Question

高中数学急求大神解答？

第三个问怎么求呀？

Answer 1

f(x)=(3x+4)/(x+1) ,g(x) =x^2-2ax+1
y
=f(g(x))
=[3(x^2-2ax+1)+4]/[(x^2-2ax+1)+1 ]
=( 3x^2-6ax+7)/( x^2-2ax +2)
=[( 3x^2-6ax+6) +1]/( x^2-2ax +2)
= 2 +1/( x^2-2ax +2)
= 2 +1/[ (x-a)^2 + 2-a^2 ]
max f(g(x)) = f(g(a))
case 1: a≤0

f(g(x)) = 2 +1/[ (x-a)^2 + 2-a^2 ]
max f(g(x))
=f(g(0))
=2 +1/[ (0-a)^2 + 2-a^2 ]
=2 + 1/2
=5/2
min f(g(x))
=f(g(1))
=2 +1/[ (1-a)^2 + 2-a^2 ]
=2 + [ 1/ (3-2a )]
case 2: 0<a<1

max f(g(x))
=f(g(a))
=2 +1/(2-a^2)
f(g(1)) =2 + [ 1/ (3-2a )]
f(g(0)) =5/2
min f(g(x)) = min { f(g(1)) , f(g(0)) }
case 3: a≥1
max f(g(x)) =f(g(1)) =2 + [ 1/ (3-2a )]
min f(g(x)) =f(g(0)) =5/2