Re: [微積] 三題微積分
F(x)=sin(x^2) g(x)=sin^2(x)
F'(x)=cos(x^2)*2x->F'(0)=1*0=0
=>F'(0)=G'(0)
G'(x)=2*sin(x)cos(x)->G'(0)=2*0*1=0
F''(x)=-sin(x^2)*2x+2*cos(x^2)->F''(pi/6)=-sin(pi/36)*pi/3+2*cos(pi/36)
G''(x)=2[cos^2(x)-sin^2(x)]->G''(pi/6)=2(3/4-1/4)=1
明顯F''(x)>G''(x)
F''(x) F'(t)-F'(0)
------=----------- > 1 => F'(t)>G'(t)
G''(x) G'(t)-G'(0)
F'(x) F(t)-F(0)
-----=----------->1 因F(0)=G(0) 所以 F(x)>G(x) => sin(x^2)>sin^2(x)
G'(x) G(t)-G(0)
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