Question

设f(x)=2x^2/x+1,求f(x)在x属于[0,1]上的值域

Answer 1

f(x)=2x^2/(x+1)
=(2x^2+2x-2x-2+2)/(x+1)
=2x-2+2/(x+1)
x属于[0,1] , f(x)=2*[x-1+1/(x+1)]=2*[(x+1)+1/(x+1)-2]>=2*[2*√(x+1)*(1/(x+1)-2]=0
x+1=1/(x+1)时最小，即x=0时，f(x)最小
x1>x2>0
f(x1)-f(x2)=2[(x1-x2)+1/(x1+1)-1/(x2+1)]
=2[(x1-x2)+(x2-x1)/[(x1+1)(x2+1)] ]
=2(x1-x2)(1-1/[(x1+1) (x2+1)]>0
f(x)增函数
x=0,f(x)最小值=-2+2=0
x=1,f(x)最大值=2-2+2/3=2/3
值域[0,2/3]

Answer 2

f(x)=2x^2/(x+1)
=(2(x+1)^2-4(x+1)+2)/(x+1)
=2(x+1)+2/(x+1)-4
在定义域内是增函数
值域是f(0)=0, f(1)=1
即[0,1]