Answer :

Condition (1):


Since, y=x(x-4) is a polynomial and we know every polynomial function is continuous for all xϵR.


y= x(x-4) is continuous on [0,4].


Condition (2):


Here, y’= (x-4)+x which exist in [0,4].


So, y= x(x-4) is differentiable on (0,4).


Condition (3):


Here, y(0)=0(0-4)=0


And y(4)= 4(4-4)=0


i.e. y(0)=y(4)


Conditions of Rolle’s theorem are satisfied.


Hence, there exist at least one cϵ(0,4) such that y’(c)=0


i.e. (c-4)+c=0


i.e. 2c-4=0


i.e. c=2


Value of c=2ϵ(0,4)


So,y(c)=y(2)=2(2-4)=-4


By geometric interpretation, (2,-4) is a point on a curve y=x(x-4),where tangent is parallel to x-axis.


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