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July 2022
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Найти наибольшее значение функции f(x) = x^2-4x+4 на отрезке [0;3]
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армения20171
F(x)=x²-4x+4 ;[0;3]
D(f)=(-oo;+oo)
f'(x)=2x-4
f'(x)=0
2x-4=0
x=2
f(2)=4-8+4=0
f(0)=4
f(3)=9-12+4=1
f(max)=4
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Answers & Comments
D(f)=(-oo;+oo)
f'(x)=2x-4
f'(x)=0
2x-4=0
x=2
f(2)=4-8+4=0
f(0)=4
f(3)=9-12+4=1
f(max)=4