已知函数fx=-cos 若方程fx=0有解

f'(x)=[2xe^x-x²e^x]/(e^x)²=(2x-x²)/(e^x)∴(-∞,0)单调递减,(0,2)单调递增；(2,+∞)单调递减∴极小值是f(0)=0极大

f(x)=cos²(x-30°)-sin²xf(15°)=cos²(15°-30°)-sin²(15°)=cos²(-15°)-sin²(1

f(x)=cos2x+根号3sin2x=2sin(2x+π/2）所以周期为π对称轴2x+π/2=π/2+kπ（k是整数）即x=kπ/2k是整数单调区间-π/2+2kπ

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设g(x)=x^2-f(x)求g'(x)=2x-1/x+a/x^2通分有g'(x)=(2x^3-x+a)/x^2考虑其在(0,+∞)上单调性若2x^3-x+a>=0则g(x)最小值满足g(x)>0即可

f'(x)=1-1/xf'(2)=1-1/2=1/2f(2)=2-1-ln2=1-ln2由点斜式得切线方程：y=1/2*(x-2)+1-ln2即y=x/2-ln2由f'(x)=0,得x=1x0因此f(

f(x)=cos²(x-30°)-sin²x=1/2[1+cos(2x-60º)]-1/2(1-cos2x)=1/2(cos2xcos60º+sin2xsin6

f(x)=1/2cos^2x+[(根号3)/2]sinxcosx+1=1/4cos2x+1/4+根号3/4sin2x+1=1/2(sin(pi/6)cos2x+cos(pi/6)sin2x)+5/4=

令t=sinx则f=(1-t^2)+2t=-t^2+2t+1=-(t-1)^2+2因为|t|

f(x)=√3sin2x+cos2x=2(sin2x*√3/2+cos2x*1/2)=2(sin2xcosπ/6+cos2xsinπ/6)=2sin(2x+π/6)所以f(π/6)=2sin(2×π/

1)f（x）=sin(x/2)cos(x/2)＋√3cos²(x/2)=(sinx)/2＋(√3cosx)/2-1/2令cos(π/3)=1/2sin(π/3)=√3/2∴f（x）=sin(

(1) 等式化简后：f(2)=±（√19／2）+3

f(x)=x^2/(e^x)因为对于任意x,e^x＞0,所以f(x)的定义域为R===>f'(x)=[2x*e^(x)-x^2*e^x]/(e^x)^2===>f'(x)=

f(x)=(ax-a)/(x+1)=a-2a/(x+1)f(f(x))=a-2a/(f(x)+1)=a-2a(x+1)/[(a+1)x+1-a]要使f(f(x))=x,则必须分母中(a+1)x=0,则

f(x)=cos(2x-π/3)-cos2x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)最小正周期T=2π/2=π(2)0

f(x)=cos²x+sinxcosx=(cos2x+1)/2+1/2sin2x=(1/2cos2x+1/2sin2x)+1/2=√2/2*(√2/2cos2x+√2/2sin2x)+1/2

函数f(x)=cos(2x-派/6)x∈R函数的最小正周期T=2π/2=π对称轴方程2x-π/6=kπx=kπ/2+π/12k∈Zcosa=-3/5sina=4/5sin2a=2sinacosa=-2

(1)、f(x)=2cos²x-（sinx-cosx）²=2cos²x-(1-sin2x)=cos2x+sin2x运用一下化一公式得f(x)=√2sin(2x+π/4),

可令x=x+1,代入,得到,f（x+2）=【1+f（x+1）】/【1-f（x+1）】=-1/f（x）令x=x+2,代入,得到,f（x+3）=【1+f（x+2）】/【1-f（x+2）】=-1/f（x+1