已知函数f(x)=e^(x-m)-ln(2x)当m＜=2时,证明f(x)>-ln2

f(x)=e^(x-m)-ln(2x)>=e^(x-2)-ln(2x),即证e^(x-2)-ln(2x)>-ln2,即e^(x-2)>ln(2x)-ln2=1nx,令g(x)=e^(x-2)-1nx,g'(x)=e^(x-2)-1/x,显然g'(x)递增,设g'(x)=e^(x-2)-1/x=0的根为x1,则g(x)在（0,x1)递减,在（x1,+8)递增...