函数y=a^(1-x) (a>0,a=/1)的图象恒过定点A,若点A在直线mx+ny-1=0 (nm>0)上,求1/m+1/n的最小值
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![函数y=a^(1-x) (a>0,a=/1)的图象恒过定点A,若点A在直线mx+ny-1=0 (nm>0)上,求1/m+1/n的最小值](/uploads/image/z/11963178-18-8.jpg?t=%E5%87%BD%E6%95%B0y%3Da%5E%281-x%29+%28a%3E0%2Ca%3D%2F1%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E6%81%92%E8%BF%87%E5%AE%9A%E7%82%B9A%2C%E8%8B%A5%E7%82%B9A%E5%9C%A8%E7%9B%B4%E7%BA%BFmx%2Bny-1%3D0+%28nm%3E0%29%E4%B8%8A%2C%E6%B1%821%2Fm%2B1%2Fn%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC)
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函数y=a^(1-x) (a>0,a=/1)的图象恒过定点A,若点A在直线mx+ny-1=0 (nm>0)上,求1/m+1/n的最小值
函数y=a^(1-x) (a>0,a=/1)的图象恒过定点A,若点A在直线mx+ny-1=0 (nm>0)上,求1/m+1/n的最小值
函数y=a^(1-x) (a>0,a=/1)的图象恒过定点A,若点A在直线mx+ny-1=0 (nm>0)上,求1/m+1/n的最小值
y=a^(1-x) (a>0,a=/1)的图象恒过定点A,
A(1,1)
点A在直线mx+ny-1=0 (nm>0)上
m+n=1,n=1-m
1/m+1/n
=(m+n)/mn
=1/mn=1/m(1-m)
=1/(m-m^2)
当m=1/2时,m-m^2有最大值1/4
则1/m+1/n=1/(m-m^2)有最大值4