设函数f(x)=log1/2(1-1/2x) 1.证明(x)=log1/2 (1-1/2 x)在(-∞,1/2)上是增函数2.解不等式:f(x)>1
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![设函数f(x)=log1/2(1-1/2x) 1.证明(x)=log1/2 (1-1/2 x)在(-∞,1/2)上是增函数2.解不等式:f(x)>1](/uploads/image/z/7839225-9-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dlog1%2F2%281-1%2F2x%29+1.%E8%AF%81%E6%98%8E%28x%29%3Dlog1%2F2+%281-1%2F2+x%29%E5%9C%A8%EF%BC%88-%E2%88%9E%2C1%2F2%EF%BC%89%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B02.%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8F%EF%BC%9Af%28x%29%3E1)
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设函数f(x)=log1/2(1-1/2x) 1.证明(x)=log1/2 (1-1/2 x)在(-∞,1/2)上是增函数2.解不等式:f(x)>1
设函数f(x)=log1/2(1-1/2x) 1.证明(x)=log1/2 (1-1/2 x)在(-∞,1/2)上是增函数
2.解不等式:f(x)>1
设函数f(x)=log1/2(1-1/2x) 1.证明(x)=log1/2 (1-1/2 x)在(-∞,1/2)上是增函数2.解不等式:f(x)>1
(1)设x1<x2<1/2
∴f(x1)-f(x2)=㏒1/2 [ (2-x1)/(2-x2)]
∵x1<x2<1/2 ∴﹣x1>﹣x2 ∴2-x1>2-x2>0 ∴ (2-x1)/(2-x2)>1
∴f(x1)-f(x2)=㏒1/2 [ (2-x1)/(2-x2)]<㏒1/2 1=0 ∴f(x1)<f(x2)
∴f(x)=log1/2 (1-1/2 x)在(-∞,1/2)上是增函数
(2)f(x)>1 ∴㏒1/2 (2-x)>1 ∴2-x>0 2-x<1/2 ∴3/2<x<2