已知奇函数f(x=2)=f(-x),且当x∈(0,1)时,f(x)=2的x次方。1.证明f(x+4)=f(x);2.求f(log2 24)的值。
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题意不清楚啊,“当x∈(0,1)时,f(x)=2的x次方”???
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由f(-x) = f(x+2)及-f(x)=f(-x),得-f(x)=f(x+2);
由f(-x) = f(x+2)得f(x)=f(-x+2)=-f(x-2);
有以上两式得f(x+2)=f(x-2),即f(x)=f(x+4)
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f(log2 24)=f(log2(16) + log2(3/2)) = f(4+log2(3/2)) = f(log2(3/2));
由于当x∈(0,1)时,f(x)=2的x次方,
又2^0 < 3/2 < 2^1 ,故0 < log2(3/2) < 1;
因此由题设f(log2(3/2)) = 2^(log2(3/2)) = 3/2
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由f(-x) = f(x+2)及-f(x)=f(-x),得-f(x)=f(x+2);
由f(-x) = f(x+2)得f(x)=f(-x+2)=-f(x-2);
有以上两式得f(x+2)=f(x-2),即f(x)=f(x+4)
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f(log2 24)=f(log2(16) + log2(3/2)) = f(4+log2(3/2)) = f(log2(3/2));
由于当x∈(0,1)时,f(x)=2的x次方,
又2^0 < 3/2 < 2^1 ,故0 < log2(3/2) < 1;
因此由题设f(log2(3/2)) = 2^(log2(3/2)) = 3/2
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