设函数f(x)=ax+1/(x+b) a b属于整数 曲线y=f(x)再点(2 f(2))处的切线方程为 y=3证明曲线
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设函数f(x)=ax+1/(x+b) a b属于整数 曲线y=f(x)再点(2 f(2))处的切线方程为 y=3证明曲线上任一点的切线与直线x=1和直线y=x所围成的三角形的面积为定值并求出此值
![设函数f(x)=ax+1/(x+b) a b属于整数 曲线y=f(x)再点(2 f(2))处的切线方程为 y=3证明曲线](/uploads/image/z/2530420-52-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dax%2B1%2F%28x%2Bb%29+a+b%E5%B1%9E%E4%BA%8E%E6%95%B4%E6%95%B0+%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%86%8D%E7%82%B9%EF%BC%882+f%282%29%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BA+y%3D3%E8%AF%81%E6%98%8E%E6%9B%B2%E7%BA%BF)
解题思路: 利用导数计算
解题过程:
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最终答案:略
解题过程:
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最终答案:略
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