还是那道题,里面的解法看不太懂已知f(x)=x^2 lnx.(1)证明:对任意的t>0,存在唯一的s,使t=f(x).(2)证明:当0<x≤1时,f(x)≤0.设t>0,令h(x)=f(x)-t,x∈[1,+∞).由(1)知,h(x)在区间(1,+∞)内单调递增
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![还是那道题,里面的解法看不太懂已知f(x)=x^2 lnx.(1)证明:对任意的t>0,存在唯一的s,使t=f(x).(2)证明:当0<x≤1时,f(x)≤0.设t>0,令h(x)=f(x)-t,x∈[1,+∞).由(1)知,h(x)在区间(1,+∞)内单调递增](/uploads/image/z/13605308-44-8.jpg?t=%E8%BF%98%E6%98%AF%E9%82%A3%E9%81%93%E9%A2%98%2C%E9%87%8C%E9%9D%A2%E7%9A%84%E8%A7%A3%E6%B3%95%E7%9C%8B%E4%B8%8D%E5%A4%AA%E6%87%82%E5%B7%B2%E7%9F%A5f%28x%29%3Dx%5E2+lnx.%EF%BC%881%EF%BC%89%E8%AF%81%E6%98%8E%3A%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84t%3E0%2C%E5%AD%98%E5%9C%A8%E5%94%AF%E4%B8%80%E7%9A%84s%2C%E4%BD%BFt%3Df%28x%29.%282%29%E8%AF%81%E6%98%8E%EF%BC%9A%E5%BD%930%EF%BC%9Cx%E2%89%A41%E6%97%B6%2Cf%28x%29%E2%89%A40.%E8%AE%BEt%EF%BC%9E0%2C%E4%BB%A4h%28x%29%EF%BC%9Df%28x%29%EF%BC%8Dt%2Cx%E2%88%88%5B1%2C%EF%BC%8B%E2%88%9E%29%EF%BC%8E%E7%94%B1%281%29%E7%9F%A5%2Ch%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%281%2C%EF%BC%8B%E2%88%9E%29%E5%86%85%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E)
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