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Katyakotovik
@Katyakotovik
July 2022
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Найти точки перегиба функции f(x) = cosx, –π < x < π
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mikael2
F(x) = cosx f'(x) =-sinx f''(x) = -cosx
-cosx =0 cosx=0 x=π/2+πk k∈Z
–π < x < π
→x1=π/2 x2=-π/2
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Answers & Comments
-cosx =0 cosx=0 x=π/2+πk k∈Z
–π < x < π →x1=π/2 x2=-π/2