求函数f(x)=x+1分之2x-1在区间{2,4}上的最大值和最小值

f(x)=x+1分之2x-1
=2-3/(x+1)
f'(x)=3/(x+1)²>0
函数单增
故x=2时,f(x)最小=f(2)=2-3/(2+1)=1
x=4时,f(x)最大=f(4)=2-3/(4+1)=2-3/5=7/5
希望能帮到你,祝学习进步O(∩_∩)Of(x)的导数啊，没学过吗？那用概念解设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)是单调增加故x=2时，f(x)最小=f(2)=2-3/(2+1)=1x=4时，f(x)最大=f(4)=2-3/(4+1)=2-3/5=7/5希望能帮到你，祝学习进步O(∩_∩)O