Question

求最值1、y=5+根号2x^2-10x+3 2、y=(x-1)/(x+2) x属于[-1,3]

Answer 1

∵根号(2x^2-10x+3)=根号[2(x-2.5)^2-9.5]≥0

y=5+根号(2x^2-10x+3）≥5,最小值5

y=(x-1)/(x+2) =(x+2-3)/(x+2)=1 - 3/(x+2)

x+2单调增

3/(x+2)单调减

1 - 3/(x+2)单调增

y=(x-1)/(x+2)=1 - 3/(x+2)单调增

x=-1时,最小值1-3/(-1+2)=-2

x=3时,最大值1-3/(3+2)=2/5