Question

Доведіть що функція f(x)=sin^2(x)-2x^2 є опуклою вгору на R

Answer 1

Ответ:

Объяснение:

f'(x)=2sinxcosx-4x=sin2x-4x

f''(x)=2cos2x-4

так как Icos2xI≤1 то  f''(x)=2cos2x-4x<0  

f''(x)<0  для любого х ⇒
график функции выпуклый вверх

График на картинке

Answer 2

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