一道高中数学函数题

Question

已知函数f(x)=x^2+ax-lnx,a属于r.(1)若函数f(x)在[1,2]上是减函数,求实数a的取值范围.(2)令g(x)=f(x)-x^2,是否存在实数a,当x属于(0,e](e是自然常数)时,函数g(x)的最小值是3,若存在,求出a的值;若不存在,说明理由.(3)当x属于(0,e]时,证明e^2*x^2-5/2>(x+1)lnx. 急!过程请写明,必将追加分!手机打上来的很不容易,大家拜托了!

Answer 1

1. f'(x) = 2x+a+(-1/x)
=> 当x属于[1,2]时，f'(x)是增函数
=> f'(1)<=f'(x)<=f'(2)
=> a+1 <= f'(x) <= a+ 7/2
函数f(x)在[1,2]上是减函数
=> 当x属于[1,2]时,f'(x)<=0
=> a+ 7/2 <=0
=> a≤-7/2

2. g(x)=f(x)-x^2 = ax-lnx
=> g'(x)=a-(1/x)
=> 当x属于(0,e]时，g'(x)是增函数
=> g'(x)<=g'(e) = a- (1/e)
=> (1)、当a<=e时,g'(x)<=0
=> g(x)在(0,e]上是减函数
=> g(x)>=g(e)=ae-1 ,函数g(x)的最小值是3
=> ae-1 = 3
=> a = 4/e
(2)、当a>e时： 当x∈(0,1/a] =>g(x)是减函数
当x∈(1/a,e] =>g(x)是增函数
=> g(x)>=g(1/a)=1-ln(1/a) =3
=> ln(1/a) = -2
=> a = e^2

3.当x属于(0,e]时,设 h(x) = e^2*x^2-5/2 - (x+1)lnx
根据单调性最值来做

Answer 2

1.对f(x)求导，可得f`(x)=2x+a-(1/x),若f(x)在[1,2]上为减，
则f`(x)在[1,2]上小于等于0，而f`(x)在[1,2]上为增，
所以只要在x=2处f`(x)≤0即4+a-(1/2)≤0,可解：a≤-7/2
(f`(x)在给定区域上为增，∵y=2x与y=-1/x都是增）

Answer 3

1.对f(x)求导，可得f`(x)=2x+a-(1/x),若f(x)在[1,2]上为减，
则f`(x)在[1,2]上小于等于0，而f`(x)在[1,2]上为增，
所以只要在x=2处f`(x)≤0即4+a-(1/2)≤0,可解：a≤-7/2
(f`(x)在给定区域上为增，∵y=2x与y=-1/x都是增） 1. f'(x) = 2x+a+(-1/x)
=> 当x属于[1,2]时，f'(x)是增函数
=> f'(1)<=f'(x)<=f'(2)
=> a+1 <= f'(x) <= a+ 7/2
函数f(x)在[1,2]上是减函数
=> 当x属于[1,2]时,f'(x)<=0
=> a+ 7/2 <=0
=> a≤-7/2

2. g(x)=f(x)-x^2 = ax-lnx
=> g'(x)=a-(1/x)
=> 当x属于(0,e]时，g'(x)是增函数
=> g'(x)<=g'(e) = a- (1/e)
=> (1)、当a<=e时,g'(x)<=0
=> g(x)在(0,e]上是减函数
=> g(x)>=g(e)=ae-1 ,函数g(x)的最小值是3
=> ae-1 = 3
=> a = 4/e
(2)、当a>e时： 当x∈(0,1/a] =>g(x)是减函数
当x∈(1/a,e] =>g(x)是增函数
=> g(x)>=g(1/a)=1-ln(1/a) =3
=> ln(1/a) = -2
=> a = e^2

3.当x属于(0,e]时,设 h(x) = e^2*x^2-5/2 - (x+1)lnx
根据单调性最值来做

Answer 4

好不好用导数的？