证明当x>0时,e^x-x>2-cosx
问题描述:
证明当x>0时,e^x-x>2-cosx
答
设f(x)=e∧x-x+cosx-2f'(x)=e∧x-sinx-1再次求导,f“(x)=e∧x-cosx∵x>0,∴e∧x>1,0<cosx<1,∴e∧x-cosx>0,即f“(x)>0∴f'(x)递增,又f'(0)=e∧0-sin0-1=0∴f'(x)>0,∴f(x)在(0,正无穷)递增,...