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设a>0为常数,函数f(x)=x^(1/2)-ln(x+a)求a=3/4时,函数f(x)的极大,
人气:377 ℃ 时间:2019-10-19 22:47:09
解答
f(x)=√x-ln(x+3/4)
保证根号有意义及真数大于0,有x≥0,x+3/4>0,联立解得x≥0
对f(x)求导得
f’(x)=(1/2)√x-1/(x+3/4)
令f’(x)≥0 以求原函数的增区间,得(1/2)√x-1/(x+3/4)≥0,整理得
(x+3/4-2√x)/[2(x+3/4)*2√x] ≥0
x+3/4-2√x≥0
(√x)²-2√x+3/4≥0
(√x)²-2√x+3/4≥0
(2√x-3)*(√x-1)≥0
0≤x<1或x>3/2
令f’(x)≥0,以求原函数的增区间,得(1/2)√x-1/(x+3/4)≥0,整理得
(x+3/4-2√x)/[2(x+3/4)*2√x] ≥0
x+3/4-2√x≥0
(√x)²-2√x+3/4≥0
(√x)²-2√x+3/4≥0
(2√x-3)*(√x-1)≥0
0≤x≤1或x≤3/2
同理令f’(x)≤0,以求原函数的减区间,得(1/2)√x-1/(x+3/4)≤0,整理得
1≤x≤3/2
所以
f(x)在x=1时有极大值,极大值为f(1)=√1-ln(1+3/4)=1-ln(7/4)
f(x)在x=3/2时有极小值,极小值为f(3/2)=√(3/2)-ln(3/2+3/4)=√(3/2)-ln(9/4)
=√6/2-2ln(3/2)
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