The known function f x loga 1 x loga x 3 0 a 1

x)=loga[(1-x)(x+3)]=0=loga(1)then (1-x)(x+3)=1

x^2-2x+3=1

x^2+2x-2=0

Defined by domains, 1-x>0, x+3>0

3=-4=loga(a^-4)

So (1-x)(x+3)<=a -4

1-x)(x+3)=-x 2-2x+3=-(x+1) 2+4, so the true maximum value = 4

So a (-4) = 4

a=4^(-1/4)=√2/2

1.-3＜x＜1 2.Let f(x)=0 give x=plus/minus(change sign 3)-1 3

According to the meaning, f(x)=loga(1-x)(x+3) -4 so we get lg(1-x)(x+3) (lga) -4, because 0 a 1 so lga 0 so multiply lga to the right to get loga(1-x)(x+3) -4lga and wait until (1-x)(x+3) 1 (a) to the fourth power, because to make (1-x) (x+3) 1 (a to the fourth power) only need 1 (a to the fourth power) greater than or equal to (1-x)(x+3) The maximum value of the solution so that 1 (a to the fourth power) 3 pushes out 0 a 1 (a quarter of 3).

So the solution of question (3) is 0 a 1 (a quarter of 3).

(1) The defined domain is: 1-x>0, x+3>0

That is, there is -3(2)f(x)=loga(1-x)+loga(x+3)=0loga(1-x)*(x+3)=0

1-x)(x+3)=1

x+3-x^2-3x=1

x^2+2x=2

x+1)^2=3

x+1 = (+ -) root number 3

So the zero point is x=-1(+-) root number 3

(1) f(x)=loga(1-x)+loga(x+3), for logarithmic functions, as long as the true number is greater than 0

That's 1-x>0

x+3>0

So the solution is x<1, x>-3

is -3, so the domain is defined as (-3,1).

2) Find the zero point, then let f(x)=0

That is, loga(1-x) + loga(x+3)=0 according to the nature.

loga(1-x)+loga(x+3)=loga(1-x)(x+3)=0

Again, 0=loga1

And so there is. loga(1-x)(x+3)=loga1

And so it is. 1-x)(x+3)=1

Simplification is. x²+2x+1=3

That is, (x+1) =3

So x=-1 - root number 3 or x = -1 + root number 3

It is also necessary to satisfy the definition domain that x belongs to (-3,1).

Apparently all satisfied.

So x=-1 - root number 3 or x = -1 + root number 3

1-x>0 and x+3>0

Then -3(2) finds the zero point of the function f(x).

That is, when f(x)=0, the solution (1-x)(x+3)=1 only needs to solve the value of x.

Solution: (1) To make the function meaningful: then there is, and the solution gets: -3 x 1, then the domain of the function is defined as: (-3,1).

2) The function can be reduced to f(x)=loga(1-x)(x+3)=loga(-x2-2x+3).

From f(x)=0, we get -x2-2x+3=1, i.e., x2+2x-2=0, and the zero point of the function f(x) is.

3) The function can be reduced to:

f(x)=loga(1-x)(x+3)=loga(-x2-2x+3)=loga[-(x+1)2+4].

3 x 1, 0 -(x+1)2+4 4,0 a 1, loga[-(x+1)2+4] loga4, i.e. f(x)mim=loga4, the plexus is loga4=-4, a=2 of the root number of the half.

1. Define the domain: (-3,1) Finger wheel analysis: 1-x>0; x+3>02, x1=-1+3 x2=-1-3 under the root:

loga(1-x)+loga(x+3)=0, loga(1-x)=-loga(x+3), 1-x=1 (x+3), and then solve the one-dimensional quadratic equation.

256 Parse disturbance: f(x)=loga(1-x)(x+3), (1-x)(x+3) increases first and then decreases at (-3,1), and then decreases.

0 a 1, so f(x) first decreases and then increases, and slowly teases the mu f(-2)=-4, and then solves.

Solution: (1)f(x)=loga(x+1)-loga(1-x), then (2) knows from (1) that the domain of f(x) is , and f(-x)=loga(-x+1)-loga(1+x)=-loga(x+1)-loga(1-x)]=f(x), so f(x) is an odd function

3) Because f(x) is an increment function in the defined domain when a 1, so f(x) 0 x+11-x 1

0 x 1

So the range of x that makes f(x) 0 is

(1) x+1>0, and 1-x>0

At 11, loga(x+1) is single-increased, -loga(-x+1) is single-increased, and f(x) is single-increased.

and f(x) > f(0).

x>0

In this case, the value range of x is as follows:

Solution: (1) The function f(x)=loga(x+1)-loga(1-x), x+1 01-x 0, we get -1 x 1, and the domain of the function is (-1,1).

2) Since the domain of the function f(x) is (-1,1), symmetrical with respect to the origin, and f(-x)=loga(-x+1)-loga(1+x)=-f(x), f(x) is an odd function

3) When a 1, the function f(x)loga(x+1)-loga(1-x)=loga1+x1-x is the increasing function, from f(x) 0, we can get 1+x1-x 1, that is, x-2x-1 0, i.e. (x-2)(x-1) 0, 1 x 2

When 0 a 1, the function f(x)loga(x+1)-loga(1-x)=loga1+x1-x is the subtraction function, which is obtained by f(x) 0.

0 1+x1-x 1, i.e., 1+x1-x 01+x1-x 1, i.e., -1 x 1x 0 or x 1, to find -1 x 0

x [0,2], the function f(x) is always meaningful, i.e., f(x)>0, 3-ax>0 , ax<3, and always holds.

x∈[0,2], ax∈[0,2a]

The maximum value of ax is 2a<3 and a<3 2

A>0 and canvas A≠1,01

And when x=1, the sedan shirt is destroyed t=2-a>0,a<2

1I hope it can help you, please ask if you don't know anything.

f(x) =log[a, a x-1 ] is meaningful, requires: a x - 1 > 0 i.e. a x >1

Discussion:1a>1, obtained by a x >1, x>0, defines the domain (0,+

f '(x) =a x a x - 1) >0, function single increment;

2.01 gets, x<0, defines the domain (-0)f'(x) =a x a x - 1) >0, function single increment;

1-x> 0

1+x>0

1 The domain of the function f(x) = f(x) + g(x): -1logaloga

g[(b+c)/(1+bc)]

It is known that the functions f (-1) (x) = loga(x-1) (x+1) (a>0, a is not equal to 1), x (-1) (1, +

3) Let g(x)=1+logax, when [m,n] is really contained in (1,+ m, find the range of values of a?

Solution: (3) Because m the range of 0 and f (-1) (x) in [m,n] is [g(n),g(m)], so f (-1)(m)=g(m), so 1-2 (m+1)=am, a=(m-1) [m(m+1)]=(m-1) [(m-1) 2+3(m-1)+2]=1 [(m-1)+2 (m-1)+3] 3-2 2, so 0

The range of f(x) in [m,n] is [g(n),g(m)], which means that f(x) is a subtraction function, so a is between 0 and 1.