Question

f(x)=sin^4x+2sinxcosx+cos^4x的最小值

Answer 1

f(x)=sin^4x+2sinxcosx+cos^4x
=sin^4x+cos^4x+2sinxcosx
=(sin^2x+cos^2x)^2-2sin^2xcos^2x+2sinxcosx
=1- 2sin^2xcos^2x+2sinxcosx
=1-(1/2)(2sinxcosx)^2+2sinxcosx
=-(1/2)(sin2x)^2+sin2x+1
=-(1/2)[sin2x-1]^2+3/2
因为 sin2x<=1
所以最大值为 sin2x=1
时 f(x)=3/2
最小值为 sin2x=-1时
f(x)=-1/2

Answer 2

f(x)=sin^4x+2sinxcosx+cos^4x
=(sin^2x+cos^2x)^2+sin2x-1/2*sin^2(2x)
=1+sin2x-1/2*sin^2(2x)
=3/2-1/2(1-sin2x)^2
-1≤sin2x≤1，当sim2x=-1时
f(x)有最小值-1/2