解：f(x)=-(sin2x+cos2x)+3sin2x-cos2x
=2sin2x-2cos2x
=2√2sin(2x-π/4)
(1)T=2π/2=π
(2) x∈[0,π/2]时 2x-π/4∈[-π/4,3π/4]
当2x-π/=-π/4 即 x=0时 f(x)有最小值-2
当2x-π/=π/2 即 x=3π/8时 f(x)有最大值2√2
(3) x∈[0,π/2]时 2x-π/4∈[-π/4,3π/4]
当 2x-π/4∈[-π/4,π/2] 即x∈[0,3π/8]时 f(x)在其上单调递增
当 2x-π/4∈[π/2,3π/4] 即x∈[3π/8,π/2]时 f(x)在其上单调递减
所以x∈[0,π/2]时
f(x) 在[0,3π/8]上单调递增,在[3π/8,π/2]上单调递减.
希望对你 有点帮助！
=2sin2x-2cos2x
=2√2sin(2x-π/4)
(1)T=2π/2=π
(2) x∈[0,π/2]时 2x-π/4∈[-π/4,3π/4]
当2x-π/=-π/4 即 x=0时 f(x)有最小值-2
当2x-π/=π/2 即 x=3π/8时 f(x)有最大值2√2
(3) x∈[0,π/2]时 2x-π/4∈[-π/4,3π/4]
当 2x-π/4∈[-π/4,π/2] 即x∈[0,3π/8]时 f(x)在其上单调递增
当 2x-π/4∈[π/2,3π/4] 即x∈[3π/8,π/2]时 f(x)在其上单调递减
所以x∈[0,π/2]时
f(x) 在[0,3π/8]上单调递增,在[3π/8,π/2]上单调递减.
希望对你 有点帮助！
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