求y=(x^2-4x+13)/(x-1) x属于[2,5]的值域

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求y=(x^2-4x+13)/(x-1) x属于[2,5]的值域
求y=(x^2-4x+13)/(x-1) x属于[2,5]的值域

求y=(x^2-4x+13)/(x-1) x属于[2,5]的值域
x=2时,y=(4-8+13)/1=9
x=5时,y=(25-20+13)/4=9/2
y=(x^2-4x+13)/(x-1)
y'=[(2x-4)(x-1)-x^2+4x-13]/(x-1)^2
=(x^2-2x-9)/(x-1)^2
=[(x-1)^2-10]/(x-1)^2
(x-1)^2<10时,函数单调递减,(x-1)^2≥10时,函数单调递增.当(x-1)^2=10时,函数取得最小值.
x=√10+1
ymin=(11+2√10-4-4√10+13)/√10
=(20-2√10)/√10
=2(√10-1)
函数的值域为[2(√10-1),9]

因x²-4x+13=（x-1)²-2(x-1)+10.故y={(x-1)+[10/(x-1)]}-2.可令t=x-1.则由2≤x≤5,===>1≤x-1≤4,即1≤t≤4,且y=[t+(10/t)]-2.由“对勾函数”的单调性可知，y≥(2√10)-2.仅当x=√10时相等。故ymin=(2√10)-2,又ymax=y(1)=9.故值域为[(2√10)-2,9].

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