Consider the curve #y = (x^2- 2x+k)(x-6)^2 #, where #k# is a real constant. The curve has a maximum point at # x =3#. What is the value of #k#?

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Answer 1 · 2016-07-17T18:06:38

#k=3#.

#y=(x^2-2x+k)(x-6)^2#

For #y_(max),# we must have

#(1)[dy/dx]_(x=3)=0#, and,

#(2) [(d^2y)/(dx)^2]_(x=3)<0#

Now, #dy/dx=(2x-2)(x-6)^2+(x^2-2x+k)*2(x-6)#

#=2(x-6){(x-1)(x-6)+x^2-2x+k}#

#=2(x-6)(2x^2-9x+k+6)#

#:. [dy/dx]_(x=3)=0#

#rArr 2(-3)(18-27+k+6)=0#

#rArr k=3#

Next, #(d^2y)/(dx)^2=d/dx{dy/dx}#

#=d/dx{2(x-6)(2x^2-9x+k+6)}#

#=2{1((2x^2-9x+k+6)+(x-6)(2x-9)}#

#=2(4x^2-30x+k+60)#

#:. [(d^2y)/(dx)^2]_(x=3)=2(36-90+3+60)=18>0#

Hence, #(2)# is verified.

#:. k=3#.