设f（x）在R上满足2f(1+x)+f（1-x）=e^x，求f(x)的二阶导数。 急！！！！谢谢！
\u8bbef(x)\u5728R\u4e0a\u7684\u5bfc\u6570\u4e3af'\uff08x\uff09\u4e142f(x)+xf'(x)>x^2\u4e0b\u9762\u4e0d\u7b49\u5f0f\u5728R\u4e0a\u6052\u6210\u7acb\u7684\u662f\u4ee4g(x)=x^2f(x)-(1/4)x^4
g'(x) = 2xf(x) + x^2f'(x) - x^3
\u6240\u4ee5\u5f53x>0\u65f6\uff0cg'(x) > x^3 - x^3 = 0
\u5f53x<0\u65f6\uff0cg'(x) < x^3 - x^3 = 0
\u6240\u4ee5g(x) \u2265 g(0) = 0
\u6240\u4ee5x^2f(x)\u2265(1/4)x^4
x\u22600\u65f6\uff0cf(x) \u2265 (1/4)x^2 > 0
x=0\u65f6\uff0c\u4ee4\u6761\u4ef6\u4e2d\u7684x=0\u53ef\u5f97f(0)>0
\u6240\u4ee5\u9009A
f`(x)>0.5>0\u8868\u793a\u51fd\u6570\u6052\u4e3a\u5355\u8c03\u589e\u51fd\u6570
\u8bb0\u5f55F\uff08x\uff09=2f(x)-x-1\uff0c\u6240\u4ee5\u4e0d\u7b49\u5f0f\u7b49\u4ef7\u4e3aF\uff08x\uff09<0
\u663e\u7136F'\uff08x\uff09=2f'(x)-1>0
F(x)\u4e3aR\u4e0a\u7684\u589e\u51fd\u6570\uff0c\u53c8F\uff081\uff09=2f(1)-1-1=0
\u6240\u4ee5F\uff08x\uff09<0\u7684\u89e3\u96c6\u4e3a{x|x<1}.
\u5373\u6ee1\u8db32f(x)<x+1\u7684\u96c6\u5408\u4e3a {x|x<1}.
2f(1-x)+f(1+x)=e^(-x) (2)
(1)*2-(2)：3f(1+x)=2e^x-e^(-x)
f(1+x)=(2e^x-e^(-x))/3
即f(t)=(2e^(t-1)-e^(1-t))/3
所以f'(x)=(2e^(x-1)+e^(1-x))/3
f''(x)=(2e^(x-1)-e^(1-x))/3
绛旓細鍗f(t)=(2e^(t-1)-e^(1-t))/3 鎵浠'(x)=(2e^(x-1)+e^(1-x))/3 f''(x)=(2e^(x-1)-e^(1-x))/3
绛旓細f(1-x)+1/2f(x)=1/2(1-x)^2 鍑忓幓鍘熷嚱鏁板緱 f(x)=1/3x^2+2/3x-1/3
绛旓細f(x)=2f(1-x)+x^2-1 f(0)=2f(1)-1 f(1)=2f(0)瑙ｅ緱锛歠(0)=4f(0)-1 3f(0)=1 f(0)=1/3
绛旓細鍙互瑙傚療鍒帮紝濡傛灉绛夊紡涓よ竟鍚屾椂闄や互x²锛屽氨寰楀埌(f(x)-xf'(x))/x²<0 灏辨槸g(x)=f(x)/x鐨勫鍑芥暟g'(x)<0 鎵浠(x)鏄釜鍗曡皟閫掑噺鍑芥暟锛屾墍浠(1)>g(2)鍗砯(1)/1>f(2)/2 鎵浠2f(1)>f(2)
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