Question

高一数学问题求解

已知f（x）=2^x/(4^x+1)。判断f（x）在（0,1）上的单调性（最好用同增异减法）

Answer 1

证明:设,X2＞X1＞0,
F(X2)-F(X1)=2^X2/(4^X2+1)-2^X1/(4^X1+1)
=[(4^X+1)*2^X2-(4^X2+1)*2^X1]/[(4^X2+1)*(4^X1+1)]
=(4^X1*2^X2-4^X2*2^X1+2^X2-2^X1)/数带山[(4^X2+1)*(4^X1+1)]
=(2^2X1*2^X2-2^2X2*2^X1+2^X2-2^X1)/[(4^X2+1)*(4^X1+1)]
=[2^X1*2^X2(2^X1-2^X2)-(2^X1-2^X2)]/[(4^X2+1)*(4^X1+1)]
=[(2^X1-2^X2)*(2^X1*2^X2-1)/[(4^X2+1)*(4^X1+1)]

因为X属于(0,1),X2＞X1＞行余0,
∴4^X＞0,2^X＞0,(4^X2+1)＞0,(4^X1+1)＞0.
2^X2＞2^X1＞0,2^X2-2^X1＞薯中0,
2^X2＞2^X1＞1＞0,不等式各项同时乘2^X1,得
2^X2*2^X1＞2^X1*2^X1＞1,∴2^X1*2^X1-1＞0,
由上式可得,
(2^X1-2^X2)＜0,(2^X1*2^X2-1)＞0.
∴F(X2)-F(X1)＜0,F(X2)＜F(X1),

∴在X2＞X1时,F(X2)＜F(X1),

F(x)=(2^x)/(4^x+1)，f(x)在（0，1）上单调递减.