We specify that when t x t 1, y f x is maximum

Solution: y x 2-x-1, then.

y＇＝2x-1＝2（x-1／2）。

When 0 x 1 2, y 0, y x 2-x-1 is a subtractive function;

When 1 2 x 2, y 0, y x 2-x-1 is the increment function;

When x 1 2, the y minimum value is -5 4;

When x 2, y max is 1

Because y=x -x-1=(, so under the condition of 0 x 2, when x=, the value of y is the smallest, is, when x=2, the value of y is the largest, is 1

This proposition is wrong. If the equal sign cannot be obtained, m is not the minimum value of the sum f.

For example, if the minimum value is n, so that m=n-1, f(x)>=m is true, but m is not the minimum value of f

The above discussion is still in the case of the existence of a minimum value, and it is possible that the minimum value does not exist, such as y=1 x, i=(0, positive infinity).

f(x)>=0, but there is no minimum caution in the absence of filial piety.

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

1+x^2≥1

So when x=0, f(x)=1 (1+x 2) and the maximum value is 1f(x)=1 (1+x 2), there is no minimum value.

y=x+1/(x-1)=1+(x-1+1/(x-1))=1-(1-x+1/(1-x));

x＜1;1-x＞0;

1-x+1/(1-x)≥bai2;

(1-x+1/(1-x))≤2;

y=1-(1-x+1/(1-x))≤1-2=-1;

Maximum value = -1;du

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y=x+[1/(x-1)]

y-1=(x-1)+[1/(x-1)]

y-1=-{-x-1)-[1 (x-1)] comes from [-(x-1)]*1 (x-1)]=-2 if and only if -(x-1)=-1 (x-1), i.e., x=0, the equal sign is y-1 -2

y≤-1

y=-x²-x+1

(x²+x)+1

(x+1/2)²+5/4

1 x 2 decreases monotonically in this interval.

So: when x=1, there is a maximum value of =-9 4+5 4=-1, and when x=2, there is a minimum value of =-25 4+5 4=-5

The vertices are in x=-1 2 and the function opening is facing down, so.

The maximum value is taken at 1 and is -1

The minimum value is taken at 2 and is -5

Answer: f(x)=2x (x +1), x> 0 numerator and denominator of Zen base are in x together.

f(x)=2/(x+1/x)

Because: g(x)=x+1 x>=2 (x*1 x)=2 is blinded by: 0

Solution: Because x 0

So -x 0

So y=(1+x) x=1 x+x=-[(1 x)+(x)] 2 [(1 x)*(x)]=-2

That is, the maximum value of y=(1+x) x is -2

When x1>0 there is: 1 x1+x1 2 so -1 x1-x1 -2, let -x1=x

then x<0, and:

1 x1-x1=1 x+x=(1+x) x=y -2, so x 0, y=(1+x) x max is -2

1.When y=1, f(x)=f(x)+f(1), so f(1)=0; Right;

2.When y=1 x, f(1)=f(x)+f(1 x)=0, so f(x)=-f(1 x), false;

3.When y=1 z, f(x z)=f(x)+f(1 z)=f(x)-f(z), so f(x y)=f(x)-f(y), right;

4.When y=x(n-1), f(x n)=f(x)+f[x(n-1)]=2f(x)+f[x(n-2)]=......=nf(x), so right;

5.Push can't be made, wrong;

Let x=y=1, get f(1)=0, right.

If you add f(1) to the right of the equation, then f(1 x) = f(x) + f(1) = f(x 1) = f(x), which is obviously incorrect.

Shift, f(x y) + f(y) = f(x y y) = f(x) pair.

Let y=x, substitute f(x)=2f(x), after replacing x with x, i.e., f(x)=f(x)+f(x)=3f(x), and so on f(x n)=nf(x) pair.

In contrast, it is clearly wrong.

f(2x) = 2f(x) is also incorrect.

The right one has