algebra help 4.docx

g(x)= -3x^2-24x-46

taking derivative, and for critical point

g'(x)=0

-6x-24=0

x=-4

taking second derivative

g''(x)=-6<0

so at x=-4 exists a maximum value.

the maximum value is

-3(-4)^2-24(-4)-46=-3(16)+96-46

=-48+96-46

=96-94

=2 = maximum value

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