Question

求f(x)＝x / sinx 在区间[0，π/2]的凸性

Answer 1

f'(x)=(sinx-xcosx)/sin²x
f''(x)=[(cos-cosx+xsinx)sin²x-2sinxcosx(sinx-xcosx)]/sin⁴x
=(xsin²x-2sinxcosx+2xcos²x)/sin³x
=(x+xcos²x-sin2x)/sin³x
令g(x)=x+xcos²x-sin2x
g'(x)=1+cos²x-xsin2x-2cos2x
=1.5-xsin2x-1.5cos2x
驻点x=2kπ,x=2kπ+π
x∈(0,π),g'(x)>0,g(x)单调递增
∴g(x)≥g(0)=0
∴f''(x)≥0
f(x)在区间为凹函数。