帮忙解答～高三函数题，谢谢～
\u9ad8\u4e09\u51fd\u6570\u9898\uff0c\u8c22\u8c22\uff01\uff01F(x)=f(x)+g(x) F(-x)=f(-x)+g(-x)=g(x)-f(x)
F(x)+F(-x)=2g(x) \u5219 g(x)=x^4+bx^2+d f(x)=ax^3+cx
g(\u6839\u53f72)=4+2b+d g(0)=d g(1)=1+b+d
g(\u6839\u53f72)+g(0)-2g(1)=4+2b+d+d-2(1+b+d)=2
1\u3001\u5148\u8bf4\u660e\u4e0b\uff0c\u6761\u4ef6"\u5bf9\u4efb\u610fx\u5c5e\u4e8e-\u6839\u53f72\u5230\u6839\u53f72\uff08\u95ed\u533a\u95f4\uff09\uff0c\u6052\u6709F\uff08x\uff09\u7edd\u5bf9\u503c\u5c0f\u4e8e0.5"\u5e94\u4e3a"\u6052\u6709F\uff08x\uff09\u7edd\u5bf9\u503c\u5c0f\u4e8e\u7b49\u4e8e0.5"
x\u5c5e\u4e8e-\u6839\u53f72\u5230\u6839\u53f72(\u95ed\u533a\u95f4)\u65f6\uff0c
\2g(x)\=\F(x)+F(-x)\\u2264\F(x)\+\F(-x)\\u22641 \g(x)\\u22640.5
\u5219\u6709 \g(\u6839\u53f72)\\u22640.5 \g(0)\\u22640.5 \g(1)\\u22640.5
\u7531\u7b2c1\u95ee\u77e5\u9053 g(\u6839\u53f72)+g(0)-2g(1)=2\u2264\g(\u6839\u53f72)\+\g(0)\+2\g(1)\\u22642
\u4e0d\u7b49\u5f0f\u7b49\u53f7\u6210\u7acb\u7684\u6761\u4ef6\u662f g(\u6839\u53f72)=g(0)=0.5 g(1)=-0.5
\u56e0\u6b64\u89e3\u51fa b=-2,d=0.5
-0.5\u2264F(1)\u22640.5 \u4e14 -0.5\u2264F(-1)\u22640.5
-0.5\u22641+a+b+c+d\u22640.5 \u4e14 -0.5\u22641-a+b-c+d\u22640.5
\u5c06b=-2,d=0.5\u4ee3\u5165\u4e0a\u9762\u4e24\u4e2a\u4e0d\u7b49\u5f97\uff1a 0\u2264a+c\u22641 \u4e14 -1\u2264a+c\u22640
\u6240\u4ee5a+c=0
\u6ce8\uff1a\u6b64\u95ee\u7684\u5173\u952e\u5728\u4e8e\u5206\u6790\u7b2c\u4e00\u95ee\u4e0e\u6240\u7ed9\u6761\u4ef6\u7684\u5173\u7cfb,2\u662f4\u500d\u76840.5\uff0c\u7b49\u5f0f\u5de6\u8fb9\u4e5f\u521a\u597d\u662f4\u4e2a\u6570\uff0c\u89e3\u6cd5\u81ea\u7136\u51fa\u73b0\u4e86\u3002(\u4e00\u822c\u8fd9\u79cd\u9898\u76ee\u7b2c\u4e00\u95ee\u90fd\u662f\u5bf9\u4e0b\u9762\u7684\u63d0\u793a\uff0c\u4ee5\u540e\u518d\u505a\u8fd9\u79cd\u9898\u76ee\u5c31\u5e94\u8be5\u6ce8\u610f\u4e86\u3002\u4f46\u4e0d\u8981\u62d8\u6ce5\u4e8e\u6b64\uff0c\u4e0b\u6b21\u5982\u679c\u8fd9\u79cd\u65b9\u6cd5\u884c\u4e0d\u901a\uff0c\u5e94\u8be5\u7acb\u523b\u60f3\u522b\u7684\u529e\u6cd5\uff0c\u8981\u4fdd\u6301\u601d\u7ef4\u7684\u53d1\u6563\uff01\uff01)
2\u3001f(x)=a(x^3-x)
a=0\u65f6\uff0cf(x)=0 \u539f\u65b9\u7a0b\u4e3a 0\uff1dx/(x^2+1) \u53ea\u6709\u4e00\u4e2a\u89e3x=0 \u4e0d\u7b26\u5408\u6761\u4ef6\uff0c\u820d\u53bb
a\u4e0d\u7b49\u4e8e0\u65f6,\u539f\u65b9\u7a0b\u4e3a\uff1aa(x^3-x)\uff1d[a(ax+1)]/(x^2+1)
(x^3-x)(x^2+1)=ax+1 \u5373 x^5-(a+1)x-1=0 \u4ee4T(x)=x^5-(a+1)x-1
T(X)\u7684\u5bfc\u51fd\u6570\u4e3aK(x)=5x^4-(a+1)
a=-1\u65f6,\u65b9\u7a0bx^5-1=0\u53ea\u6709x=1\u4e00\u4e2a\u89e3\uff0c\u4e0d\u7b26\u5408\uff0c\u820d\u53bb
a0,T(x)\u5728[-2,\u6b63\u65e0\u7a77)\u4e0a\u4e3a\u589e\u51fd\u6570\uff0c\u65b9\u7a0bT(x)=0\u6700\u591a\u6709\u4e00\u4e2a\u89e3\uff0c\u4e0d\u7b26\u5408\uff0c\u820d\u53bb
a>-1\u4e14a\u4e0d\u7b49\u4e8e0\u65f6\uff0c\u7531K(x)=0\u89e3\u5f97X=[(a+1)/5]^(1/4)\u6216X=-[(a+1)/5]^(1/4)
\u5219\u53ef\u77e5T(x)\u5728[-2,-{(a+1)/5}^(1/4)]\u4e0a\u4e3a\u589e\u51fd\u6570\uff0c
\u5728[-{(a+1)/5}^(1/4)\uff0c{(a+1)/5}^(1/4)]\u4e0a\u4e3a\u51cf\u51fd\u6570\uff0c
\u5728[{(a+1)/5}^(1/4)\uff0c\u6b63\u65e0\u7a77)\u4e0a\u4e3a\u589e\u51fd\u6570\u3002
\u65b9\u7a0bT(x)=0\u7684\u4e09\u4e2a\u89e3\u5fc5\u5728T(x)\u7684\u4e09\u4e2a\u5355\u8c03\u533a\u95f4\u4e0a
\u5219\u6709:T(-2)\u22640,T(-[(a+1)/5]^(1/4))>0,T([(a+1)/5]^(1/4))<0,
\u7531-2<-[(a+1)/5]^(1/4)\u5f97a<79
\u75312(a+1)-33\u22640,[(a+1)/5]^(5/4)>1/4,[(a+1)/5]^(5/4)>-1/4
\u5f97a\u226431/2,[(a+1)/5]^(5/4)>1/4 \u5373 a>-1+5/256^(1/5),
\u6240\u4ee5a\u7684\u53d6\u503c\u8303\u56f4\u662f(-1+5/256^(1/5),0)\u5e76(0,31/2)
\u89e3\uff1a
\u7531\u9898\u8bbe\u53ef\u77e5\uff0c\u5728\u533a\u95f4[1,2]\u4e0a\uff0c\u6052\u6709\uff1a
x|x-a|+2x-3\uff1c2x-2
\u2234\u6052\u6709\uff1ax|x-a|\uff1c1
|x-a|\uff1c1/x
-1/x\uff1cx-a\uff1c1/x
\u2234\u5f531\u2264x\u22642\u65f6\uff0c\u6052\u6709x-(1/x)\uff1ca\uff1cx+(1/x)
\u6784\u9020\u51fd\u6570g(x)=x+(1/x), h(x)=x-(1/x). 1\u2264x\u22642
\u6613\u77e5\uff0c\u6052\u6709\uff1a2\u2264g(x)\u22645/2. 0\u2264h(x)\u22643/2
\u2234\u7531\u9898\u8bbe\u53ef\u5f97\uff1a3/2\uff1ca\uff1c2
T/2=π/2==>T=π==>w=2；
所以，f(x)=Asin(2x+φ)
因为，f(x)=Asin(2x+φ)过最低点M(2π/3,-2)
所以，f(2π/3)=2sin(4π/3+φ)=-2==>sin(4π/3+φ)=-1==>4π/3+φ=3π/2==>φ=π/6
所以，f(x)=2sin(2x+π/6)
(2)因为，在区间[π/12,π/2]上
f(x)单调减区间：2kπ+π/2<=2x+π/6<=2kπ+3π/2==>kπ+π/6<=x<=kπ+2π/3
所以，最大值：f(π/6)=2；最小值：f(π/2)=2sin(π+π/6)=-1
相邻两个交点为½π所以周期为π。ω为2。最低点纵坐标即为A值。最低点横坐标带入括号中的式子并令其等于2πk－½π算出φ
额，哥哥我才初一，看都没看懂
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