设函数f(x)=a2^x-1/1+2^x(a属于R),f(x)在在R上是奇函数,1.求f(x)的值域;2解不等式0
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![设函数f(x)=a2^x-1/1+2^x(a属于R),f(x)在在R上是奇函数,1.求f(x)的值域;2解不等式0](/uploads/image/z/3757633-25-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Da2%5Ex-1%2F1%2B2%5Ex%28a%E5%B1%9E%E4%BA%8ER%29%2Cf%28x%29%E5%9C%A8%E5%9C%A8R%E4%B8%8A%E6%98%AF%E5%A5%87%E5%87%BD%E6%95%B0%2C1.%E6%B1%82f%28x%29%E7%9A%84%E5%80%BC%E5%9F%9F%EF%BC%9B2%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8F0)
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设函数f(x)=a2^x-1/1+2^x(a属于R),f(x)在在R上是奇函数,1.求f(x)的值域;2解不等式0
设函数f(x)=a2^x-1/1+2^x(a属于R),f(x)在在R上是奇函数,1.求f(x)的值域;2解不等式0
设函数f(x)=a2^x-1/1+2^x(a属于R),f(x)在在R上是奇函数,1.求f(x)的值域;2解不等式0
f(-x)=[a*2^(-x)-1]/[1+2^(-x)]=[a-2^x]/[2^x+1],---(1)
f(x)在R上是奇函数,f(-x)=-f(x)=[-a*2^x+1]/[1+2^x],---(2)
比较(1)(2)知 a=1,f(x)=(2^x-1)/(1+2^x)=1-2/(1+2^x).
1.由于分母 1+2^x>1,故 -1