Question

求函数f（x）=lgx/x在区间[1/2,1 ]的值域

Answer 1

易知,(lgx)′=1/(x㏑10).∴对函数f(x)=(lgx)/x求导得f'(x)=[(1/㏑10)-lgx]/x².∵1/2≤x≤1,===>lg(1/2)≤lgx≤lg1.===>-lg2≤lgx≤0.===>0≤-lgx≤lg2.===>0＜1/㏑10≤(1/㏑10)-lgx≤(1/㏑10)+lg2.即在[1/2,1]上,有f'(x)＞0,∴在[1/2,1]上,函数f(x)=(lgx)/x是增函数,∴f(1/2)≤f(x)≤f(1).即-2lg2≤f(x)≤0,∴值域为[-2lg2,0].