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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