lzth.net
当前位置:首页 >> 函数F(x)=sin2x+Asin( π 2 %2x )的... >>

函数F(x)=sin2x+Asin( π 2 %2x )的...

解由f(x)=asinxcosx+sin(π/2-2x)=1/2a*2sinxcosx+cos2x=1/2asin2x+cos2x又由f(π/8)=√2即1/2asin2(π/8)+cos2(π/8)=√2即1/2asin2(π/8)+√2/2=√2即1/2asin2(π/8)=√2/2即asin2(π/8)=√2即a*√2/2=√2即a=2故f(x)=sin2x+cos2x=√2(√2/2sin2x+√2/2cos2x)=√2sin(2x+π/4)≤√2.故函数周期T=2π/2=π最大值为√2

偶函数

f(x)=sin2x-xf'(x)=2cos2x-1=0cos2x=1/22x=±π/3x=±π/6因为函数是奇函数,所以先求出[0,π/2]上的最值可疑值f(0)=0,f(π/6)=sin2*π/6-π/6=√3/2-π/6约等于0.342.f(π/2)=sin2*π/2-π/2=-π/2=-1.57.所以f(-π/

f(x)=sin2x+2cosx-1+1=sin2x+cos2x+1=√2(√2/2*sin2x+√2/2cos2x)+1=√2(sin2xzosπ/4+cos2xsinπ/4)+1=√2sin(2x+π/4)+1所以T=2π/2=π -1≤sin(2x+π/4)≤1-√2≤√2sin(2x+π/4)≤√2再加上1所以值域是[-√2+1,√2+1]

f(x)=sin2x-x f'(x)=2cos2x-1=0 cos2x=1/2 2x=±π/3 x=±π/6 因为函数是奇函数,所以 先求出[0,π/2]上的最值可疑值 f(0)=0,f(π/6)=sin2*π/6-π/6=√3/2-π/6约等于0.342.. f(π/2)=sin2*π/2-π/2=-π/2=-1.57. 所以 f(-π/6)=π/6-√3/2 f(-π/2)=π/2 所以 最小值=-π/2 最大值=π/2.

f(x)=sin2x-2sinx=sin2x+(1-2sinx)-1=sin2x+cos2x-1=√2sin(2x+π/4)-1(1)最小正周期=2π/2=π(2)最大值=√2-1此时2x+π/4=2kπ+π/2 x=kπ+π/8 {x|x=kπ+π/8,k∈N}

f(x)=ln2x-sin2xf'(x)=1/(2x)*(2x)'-cos2x*(2x)'=1/(2x)*2-cos2x*2=1/x-2cos2xf'(π/4)=1/(π/4)-2cos(2π/4)=4/π-2cosπ/2=4/π-0=4/π

提示:利用倍角公式和y=asinx+bcosx 型的化简方法化简原函数:f(x)=(sin2x+cos2x)^2-2(sin2x)^2=(sin2x)^2+2sin2xcos2x+(cos2x)^2-2(sin2x)^2=(cos2x)^2-2(sin2x)^2+2sin2xcos2x=cos4x+sin4x=√2sin(4x+π/4)然后就很好做了.

f(x)=√(1+a^2)[1/√(1+a^2)sin2x+a/√(1+a^2)cos2x] f(x)=sin(2x+b) 其中tanb=a 所以2*π/6+b=π/2 所以:b=π/6 所以a=tanπ/6=√3/3 所以:g(x)=√3/3asin2x-cos2x=2√3/3[sin2xcosπ/3-sinπ/3cos2x] =2√3/3sin(2x-π/3) 中心对称点:2x-π/3=kπ(与X轴的交点) x=kπ/2+π/6(对称中心:(kπ/2+π/6,0)) 当k=-1时 x=-π/3 点坐标:(-π/3,0)

相关文档
网站首页 | 网站地图
All rights reserved Powered by www.lzth.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com