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f(x)=xsinx-3/2 f'(x)=sinx+xcosx 令f'(x)=0得tanx=-x,解為x0,x0∈(π/2,π) ∴(0,x0),f(x)遞增,(x0,π),f(x)遞減 f(x)max=f(x0)>f(π/2)>0 f(0)=-3/2
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