作业帮高等数学设f(x)=(1 cosx)x 1sin(x²-3x)
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/15 15:09:31
f(x)=e^(3x-1)×cos(2πx+π/3),f'(x)=e^(3x-1)*(3x-1)'cos(2πx+π/3)-e^(3x-1)*sin(2πx+π/3)*(2πx+π/3)'=3e^(3
f(x)=cos(2x+π/3)+1/2-1/2cos2xf(C/2)=cos(C+π/3)+1/2-cosC/2=-1/4,cosCcosπ/3-sinCsinπ/3-cosC/2=-3/4,cos
因为2f(x)cosx=d/dx[f(x)]²=2f(x)f'(x),所以2f(x)[f'(x)-cosx]=0,有f'(x)=cosx得：f(x)=sinx+C因为f(0)=1,所以f(x
1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2）c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a
帮楼上补充的f(x)为了好表示肯定可以用y等价表示咯,
f(x)=2-2x^2f(cosx/2)=1-cosx
f(x)=cos(2x+π/3)+sin^2x-1/2=cos(2x+π/3)+(1-cos2x)/2-1/2=cos2xcos(π/3)-sin2xsin(π/3)-cos2x*1/2=-√3/2*
f(x+y)=f(x)+f(y)/1-f(x)f(y),则f(x)=tan(ax)怎么证明?令x=yf(2x)=f(x)+f(x)/[1-f(x)]^2tan2x=tanx+tanx/1-[tanx]
f(x)=sin2x+2cos²x+1=sin2x+2cos²x-1+2=sin2x+cos2x+2=√2(sin2xcosπ/4+cos2xsinπ/4)+2=√2sin(2x+
f(x)=sin2x+2cos²x+1=sin2x+cos2x+2=√2(√2/2sin2x+√2/2cos2x)+2=√2cos(2x-π/4)+2∴2x-π/4=2kπ时有最大值2+√2
f(x)=cos(2x-π/3）-cos2x-1=cos2x*cos(π/3）+sin2x*sin(π/3）-2cos2x*cos(π/3）-1=-[cos2x*cos(π/3）-sin2x*sin(
f(sinx)=cos2x+1=1-2sin^2x+1=2-2sin^2xf(cosx)=2-2cos^2x=2(1-cos^2x)=2sin^2x很高兴为您解答,【the1900】团队为您答题.请点
cosx=1-2(sinx/2)^2f=[sin(2/x)]=1+cosx=2-2(sinx/2)^2f(x)=2-2x^2f[cos(2/x)]=2-2[cos(2/x)]^2
/>g(1/4)+f(1/3)+g(5/6)+f(3/4)=cos(兀/4)+f(-2/3)+1+g(-1/6)+1+f(-1/4)+1=√2/2-√3/2+1+√3/2+1-√2/2+1=3.谢谢!
这个用cos（α-β）好想可以做出来,最好问老师
x→0时,1/2√x→∞.要把sin√x与1/√x合在一起讨论,这是个等价无穷小再问：为什么趋于无穷啊？不好意思我高数刚学很多不明白，能解释详细点吗谢谢再答：分子是1，分母趋向于0，分式不就是趋向于∞
f(x)=-sin²x+sinx+a=-(sinx-1/2)²+a最大值=aa=1a>=3所以a的取值范围：3≤a≤17/4
f(x)=cos²xsin²x/cos²x=sin²x所以f(π/6)=(1/2)²=1/4
最小正周期为π,当x=π/6时有最大值为根号三-1