tangent.

Calculate the value of the function in x=2
#f(2)=(2-1)^3=1^3=1#

The tangent intersects the point (2,1)

the equation of the tangent
#y = k*x +m#
#k=3#

#y = 3*x + m#
#1 = 3*2 + m#
#m =1-6=-5#

#y =3*x -5#