如果f（x）是奇函数，则f（x2）是偶函数，正确吗，举例说明
\u8bbef\uff08x\uff09\u4e3a\u5947\u51fd\u6570\uff0cg\uff08x\uff09\u4e3a\u5076\u51fd\u6570\uff0c\u4e14f\uff08x\uff09-g\uff08x\uff09=x2-x\uff0c\u6c42f\uff08x\uff09\u56e0\u4e3af(x)\u4e3a\u5947\u51fd\u6570 \u6240\u4ee5f(x)=-f(-x)
\u56e0\u4e3ag(x)\u4e3a\u5076\u51fd\u6570 \u6240\u4ee5g(x)=g(-x)
f(x)-g(x\uff09=x^2-2x
\u53d6x\u4e3a-x
f(-x)-g(-x)=(-x)^2-2(-x)
-f(x)-g(x)=x^2+2x
\u4e24\u5f0f\u76f8\u52a0\u5f97-2g(x)=2x^2 g(x)=-x^2
\u4ee3\u56def(x)-g(x\uff09=x^2-2x\u5f97 f(x)=-x
\u56e0\u4e3a:f(x)\u662f\u5076\u51fd\u6570
\u6240\u4ee5:f(-x)=f(x)
\u56e0\u4e3a:g(x)\u662f\u5947\u51fd\u6570
\u6240\u4ee5:f(-x)=-f(x)
\u53c8\u56e0\u4e3a:f(x)+g(x)=x2+x-2 \u00b7\u00b7\u00b7\u00b7\u00b7 A
\u6240\u4ee5:f(-x)+g(-x)=f(x)+[-g(x)]=f(x)-g(x)=(-x)2+(-x)-2=x2\uff0dx-2\u00b7\u00b7\u00b7\u00b7\u00b7\u00b7B
\u6240\u4ee5:A\uff0bB\u4e3a
[f(x)+g(x)]\uff0b[f(-x)+g(-x)]=[x2+x-2]+[x2-x-2]
\u5316\u7b80\u4e3a\uff1a
2f\uff08x\uff09\uff1d2x2-4
\u6240\u4ee5:f\uff08x\uff09\uff1dx2-2
\u56e0\u4e3a\uff1af(x)+g(x)=x2+x-2
\u6240\u4ee5\uff1ag(x)=x
正确
？
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