求y=3sin(3x 4-π 4)的最值和定义域
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函数的周期T=2πω=2π2=π,由-π2+2kπ≤2x+π3≤π2+2kπ,解得−5π12+kπ≤x≤π12+kπ,即函数的递增区间为[−5π12+kπ,π12+kπ],k∈Z,由2x+π3=π2+
∵y=sin(2x+π3),∴由2kπ−π2≤2x+π3≤2kπ+π2,k∈Z.得kπ-5π12≤x≤kπ+π12,k∈Z.∴当k=0时,递增区间为[0,π12],当k=1时,递增区间为[7π12,π
π/2+2kπ再问:换元法有没有?再答:令3x+π/4=t,y=2sint的递减区间是:π/2+2kπ
y=-1/2sin(2/3x-π/4)所以y和sin(2/3x-π/4)单调性相反sinx的增区间是(2kπ-π/2,2kπ+π/2)减区间是(2kπ+π/2,2kπ+3π/2)所以sin(2/3x-
解题思路:三角函数图像解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.
y=3sin(3x+π/4)单调增区间是:2kPai-Pai/2
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
y=2sin(3x+π/4)依题意-π/2+2kπ
用点斜式,首先求斜率K,在任意一点斜率K(x)=y‘=4x3-4x当x=2,k=24,所以直线方程就是y-11=24(x-2).
∵(π3+4x)+(π6-4x)=π2,∴cos(4x-π6)=cos(π6-4x)=sin(π3+4x),∴原式就是y=2sin(4x+π3),这个函数的最小正周期为2π4,即T=π2.当-π2+2
x²+1=-3x两边平方x^4+2x²+1=9x²x^4+1=7x²两边平方x^8+2x^4+1=49x^4x^8+1=47x^4两边除以x^4x^4+1/x^
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
y∈[1,3]当y=1时,sin(x+π/3)=-1,x+π/3=2kπ-π/2,x=kπ-5π/12,k∈Z当y=3时,sin(x+π/3)=1,x+π/3=2kπ+π/2,x=kπ+π/12,k∈
[3/2,13/4]
sinx的减区间是(2kπ+π/2,2kπ+3π/2)所以这里2kπ+π/2
1、y=(cos^2x+sin^2x)^2-2cos^2xsin^2x=1-1/2(sin2x)^2=1-1/4(1-cos4x)=3/4+1/4cos4x周期T=2pi/4=pi/22、y=(根3/
y=(sinxcosπ/3-cosxsinπ/3)sinx=(1/2*sinx-√3/2*cosx)sinx=1/2*sin²x-√3/2*sinxcosx=1/2*(1-cos2x)/2-
x平方-3x+1=0二边同除以xx-3+1/x=0x+1/x=3x^2+1/x^2=(x+1/x)^2-2=3^2-2=7x^4+1/x^4=(x^2+1/x^2)^2-2=7^2-2=47
y′=12x3-12x2,y″=36x2-24x=12x(3x-2)令y″=0解得,x=0或x=23.所以曲线的拐点为(0,1),(23,1127).当x<0或x>23时,y″>0,则曲线的凹区间为(
y=sinx增区间[2kπ-π/2,2kπ+π/2]所以本题,2kπ-π/2≤π/4+2x≤2kπ+π/2kπ-3π/8