1）y=(1/2)^x為單調(diào)遞減的指數(shù)函數(shù),y=(1/4)^x也為單調(diào)遞減的指數(shù)函數(shù),因此兩個嚴格單調(diào)遞減 函數(shù)與常函數(shù)y=-2的和函數(shù)必單調(diào)遞減,因此原函數(shù)在定義域內(nèi)單調(diào)遞減.
2）f(x)=(1/2)^x+(1/4)^x-2
不妨令t=(1/2)^x,易知t∈[0,∞]
則原方程為f(x)=t+t^2-2=(t+1/2)^2-9/4
因為t∈[0,∞],所以t+1/2∈[1/2,∞],所以f(x）=(t+1/2)^2-9/4∈[-2,∞]
3）由2）知,f(x)=t+t^2-2=(t+1/2)^2-9/4>0 => (t+1/2)^2>9/4 =>(t+1/2)>3/2 (因為由2）知t∈[0,∞]）,所以 t>1
又t=(1/2)^x,所以(1/2)^x>1 => x