已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)讨论h(x)的奇偶性f(x)=log(2)[(x-1)/(x+1)], g(x)=2ax+1-a, h(x)=f(x)+g(x)1、f(-x)=log(2)[(-x-1)/(-x+1)]=log(2)[(x+1)/(x-1)]=-log(2)[(x-1)/(x+1)]=-f(x)g(-x)=-2ax+1-a,若1-a=0,即a=1,
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![已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)讨论h(x)的奇偶性f(x)=log(2)[(x-1)/(x+1)], g(x)=2ax+1-a, h(x)=f(x)+g(x)1、f(-x)=log(2)[(-x-1)/(-x+1)]=log(2)[(x+1)/(x-1)]=-log(2)[(x-1)/(x+1)]=-f(x)g(-x)=-2ax+1-a,若1-a=0,即a=1,](/uploads/image/z/5191751-47-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlog2%28%28x-1%29%2F%28x%2B1%29%29%2Cg%28x%29%3D2ax%2B1-a%2C%E5%8F%88h%28x%29%3Df%28x%29%2Bg%28x%29%E8%AE%A8%E8%AE%BAh%28x%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7f%28x%29%3Dlog%282%29%5B%28x-1%29%2F%28x%2B1%29%5D%2C+g%28x%29%3D2ax%2B1-a%2C+h%28x%29%3Df%28x%29%2Bg%28x%291%E3%80%81f%28-x%29%3Dlog%282%29%5B%28-x-1%29%2F%28-x%2B1%29%5D%3Dlog%282%29%5B%28x%2B1%29%2F%28x-1%29%5D%3D-log%282%29%5B%28x-1%29%2F%28x%2B1%29%5D%3D-f%28x%29g%28-x%29%3D-2ax%2B1-a%2C%E8%8B%A51-a%3D0%2C%E5%8D%B3a%3D1%2C)
已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)讨论h(x)的奇偶性f(x)=log(2)[(x-1)/(x+1)], g(x)=2ax+1-a, h(x)=f(x)+g(x)1、f(-x)=log(2)[(-x-1)/(-x+1)]=log(2)[(x+1)/(x-1)]=-log(2)[(x-1)/(x+1)]=-f(x)g(-x)=-2ax+1-a,若1-a=0,即a=1,
已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)讨论h(x)的奇偶性
f(x)=log(2)[(x-1)/(x+1)], g(x)=2ax+1-a, h(x)=f(x)+g(x)
1、f(-x)=log(2)[(-x-1)/(-x+1)]=log(2)[(x+1)/(x-1)]=-log(2)[(x-1)/(x+1)]=-f(x)
g(-x)=-2ax+1-a,若1-a=0,即a=1,则g(-x)=-g(x),
∴h(-x)=-f(x)-g(x)=-[f(x)+g(x)]=-h(x),则h(x)为奇函数
若a={-log(2)[(x-1)/(x+1)]}/(2x)=-f(x)/(2x),则g(x)=-f(x)+1+f(x)/(2x)
∴h(x)=f(x)+g(x)=1+f(x)/(2x),此时,h(-x)=1+f(-x)/(-2x)=1-f(x)/(-2x)=1+f(x)/(2x)=h(x)
∴ 此时h(x)为偶函数
若a取上述两种情况之外的值,则h(x)为非奇非偶函数
为什么a={-log(2)[(x-1)/(x+1)]}/(2x)=-f(x)/(2x)?老师只讲了两种.
已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)讨论h(x)的奇偶性f(x)=log(2)[(x-1)/(x+1)], g(x)=2ax+1-a, h(x)=f(x)+g(x)1、f(-x)=log(2)[(-x-1)/(-x+1)]=log(2)[(x+1)/(x-1)]=-log(2)[(x-1)/(x+1)]=-f(x)g(-x)=-2ax+1-a,若1-a=0,即a=1,
一般情况下呢,大家都把a当作常数,若把a当作常数呢,当然就只有两种情况
a={-log(2)[(x-1)/(x+1)]}/(2x)=-f(x)/(2x)这种情况下,a含有x变量,当然是不存在的
但是,原题目并没有限定a是否为常数或变量,全面讨论的情况下,当然要考虑a作为变量的可能
至于a是怎么来的,当然是先假定h(x)是偶函数,然后用h(-x)=h(x)倒推回来的