f(x)=2sinπx与函数g(x)=三次根号(x-1)的图象所有交点横坐标之和
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g(x)=tg(x/2)周期T=π/(1/2)=2πf(x)=2+sin^2ax=2+(1-cos2ax)/2=-cos2ax/2+5/2T=2π/2a=2πa=1/2
∵f(x)=cos(2x-π/3)+(sinx)^2-(cosx)^2=cos(2x-π/3)-cos2x=2sin(2x-π/6)sin(π/6)=sin(2x-π/6).∴g(x)=[sin(2x
化简就可以啦,f(x)+g(x)=3sin(2x-π/3)+4sin(2x+π/3)=3(sin2x*1/2-cos2x*根号3/2)+4(sin2x*1/2+cos2x*根号3/2)=7/2sin2
f(x)=cos(x-π/)+sin^2x-cos^2x=-cosx+sin^2x-cos^2x=-2cos^2x-cosx+1最小正周期2π
函数g(x)与f(x)的图像关于直线对称,故可设g(x)=2sin(πx/4+a);(周期和振幅相等);当x=2时,f(2)=g(2);故sin(π/2+a)=sinπ3/4,所以a=π/4+2kπ(
f(x)=sin(2x+π/2)=cos2xg(x)=f(x)+f(π/4-x)=cos2x+cos(π/2-2x)=cos2x+sin2x=√2sin(2x+π/4)单增区间2x+π/4∈[2kπ-
函数g(x)=sin(x+π6),f(x)=2cosx•g(x)−12=32sin2x+12cos2x=sin(2x+π6).(1)函数f(x)的最小正周期T=π,因为2x+π6=kπ,所以对称中心坐
(1)f(x)=sin(x-π/6)+cos(x-π/3),g(x)=2sin²x/2f(x)=sinxcosπ/6-cosxsinπ/6+cosxcosπ/3+sinxsinπ/3=√3s
f(x)=sin(πx/2-π/6)-2cos²πx/4+1-3=sinπx/2cosπ/6-cosπx/2sinπ/6-cosπx/2-3=√3/2sinπx/2-1/2cosπx/2-c
(1)单调递增,∴-π/2+2nπ
f(x)=sin(x/2)-√3[1-cos(x/2)]+√3=2[(1/2)sin(x/2)+(√3/2)cos(x/2)]=2sin(x/2+π/3)(1)g(x)=f(x+π/3)=2sin[(
这种题貌似只能用求导做了F(x)=x^2-1-2lnx,注意到F在x>0上定义F'(x)=2x-2/x解F'(x)=2(x-1/x)=0得x=1又当00,则F单调增故x=1为F的最小值点,F(1)=0
y'=f'(sin^2x)*(sin^2x)'=g(sin^2x)*2sinxcosx
(1)f(x)可化简为-Sin(2x+π/4)剩下的问题可自行解决(2)关于x=π/4对称,由对称性可知,两函数横坐标之和x1+x2=π/2,纵坐标不变,因此g(x)=-Sin(2(π/2-x)+π/
当w=2时∵f(x)g(x)=sin2xsin(2x+π2)=sin2xcos2x=12sin4x,T=2π4=π2,故①正确;当w=1时∵f(x)+g(x)=sinx+sin(2x+π2)=sinx
t=f(x)再问:我想要具体的答案,谢谢.再答:f(x)=sin(2x-π/6)+1t=f(x)t范围[0,2]设h(t)=h(f(x))=g(x)h(t)=t^2+t,t范围[0,2]值域[0,6]
1增加的时间间隔2kπ-π/2
函数g(x)=3x−1关于(1,0)对称,函数g(x)单调递增,且函数f(x)=2sinπx也关于(1,0)对称,由3x−1=2,解得x-1=8,即x=9,由3x−1=−2,解得x-1=-8,即x=-
因为(1,0)是f(x)=2sinπx的中心之一而(1,0)是g(x)=³√(x-1)的中心∴所有左右交点关于(1,0)对称∴x1+x2=2*1=2而有7对,加上本身H点∴横坐标之和=2*7
f(x1)的值域A=[-3/2,1/2],g(x2)的值域B=[-3-m,-m]A是B的子集-3-m1/2-3/2