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y=x^3-32x^2+3
interval -1<=x<=2
nilai maksimum
y=3x^2-64x
f(-1)=3(-1)^2-64(-1)
=3(-1)^2+64
=3(1)+64
=67
f(0)=3(0)^2-64(0)
=3(0)^2-0
=0
f(1)=3(1)^2-64(1)
=3(1)-64
=3-64
=-61
f(2)=3(2)^2-64(2)
=3(4)-128
=12-128
=-116
Nilai maksimum pada x=-1, menghasilkan y=67.
Pilihan D.
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