f(x)=log(a)^[(1-mx)/(x-1)] (a>1),的图像关于原点对称,问:求m的值;利用函数单调性判断函数f(x)的增减性
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![f(x)=log(a)^[(1-mx)/(x-1)] (a>1),的图像关于原点对称,问:求m的值;利用函数单调性判断函数f(x)的增减性](/uploads/image/z/5166992-56-2.jpg?t=f%28x%29%3Dlog%28a%29%5E%5B%281-mx%29%2F%28x-1%29%5D+%28a%3E1%29%2C%E7%9A%84%E5%9B%BE%E5%83%8F%E5%85%B3%E4%BA%8E%E5%8E%9F%E7%82%B9%E5%AF%B9%E7%A7%B0%2C%E9%97%AE%3A%E6%B1%82m%E7%9A%84%E5%80%BC%3B%E5%88%A9%E7%94%A8%E5%87%BD%E6%95%B0%E5%8D%95%E8%B0%83%E6%80%A7%E5%88%A4%E6%96%AD%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%A2%9E%E5%87%8F%E6%80%A7)
f(x)=log(a)^[(1-mx)/(x-1)] (a>1),的图像关于原点对称,问:求m的值;利用函数单调性判断函数f(x)的增减性
f(x)=log(a)^[(1-mx)/(x-1)] (a>1),的图像关于原点对称,问:求m的值;利用函数单调性判断函数f(x)的增减性
f(x)=log(a)^[(1-mx)/(x-1)] (a>1),的图像关于原点对称,问:求m的值;利用函数单调性判断函数f(x)的增减性
1:
f(x)=log(a)^[(1-mx)/(x-1)] (a>1),的图像关于原点对称;
即函数为奇函数;
f(-x)=log(a)^[(1+mx)/(-x-1)]
-f(x)=-log(a)^[(1-mx)/(x-1)]=log(a)^[(x-1)/(1-mx)];
f(-x)=-f(x);
所以:
(1+mx)/(-x-1)=(x-1)/(1-mx);
(1+mx)(1-mx)=-(x-1)(x+1)
1-m^2x^2=1-x^2;
m=1,(不和定义域要求,舍去)或m=-1;
所以m=-1;
函数为f(x)=log(a)[(1+x)/(x-1)];
2:
f(x)=log(a)[(x+1)/(x-1)]=log(a)^[1+2/(x-1)];
为符合函数;
a>1;
log(a)^x,单调递增;
1+2/(x-1),单调递减;
所以f(x)=log(a)[(x+1)/(x-1)]单调递减;
f(-x)=-f(x),得m=-1(m=1舍)
f(x)=log(a)^[(1+x)/(x-1)] =log(a)^[(1+2/(x-1)]
易知 u=1+2/(x-1)在(-∞,-1),(1,∞)上都是减函数,又a>1,
故知f(x)在(-∞,-1),(1,∞)上都是减函数.