Question

高数 选择题求解（极值、拐点）？

如图

Answer 1

f'(x)>0
F(x)
=∫(0->x) (2t-x) f(t) dt
=2∫(0->x) t.f(t) dt - x.∫(0->x) f(t) dt
F'(x)
=2xf(x) - xf(x) - ∫(0->x) f(t) dt
=xf(x) - ∫(0->x) f(t) dt
F'(0) =0
F''(x)
=xf'(x)+f(x) -f(x)
=xf'(x)
F''(0) =0
F'''(x)
=xf''(x) + f'(x)
F'''(0) =f'(0) >0
ans : C

Answer 2

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Answer 3

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Answer 4

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Answer 5

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