已知函数f(x)=loga[(x-3)/(x+3)],(a>0且a≠1) (a为底数)(1) 判定f(x)的单调性(2) 设g(x)=1+logа(x-1),若方程f(x)=g(x)有实根,求a的取值范围 (a为底数)(3) 求函数h(x)=[f(x)*lna+ln(x+3)]-(x²/8)在[4,6]上的最大值和最
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![已知函数f(x)=loga[(x-3)/(x+3)],(a>0且a≠1) (a为底数)(1) 判定f(x)的单调性(2) 设g(x)=1+logа(x-1),若方程f(x)=g(x)有实根,求a的取值范围 (a为底数)(3) 求函数h(x)=[f(x)*lna+ln(x+3)]-(x²/8)在[4,6]上的最大值和最](/uploads/image/z/15084231-15-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dloga%5B%28x-3%29%2F%28x%2B3%29%5D%2C%28a%3E0%E4%B8%94a%E2%89%A01%29+%28a%E4%B8%BA%E5%BA%95%E6%95%B0%29%281%29+%E5%88%A4%E5%AE%9Af%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E6%80%A7%282%29+%E8%AE%BEg%28x%29%3D1%2Blog%D0%B0%28x-1%29%2C%E8%8B%A5%E6%96%B9%E7%A8%8Bf%28x%29%3Dg%28x%29%E6%9C%89%E5%AE%9E%E6%A0%B9%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4++%28a%E4%B8%BA%E5%BA%95%E6%95%B0%29%283%29+%E6%B1%82%E5%87%BD%E6%95%B0h%28x%29%3D%5Bf%28x%29%2Alna%2Bln%28x%2B3%29%5D-%28x%26%23178%3B%2F8%29%E5%9C%A8%5B4%2C6%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80)
已知函数f(x)=loga[(x-3)/(x+3)],(a>0且a≠1) (a为底数)(1) 判定f(x)的单调性(2) 设g(x)=1+logа(x-1),若方程f(x)=g(x)有实根,求a的取值范围 (a为底数)(3) 求函数h(x)=[f(x)*lna+ln(x+3)]-(x²/8)在[4,6]上的最大值和最
已知函数f(x)=loga[(x-3)/(x+3)],(a>0且a≠1) (a为底数)
(1) 判定f(x)的单调性
(2) 设g(x)=1+logа(x-1),若方程f(x)=g(x)有实根,求a的取值范围 (a为底数)
(3) 求函数h(x)=[f(x)*lna+ln(x+3)]-(x²/8)在[4,6]上的最大值和最小值
已知函数f(x)=loga[(x-3)/(x+3)],(a>0且a≠1) (a为底数)(1) 判定f(x)的单调性(2) 设g(x)=1+logа(x-1),若方程f(x)=g(x)有实根,求a的取值范围 (a为底数)(3) 求函数h(x)=[f(x)*lna+ln(x+3)]-(x²/8)在[4,6]上的最大值和最
1 讨论(x-3)/(x+3)的范围,然后在讨论a的范围,可以出单调行
2 利用g(x)=f(x),可以得到a=(x-3)/【(x-1)*(x+3)},然后去倒数,变为1/a==[ (x-1)*(x+3)]=(x*x+2x-3)/(x-3)
分子=[ (x-1)*(x+3)]=(x*x+2x-3)=x*x-9+2x-6+12=(x-3)*(x+3)+2(x-3)+12
分母=(x-3)
分子/分母=(x+3)+2+12/(x-3)
=(x-3)+12/(x-3)+8
有因为x>3 x-3>0
所以上式 可以利用均值不等式求出答案,
第三问,没做,请原谅