求导得:f'(x)=ln(e^x+1)+[xe^x/(e^x+1)]-x=ln(e^x+1)-x/(e^x+1)=[1/(e^x+1)][(e^x)ln(e^x+1)+ln(e^x+1)-ln(e^x)]又因为当x∈[-t,t]时:e^x+1>1>0,又因为ln(e^x+1)-ln(e^x)>0故f'(x)>0恒成立故该函数在[-t,t]上单调增,...