f(x)=x+1分之2x-1
=2-3/(x+1)
f'(x)=3/(x+1)²>0
函數(shù)單增
故x=2時,f(x)最小=f(2)=2-3/(2+1)=1
x=4時,f(x)最大=f(4)=2-3/(4+1)=2-3/5=7/5
希望能幫到你,祝學(xué)習(xí)進步O(∩_∩)Of(x)的導(dǎo)數(shù)啊，沒學(xué)過嗎？那用概念解設(shè)x1>x2∈[2,4]f(x1)-f(x2)=[2-3/(x1+1)]-[2-3/(x2+1)]=3[1/(x2+1)-1/(x1+1)]=3(x1-x2)/(x1+1)(x2+1)>0所以f(x)是單調(diào)增加故x=2時，f(x)最小=f(2)=2-3/(2+1)=1x=4時，f(x)最大=f(4)=2-3/(4+1)=2-3/5=7/5希望能幫到你，祝學(xué)習(xí)進步O(∩_∩)O