Knowing the function f x lnx a 2 x 2 ax a R . 1 If the function f x is in the interval

Solution: (1): When a=0, <>

f(x) is an increasing function on the interval (1,+, which is not in place;

When a≠0, to make the function f(x) on the interval (1,+ is a subtraction function, just <> in the interval (1,+ is constant, x>0, as long as <> holds, <>

The solution is <>

Or <> sum up, the range of values of the real number a is <>

2) Function <>

The defined domain is (0,+.)

When a=0, <>

The increase interval of f(x) is (0,+ f(x) has no extremum;

When a>0, the order <>

<> or <> rounded), f(x) is <>

The minus interval is <>

So at this point f(x) has a maximum value of <>

No minima; When a>0, the order <>

<> to be discarded) or <>

f(x) increases by an interval of <>

The minus interval is <>

1) f(x)= 1x+2x-a,x 0, which is known to be a 1x+2x, x 0, since 1x+2x 2 1x 2x=22, so a 22

2) It is known that f(x)=0 has a zero point of the traversal type in (0,+, i.e., 2x2

ax+1=0 has the zero point of the traversal type in (0,+, and denotes g(x)=2x2

ax+1, since g(0)=0, so a2?8 0a4 0, solution a 22

Let the two extreme points of f(x) be x1x2, then x1

x2 a2，x1

x2 12，∴f（x1

f（x2=（lnx1x1

ax1+（lnx2x2

ax2lnx1

x2a（x1

x2+（x1x2

2x1x2ln 12- a22+ a24-1=- a24-1+ln 12 -3+ln 12, so the sum of all extrema is less than -3+ln 12;

3) Let a=3, then f(x)=lnx+x2

3x，x＞1，f′（x）= 2x3?3x+1x= (x?1)(2x?1) x 0, i.e. f(x) is an increasing function on (1,+, so f(x) f(1)=-2, i.e., lnx+x2

3x＞-2，3x-x2

lnx+2，3（a1

a2…+an-（a1

a2…+an

ln（（a1a2an

2n=ln（n+1）+2n

f'(x)=[1 x*x-(1+lnx)*1] (x 2)=-(lnx) (x 2), let f'(x)=0, there is lnx=0, so the extreme copy point bai is x=0 in the interval (a, a+1 3), so there is.

The problem of dua<0 is generally a few steps: find the defined domain, find the derivative, make the derivative value 0, find the extreme point, and discuss the root of the equation in the previous step.

f'(x)=(1-1-lnx) x 2=-lnx x 2=0, the maximum point: x=1, which must be in the interval (a,a+1 3).

Thus there is a<1: 2 3

In fact, it is easy to know that a<=1 2 is the only way to satisfy the problem, because if a > 1 2, when x tends to positive infinity. There is always f(x)=(a-1 2)x2+lnx>(a-1 2)x2

And limx + a-1 2)x2 2ax=+ so the question is not satisfied.

Let g(x)=f(x)-2ax=(a-1 2)x2+lnx-2ax,x>1

The derivative of g(x) dy dx=[(2a-1)x-1](x-1) x if a=1 2. Apparently easy to know that dy dx "0 is true for all x>1. Therefore, we know that the function g(x) decreases monotonically on (1,+), so g(x)1 is true, so a=1 2 is in line with the problem.

If the root of the derivative of a<1 is x1=1 (2a-1)<0, x2=1 is obviously x1=-1 2

To sum up, the range of values for a is [-1 2,1 2].

a=1f(x)=lnx+x^2/2

f'(x)=1/x+x

1, e] on f'(x)>0

f(x) single increase.

f(1)=1 2 minimum.

f(e)=1+e2 2 maximum.

g(x)=lnx+(a-1/2)x^2-2axg'(x)=1/x+2(a-1/2)x-2ag'(x)=0

1/x-x=0

x=1 takes x=1

g(1)=-1/2-a<0

a>-1/2

f'(x)=[1-(1+lnx)] x 2=-lnx x 2 by f'(x)=0: x=1 is the extreme point.

From the title, there is a<1: 1 2

No. The minimum value of the length of i (1-k) [1+(1-k)] is a function of k, and k (0,1) is the domain that defines it, which has been given, and k cannot take 0.

(1) Let f(x)=ax-(1+a)x = -x[(1+a)x-a]=0, get x=0 or x=a (1+a) because a 0,-(1+a) 0 so i= its length is a (1+a)(2) length a(1+a)=1 (a+1 a) 1 [2 (a·1 a)]=1 2 When a=1 a, i.e., a=1, the maximum length is 1 2, so a (1+a) increases monotonically at (0,1), At (1,+ monotonically decreases, and when k (0,1), 1-k 1 and 1+k 1, so a (1+a ) takes a minimum at a=1-k or a=1+k. and (1-k) [1+(1-k) ]1+k) [1+(1+k) ]= -3k 0 So the minimum value of the length of i is (1-k) [1+(1-k)].

First question:

f′(x)=(a+1/a)/x-1/x²-1=-(x-a)(x-1/a)/x²

Let f(x)=0 give x=a or x=1 a

Because a>1

Then 0<1 a<1 is easy to obtain monotonically increasing monotonically on the f(x) interval (0,1 a) (0,1 a).

Let's give a good review first, I'm working on the second question!

According to the meaning of the problem, the domain of f(x) is (0,+

When a=0, f(x)=2lnx+1x, f(x)=2x-1x2=2x-1x2

Order f'(x)=0, the solution is x=12

When 0 x 12, f'（x）＜0；When x 12, f'（x）＞0．

and f(12)=2-2ln2, so the minimum value of f(x) is 2-2ln2, and there is no maximum

2）f′(x)=2-ax-1x2+2a=2ax2+(2-a)x-1x2．

Order f'(x)=0, the solution is x1=-1a, x2=12

If a 0, let f'(x) 0, resulting in 0 x 12;Order f'(x) 0, get x 12

If a 0, when a -2, -1a 12, let f'(x) 0, 0 x -1a or x 12;

Order f'(x) 0, resulting in -1a x 12

When a=-2, f(x)=-(2x-1)2x2 0

When -2 a 0, -1a 12 is obtained, and f is obtained'(x) 0, resulting in 0 x 12 or x -1a; Order f'(x) 0, resulting in 12 x -1a

In summary, when a 0, the decreasing interval of f(x) is (0,12) and the increasing interval is (12,+).

When a -2, the decreasing interval for f(x) is (0,-1a) and (12,+ the increasing interval is (-1a,12).

When a=-2, f(x) decreases in the interval (0,+).

When -2 a 0, the decreasing interval of f(x) is (0,12), and the (1a,+ increasing interval is (12,-1a).

(1): Extremum 2in2+1 2

2): Calculate the difficult ones below, do it yourself.

Categorize A.

When a>0

When a>0