考點(diǎn)1 移項(xiàng)作差構(gòu)造函數(shù)證明不等式——綜合性
已知函數(shù)f(x)=ex-ax(e為自然對(duì)數(shù)的底數(shù),a為常數(shù))的圖象在點(diǎn)(0,1)處的切線斜率為-1.
(1)求a的值及函數(shù)f(x)的極值;
(2)求證:當(dāng)x>0時(shí),x2x.
(1)解:f′(x)=ex-a,因?yàn)?i>f′(0)=-1=1-a,所以a=2,
所以f(x)=ex-2x,f′(x)=ex-2.
令f′(x)=0,解得x=ln 2.
當(dāng)xf′(x)<0,函數(shù)f(x)單調(diào)遞減;
當(dāng)x>ln 2時(shí),f′(x)>0,函數(shù)f(x)單調(diào)遞增.
所以當(dāng)x=ln 2時(shí),函數(shù)f(x)取得極小值,為f(ln 2)=2-2ln 2,無(wú)極大值.
(2)證明:令g(x)=ex-x2,則g′(x)=ex-2x.
由(1)可得g′(x)=f(x)≥f(ln 2)>0,
所以g(x)在R上單調(diào)遞增,
因此,當(dāng)x>0時(shí),g(x)>g(0)=1>0,所以x2x.