f(x)=(1/(2的x次方+1))+m是奇函數(shù),a=f(log2*5).求m,a.
f(x)=(1/(2的x次方+1))+m是奇函數(shù),a=f(log2*5).求m,a.
數(shù)學(xué)人氣:924 ℃時間:2020-06-06 04:56:06
優(yōu)質(zhì)解答
log2*5中的“*”號是不是用錯了,我想它的意思應(yīng)當(dāng)是“以2為底5的對數(shù)”吧!如果這樣的話就有確定的解了.按我的理解給出一個解,你看對嗎?依題意,f(-x)=1/[2^(-x)+1]+m=(2^x)/(1+2^x)+m,而奇函數(shù)的特征是:f(-x)=-f(x)所以有:1/(2^x+1)+m=(-2^x)/(1+2^x)-m,解之得:m=-1/2.于是:f(x)=1/(2^x+1)-1/2=[2-(1+2^x)]/2(2^x+1)=(1-2^x)/2(1+2^x) .又a=f(log2^5)=[1-2^(log2^5)]/2[1+2^(log2^5)]=(1-5)/2(1+5)=-1/3.
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