Question

已知函数f(x)=sin^2x+2sinx*cosx+3cos^2,x属于R，求函数f(x)的最值

Answer 1

f(x)=sin^2x+2sinx*cosx+3cos^2
=1+sin2x+2cos^2
=1+sin2x+cos2x+1
=2+√2sin(2x+π/4)
当2x+π/4=2kπ+π/2,即x=kπ+π/8时，(k∈Z) f(x)max=2+√2
当2x+π/4=2kπ-π/2,即x=kπ-3π/8时，(k∈Z) f(x)min=2-√2
友情提醒，求最值，必须写条件，否则会莫名其妙地失分！！！
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Answer 2

f(x)=sin^2x+2sinx*cosx+3cos^2x
=(sinx)^2+2sinx*cosx+(cosx)^2+2*(cosx)^2
=1+sin2x+cos2x+1
=2+sin2x+cos2x
=2+√2sin(2x+1/4*π)
因为x属于R
所以√2sin(2x+1/4*π)取值[-√2,√2]
所以f(x)取值[-√2+2,√2+2]

Answer 3

f(x)=sin^2x+2sinx*cosx+3cos^2
=sin^2x+2sinx*cosx+cos^2+2cos^2
=1+sin2x+2cos^2
=1+sin2x+cos2x+1
=2+根2sin(2x+π/4)
f(x)max=2+根2
f(x)min=2-根2