证明下列不等式:
(1)当x>1时,ex>ex;
(2)当x>1时,2√x>3-1/x.
【正确答案】:(1)令f(x)=ex-ex,则f(x)的定义域为(-∞,+∞),而
f'(x)=ex-e,
令f''(x)=0,得x=1.
当x>1时,f'(x)>0,故f(x)在(1,+∞)上单调增加,故有
f(x)>f(1),即ex-ex>0,
故ex>ex.
(2)令f(x)=2√x+1/x-3,则f(x)的定义域为(0,+∞),在(0,+∞)上,
f'(x)=1/√x-1/x2=x2-√x/x2√x,
当x>1时,f'(x)=√x(x√x-1)/x2√x>0,故f(x)在(1,+∞)上为增函数,因此有
f(x)=2√3-(3-1/x)>f(1)=0,
即2√x>3-1/x.
证明下列不等式: (1)当x>1时,ex>ex; (2)当x>1时,2√x>3-1/x.
- 2024-11-07 09:11:28
- 高等数学(经管类)(13125)