求函数y=sinx-cosx+sinxcosx,x属于[0,π]的最大值和最小值

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设sinx-cosx=t,则慧槐sinxcosx=(1-t*t)/2
0<=x<=π，则-1<=t<=根氏碧含号2
y=-t*t/2+t+1/歼笑2=-1/2(t-1)(t-1)+1

最小值为-1，t=-1,即x=0
最大值为1，t=1,即x=π/2或π

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求导后得仿败陆cosx+sinx+cos^2x-sin^2x令其为枯判0得cosx+sinx+cos^x-sin^2x=0所以cosx+sinx=sin^2x-cos^2x=(sinx+cosx)(sinx-cosx)所以1=sinx-cosx因为x在0和Pi之间所以解得x=Pi/2和Pi 所以最大值为备顷1

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设sinx-cosx=t,则sinxcosx=(1-t*t)/2
0<氏碧含=x<=π，则-1<=t<=根号2
y=-t*t/2+t+1/2=-1/2(t-1)(t-1)+1
最小值为-1，t=-1,即歼笑x=0
最大慧槐值为1，t=1,即x=π/2或π

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