f(x)=2^x/(2x+m)
f(1)=2/3
2/3 = 2/(2+m)
m=1
f(x)=2^x/(2x+1)
定義域x≠-1/2
f'(x) = {(2x+1)*2^xln2-2*2^x}/(2x+1)^2=2ln2*2^x{x+1/2-1/ln2}/(2x+1)^2
x＜1/ln2-1/2時單調(diào)減；x＞1/ln2-1/2時單調(diào)增
x=1/ln2-1/2時有極小值：2^(1/ln2-1/2)/(2/ln2) = √2*ln2*2^(1/ln2)/4
值域【√2*ln2*2^(1/ln2)/4,+∞）
f(x)=2^x/(2x+m)
f(2)=2/3
2/3 = 2^2/(2*2+m)
m=2
f(x)=2^x/(2x+2)
定義域x≠-1
f'(x) = {2(x+1)*2^xln2-2*2^x}/(2x+2)^2=ln2*2^x{x+1-1/ln2}/{2(x+1)^2}
x＜1/ln2-1時單調(diào)減；x＞1/ln2-1時單調(diào)增
x=1/ln2-1時有極小值：2^(1/ln2-1)/(2/ln2) = 2^(1/ln2)ln2/4
值域【2^(1/ln2)ln2/4,+∞）題打錯了，應(yīng)該是
已知函數(shù)f(x)=2的x次方/2的x次方+m,且f(1)=2/3,(1)求m的值(2)求函數(shù)f(x)的值域f(x)=2^x/(2^x+m)
f(1)=2/3
2^1/(2^1+m)=2/3
m=1
f(x)=2^x/(2^x+1)=(2^x+1-1)/(2^x+1)=1-1/(2^x+1)
2^x單調(diào)增
1/(2^x+1)單調(diào)減
2^x＞0
2^x+1＞1
0＜1/(2^x+1)＜1
-1＜-1/(2^x+1)＜0
0＜1-1/(2^x+1)＜1
值域（0，1）