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If x is real then find the maximum value of #(3x^2+9x+17)/(3x^2+9x+7)#.

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Answer 1 · 2018-06-29T03:15:52

Let

#y=(3x^2+9x+17)/(3x^2+9x+7)#

#=>y(3x^2+9x+7)=(3x^2+9x+17)#

#=>3(y-1)x^2+9(y-1)x+(7y-17)=0#

As #x# is real we can write

#(9(y-1))^2 -4*3(y-1)(7y-17)>=0#

#=>81(y-1)^2 -4*3(y-1)(7y-17)>=0#
#=>27(y-1)^2 -4(y-1)(7y-17)>=0#

#=>27y^2-54y+27 -4(7y^2-24y+17)>=0#

#=>27y^2-54y+27 -28y^2+96y-68>=0#

#=>-y^2+42y-41>=0#

#=>y^2-42y+41<=0#

#=>y^2-41y-y+41<=0#

#=>y(y-41)-1(y-41)<=0#

#=>(y-41)(y-1)<=0#

This inequality holds for

#1 <=y<=41#

Hence maximum value of the given expression for real values of x will be #41#

Answer 2 · 2018-06-29T05:18:19

41

#(3x^2+9x+17)/(3x^2+9x+7)=1+10/(3x^2+9x+7)#

Hence, to maximize #(3x^2+9x+17)/(3x^2+9x+7)# we need to minimize #3x^2+9x+7#. Now

#3x^2+9x+7 = 3(x^2+3x)+7#
#qquad= 3(x^2+2*x*3/2+(3/2)^2)+7-3(3/2)^2#
#qquad = 2(x+3/2)^2+1/4#

Since #(x+3/2)^2 >= 0# we have

#3x^2+9x+7>= 1/4 implies#
#10/(3x^2+9x+7) <= 40 implies#

#(3x^2+9x+17)/(3x^2+9x+7)=1+10/(3x^2+9x+7)<= 41#

Thus, the maximum value is 41 (attained for #x=-3/2#).