Question

设二次函数f(x)=ax^2-2ax+c

在区间[0,1]上单调递减，且f(m)<=f(0),则实数m的取值范围。。。详细过程

Answer 1

对f(x)求导，得到2ax-2a=2a(x-1),当x在[0,1]时，2a(x-1)<=0,因此a>0.
f(m)-f(0)=am^2-2am+c-c=am(m-2)<=0,又因为a>0,
则m(m-2)<=0.
则0<=m<=2.

Answer 2

二次函数f(x)在区间[0,1]上单调递减，f(1)<f(0)即f(1)=c-a<f(0)=c所以a>0
f(m)=am^2-2am+c<=f(0)=c，即am^2-2am<=0，a>0
所以am^2-2am<=0等价于m^2-2m<=0，0<=m<=2