Logarithmic functions of high school mathematics

The first big question: 1Solution: If the function makes sense, then there is 1+x≠0, (1-x) (1+x) 0Solution: -1 x 1That is, the domain is defined as.

2.The domain of the function is defined as , so with respect to the origin symmetry.

then f(-x)=lg[(1+x) (1-x)= -lg[(1-x) (1+x)=-f(x)

The function is odd.

3.Set -1 x1 x2 1

f（x1）-f（x2）

lg[(1-x1)/(1+x1)]-lg[(1-x2)/(1+x2)]=lg[（1-x1)(1+x2)/(1+x1)(1-x2)]

1-x1 0,1+x2 0,1+x1 0,1-x2 0lg[(1-x1)(1+x2) (1+x1)(1-x2)] 0, i.e. f(x1) f(x2).

So the function is a subtraction function.

The second question: What is the base number of logsquare2 x-2log2 x?

1, 1+x<>0 so x<>-1

2. (1-x) (1+x)>0 So 1-x>0 and 1+x>0 so -1x>1 has no solution.

So this problem is solved as -1

Substituting x=-1 gives -2=1-(lg a +2)+lg b to get 1=lg a-lg b then, lg a b =1, so a=0

So f(x)=x +3x+lg b=(x+ b, because when x belongs to r, f(x)) 2x is constant, so lg b-9 4=2 is solved, and lg b=17 4

So f(x)=x +3x+17 4

The minimum value is 2 at this point.

First, substitute f(-1)=1-lg a-2+lgb=lgb-lg a-1=-2

i.e. lgb-lg a=-1

When x belongs to r, f(x) 2x is constant.

f(x)=x +(lg a +2)x+lgb 2x, i.e. x +lg a x+lgb 0

Think of it as a quadratic function, then δ 0

i.e. δ = (LGA) -4LGB 0

lgb-lg a=-1 has been found

Substituting the above inequality, we can solve for lga=2 so a is 100

Then LGB=1

Then substitute these two values into f(x)=x +(lg a +2)x+lgb, i.e., f(x)=x 4x, 1=(x-2) 3, and when x is 2, f(x) has a minimum value of -3

So the real number a is 100 and f(x) has a minimum value of -3

f(x)=f(x)-2x=x +lga x+lgb 0 is constant, then δ=(lga) -4lgb 0

f(-1)=-2, then lgb-lga=-1, which is brought into the above equation to get (lga) -4lga+4 0

So LGA=2, A=100, LGB=1, B=10F(x)=X, 4X, 1=(X-2), 3 -3, and the minimum value is -3

Analytical: y=to the 16th power.

When y=, to the power of 16 =

1 + x) to the power of 16 = 167 137

1+x x = so x

A (1,3 2) can be thought of as a composite function of y=logau and u=3-ax.

The bottom a is greater than zero, so u=3-ax is a subtraction function, so it is necessary to subtract the original letter leakage orange number before the [0,2] signal.

y=logau must be an increasing function, so the bottom a>1 is also a true number.

u=3-ax is all positive on [0,2], and only the minimum value needs to be positive, and it is a return group subtraction function.

The minimum value is u(2)=3-2a>0

A<3 2

Composite 2 available.

a (1,3 2), I don't know if you understand.

Because y=loga(3-ax) is a subtractive function on the lead [0,2], 0 "pure a<1, and because 3-ax is greater than 0, a belongs to [0,1].

Select b to draw the artwork to know.

Because loga(x) decreases early on x>1, the land bucket is 0 so 0

x>-1 2 is the definition domain.

A logarithmic function is an increasing function within the defined domain.

So the base is greater than 1

a-1>1

a>2

Solution: There is a problem to know: x+2y>0;x-4y>0；x>0；y>0 yields: x y>4

The original formula is simplified as: LG(x+2y)(x-4y)=LG2xySo:(x+2y)(x-4y)=2xy

Simplify: x 2-4xy-8y 2=0 Divide by y 2 to get: (x y) 2-4x y-8=0, solution: x y=2+2 3, or x y=2-2 3

Because x y>4, x y = 2 + 2 3

Solution: LG(x+2y)+LG(x-4y)=LG2+LGX+LGYLG(x+2y)(x-4y)=LG2xy

x+2y)(x-4y)=2xy

x^2-4xy+2xy-8y^2=2xy

x^2-8y^2-4xy=0

Divide by y 2 on both sides at the same time

x/y)^2-4(x/y)-8=0

x/y)-2)^2=12

x y) = root number 12 + 2

x and y are greater than 0, and the answer is 4+2 root number 6

The general formula y=ax2+bx+c

Vertex y=a(x+h)2+k

The general formula of the Sun family formula y=a(x+2a b)2+4ac-b2 4ah=2a b k=4ac-b2 4a The top shouting slag point is (-2a b,4ac-b2 4a).

No thanks: The 2 after the letter is what is the square of what.