当x∈[-6,-2/3]时,判断函数y=fx+f(x+2)的单调性
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当x∈[-6,-2/3]时,判断函数y=fx+f(x+2)的单调性
fx=2sin(π/4x+π/4) 我想看将fx带入后化简的步骤~
fx=2sin(π/4x+π/4) 我想看将fx带入后化简的步骤~
f(x)=2sin(π/4x+π/4)
y=f(x)+f(x+2)
=2sin(π/4x+π/4)+2sin[π/4(x+2)+π/4]
=2sin(π/4x+π/4)+2sin[π/4x+π/2+π/4]
=2sin(π/4x+π/4)+2cos(π/4x+π/4)
=2√2[√√2/2*sin(π/4x+π/4)+√2/2*cos(π/4x+π/4)]
=2√2sin[(π/4*x+π/4)+π/4]
=2√2sin(π/4*x+π/2)
=2√2cos(π/4*x)
∵x∈[-6,-2/3]
∴-3π/2≤π/4x≤-π/6
∴当-3π/2≤π/4x≤-π,即x∈[-6,-4]时,函数递减,
当-π≤π/4x≤-π/6,即x∈[-4,-2/3]时,函数递增
y=f(x)+f(x+2)
=2sin(π/4x+π/4)+2sin[π/4(x+2)+π/4]
=2sin(π/4x+π/4)+2sin[π/4x+π/2+π/4]
=2sin(π/4x+π/4)+2cos(π/4x+π/4)
=2√2[√√2/2*sin(π/4x+π/4)+√2/2*cos(π/4x+π/4)]
=2√2sin[(π/4*x+π/4)+π/4]
=2√2sin(π/4*x+π/2)
=2√2cos(π/4*x)
∵x∈[-6,-2/3]
∴-3π/2≤π/4x≤-π/6
∴当-3π/2≤π/4x≤-π,即x∈[-6,-4]时,函数递减,
当-π≤π/4x≤-π/6,即x∈[-4,-2/3]时,函数递增
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