求好汉帮帮忙.已知函数f(x)=ln(ax+b)在x=1处的切线方程为y=1|2x-1|2+ln2.1)证明:方程f(x)-x=0有且只有一个实根.2)若s,t∈(0,正无穷),且s(1+t)^[e^f(s-1)].
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![求好汉帮帮忙.已知函数f(x)=ln(ax+b)在x=1处的切线方程为y=1|2x-1|2+ln2.1)证明:方程f(x)-x=0有且只有一个实根.2)若s,t∈(0,正无穷),且s(1+t)^[e^f(s-1)].](/uploads/image/z/6944166-54-6.jpg?t=%E6%B1%82%E5%A5%BD%E6%B1%89%E5%B8%AE%E5%B8%AE%E5%BF%99.%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dln%28ax%2Bb%29%E5%9C%A8x%3D1%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BAy%3D1%7C2x-1%7C2%2Bln2.1%EF%BC%89%E8%AF%81%E6%98%8E%EF%BC%9A%E6%96%B9%E7%A8%8Bf%EF%BC%88x%EF%BC%89-x%3D0%E6%9C%89%E4%B8%94%E5%8F%AA%E6%9C%89%E4%B8%80%E4%B8%AA%E5%AE%9E%E6%A0%B9.2%EF%BC%89%E8%8B%A5s%2Ct%E2%88%88%EF%BC%880%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%EF%BC%89%2C%E4%B8%94s%281%2Bt%29%5E%5Be%5Ef%28s-1%29%5D.)
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