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BetB
@BetB
July 2022
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Найти наибольшее и наименьшее значение функции на заданном отрезке
f(x)=2x+(1/(2x^2)) xє[1/2;2]
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Вотача200407
f`(x)=2cos2x-1=0
cos2x=1/2
2x=+ -π/3+2πn
x=+- π/6+πn
x=π/6∈[0;π]
f(0)=xin0-0=0
f(π/6)=sinπ/3-π/6=√3/2-π/6 -наиб
f(π)=sin2π-π=0-π=-π-наим
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Answers & Comments
cos2x=1/2
2x=+ -π/3+2πn
x=+- π/6+πn
x=π/6∈[0;π]
f(0)=xin0-0=0
f(π/6)=sinπ/3-π/6=√3/2-π/6 -наиб
f(π)=sin2π-π=0-π=-π-наим