Question

已知函数fx＝2sinx（cosx-sinx）

当0＜x＜派，求fx的最大值及相应的x值

Answer 1

f(x)＝2sinx（cosx-sinx）=2sinxcosx-2(sinx)^2=sin2x+(cos2x-1)=√2sin(2x+π/4)-1
∵0＜x＜π
∴0＜2x＜2π
∴π/4<2x+π/4<9π/4
因为y=sint在(π/4,3π/2)上是减函数，在(3π/2,9π/4)上是增函数
当2x+π/4=π/2即x=π/8时，f(x)取最大值，f(x)max=√2-1;
当2x+π/4=3π/2即x=5π/8时，f(x)取最小值，f(x)max=-√2-1;

您好，很高兴为您解答 希望能够帮助您
如果本题有什么不明白欢迎追问
祝你学习进步！

Answer 2

f(x)=2sinx(cosx+sinx)
=2sinxcosx+2(sinx)^2
=sin2x-[1-2(sinx)^2]+1
=sin2x-cos2x+1
=√2sin(2x-派/4)+1
故最小正周期T=2Pai/2=Pai
最大值是:根号2*1+1=根号2+1.
(2)图像你只要把在-Pai/2,-Pai/4,0,Pai/4,Pai/2五个点描上就行了.

Answer 3

f(x)=2sinxcosx-2sin^2x = sin2x - (1-cos2x) = sin2x + cos2x -1 = √2sin(2x+π/4)-1
因为0<x<π,所以π/4<2x+π/4<9π/4，所以2x+π/4=π/2，即x=π/8时，f最大值为√2-1