.对于函数f(x)=log1/2(x2-2ax+3),若函数f(x)在(-∞,1]上为增函数,求实数a的取值范围我的问题是:由g(x)在(-∞,1]上为减函数是怎样推出a≥1的?为什么g(1)>0,而且a<2是怎么得出的,
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![.对于函数f(x)=log1/2(x2-2ax+3),若函数f(x)在(-∞,1]上为增函数,求实数a的取值范围我的问题是:由g(x)在(-∞,1]上为减函数是怎样推出a≥1的?为什么g(1)>0,而且a<2是怎么得出的,](/uploads/image/z/5260718-38-8.jpg?t=.%E5%AF%B9%E4%BA%8E%E5%87%BD%E6%95%B0f%28x%29%3Dlog1%2F2%28x2-2ax%2B3%29%2C%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%28-%E2%88%9E%2C1%5D%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%88%91%E7%9A%84%E9%97%AE%E9%A2%98%E6%98%AF%EF%BC%9A%E7%94%B1g%28x%29%E5%9C%A8%28-%E2%88%9E%2C1%5D%E4%B8%8A%E4%B8%BA%E5%87%8F%E5%87%BD%E6%95%B0%E6%98%AF%E6%80%8E%E6%A0%B7%E6%8E%A8%E5%87%BAa%E2%89%A51%E7%9A%84%3F%E4%B8%BA%E4%BB%80%E4%B9%88g%281%29%EF%BC%9E0%2C%E8%80%8C%E4%B8%94a%EF%BC%9C2%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E5%87%BA%E7%9A%84%2C)
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