y =【arcsin(1-x平方)】三次方是由哪些函数复合而成的 具体解题过程

\u5df2\u77e5|a+1|+(b-2)\u7684\u5e73\u65b9=0\uff0c\u5219\u51fd\u6570y=(b+3)x\u7684-a\u6b21\u65b9+b\u7684\u5e73\u65b9-8b+16\u662f\u4ec0\u4e48\u51fd\u6570\uff1f\u5f53X=-5\u5206\u4e4b1\u65f6\uff0c\u51fd\u6570y\u503c\u591a\u5c11

\u2235|a+1|+(b-2)\u7684\u5e73\u65b9=0
\u2234a+1=0
b-2=0
\u2234a=-1
b=2
y=(b+3)x\u7684-a\u6b21\u65b9+b\u7684\u5e73\u65b9-8b+16
=5x+4
\u662f\u4e00\u4e2a\u4e00\u6b21\u51fd\u6570
\u5f53x=-1/5\u65f6
y=5x+4=-1+4=3

f'(x)=2(x+2)(x-1)^3 + 3(x+2)^2(x-1)^2
=(x+2)(x-1)^2( 2(x-1) + 3(x+2))
=(x+2)(5x+4)(x-1)^2
x1=-2, x2=-4/5, x3=1
\u5f53x=x1-\u65f6\uff0cf'(x)>0\uff1b\u5f53x=x1+\u65f6\uff0cf'(x)<0\u3002\u6240\u4ee5\u5f53x=x1\u65f6\u51fd\u6570\u53d6\u5f97\u6781\u5927\u503cf(x1)=0
\u5f53x=x2-\u65f6\uff0cf'(x)0\u3002\u6240\u4ee5\u5f53x=x2\u65f6\u51fd\u6570\u53d6\u5f97\u6781\u5c0f\u503cf(x2)=-26244/3125
\u5f53x=x3-\u65f6\uff0cf'(x)>0\uff1b\u5f53x=x3+\u65f6\uff0cf'(x)>0\u3002\u6240\u4ee5\u5f53x=x2\u65f6\u51fd\u6570\u4e0d\u5b58\u5728\u6781\u503c
\u53e6\u5916\uff0c\u697c\u4e3b\u6df7\u6dc6\u4e86\u6700\u5927\u6700\u5c0f\u503c\u4e0e\u6781\u503c\u3002\u6700\u5927\u6700\u5c0f\u503c\u662f\u6574\u4e2a\u5b9a\u4e49\u57df\u5185\uff0c\u800c\u6781\u503c\u6307\u7684\u662f\u51fd\u6570\u5728\u8be5\u70b9\u4e0a\u4f53\u73b0\u51fa\u6765\u7684\u5cf0\u503c\uff0c\u5728\u8be5\u70b9\u5de6\u53f3\u7684\u51fd\u6570\u503c\u90fd\u6bd4\u8be5\u70b9\u7684\u51fd\u6570\u503c\u5927\uff08\u6781\u5c0f\u503c\uff09\u6216\u8005\u5c0f\uff08\u6781\u5927\u503c\uff09

从内到外一层一层地剥t=1-x;u=arcsint;y=u^3;

u=x�0�5v=1-az=sinby=arcz

u=x^2,v=1-u,t=arcsinv,y=t^3.

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