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

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Answer 1 · 2016-01-10T11:07:49

The Normal Line is

#y=x/12-49/6# or #x-12y=98#

the given: #f(x) =(x-2)^(3/2)-x^3# and #x=2#

#f(2) =(2-2)^(3/2)-(2)^3#

#f(2)= -8#

the point on the curve : #(2, -8)#

compute slope #m# then use #-1/m# for the normal line

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

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

#f' (2) = -12#

The slope to be used to find the normal line is #-1/m#

#-1/m=-1/-12=1/12#

Determine now the Normal Line using

#y-y_1=(1/12)(x-x_1)#

#y-(-8)=(1/12)(x-2)#

after simplification the final answer:

#x-12y=98# or #y=x/12-49/6#