Question #9a98d

1 Answer
NickTheTurtle
Nov 2, 2017

#f'(x)=(x-3)/|x-3|#

Explanation:

#f(x)=|x-3|#

Since #|x|=sqrt(x^2)#, we can rewrite as #f(x)=sqrt((x-3)^2)#.

Use the chain rule #dy/dx=dy/(du)*(du)/(dv)*(dv)/dx# where #y=sqrt((x-3)^2),u=(x-3)^2,v=x-3#.

This gives #dy/(du)=1/(2sqrt((x-3)^2)),(du)/(dv)=2(x-3),(dv)/(dx)=1#.

Thus, #f'(x)=1/(2sqrt((x-3)^2))*2(x-3)*1=(x-3)/sqrt((x-3)^2)=(x-3)/|x-3|#.

So, when #x-3>0# (i.e. #x>3#), #f'(x)=1#. When #x-3<0# (i.e. #x<3#), #f'(x)=-1#. Else, when #x=3#, #f'(x)# is undefined.

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