Finding a summary of the knowledge of sets and functions such as monotonicity, increase and decreas

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The monotonicity of high school math functions can also be called the increase or decrease of functions. When the independent variable of the function f(x) increases (or decreases) within its defined interval, and the value of the function f(x) also increases (or decreases), the function is said to be monotonionic in the interval. Next, I will specifically introduce the high school mathematics knowledge points:

Knowledge of the monotonicity of functions.

High School Mathematics Knowledge: The Motonicity of Functions

In general, let the function f(x) be defined in the domain of i:

If for the values x1 and x2 that belong to any two independent variables on an interval within i, when x1 if for the values x1 and x2 that belong to any two independent variables on an interval within i, when x1 f(x2)Then f(x) is a subtractive function in this interval.

High school Gao Kaihong math knowledge points: monotonic intervals of functions

The monotonic interval refers to the value y of the Sun-headed function of the function in a certain interval, which increases (or decreases) with the increase of the independent variable x. If the function y=f(x) is an increasing or decreasing function in a certain interval. Then let's say that the function y=f(x) has a (strict) monotonicity in this interval, and this interval is called the monotonicity interval of y=f(x).

High School Math Knowledge Points: Monotonic Images of Functions

High School Mathematics Knowledge: Application of Monotonicity of Functions

This question is to test the mastery of quadratic functions! If the coefficient of the quadratic term is less than zero, it means that the opening is downward! Just determine whether the interval you are looking for is to the left or right of the axis of symmetry! Apparently the axis of symmetry x = 1

Problem 1: On x>1, the image of the function is from high to low, which belongs to the subtraction function Problem 2: Find the maximum and minimum values on 2 x 5.

This requires the conclusion of the first question. It is a subtraction function on 1 to positive infinity, and a subtraction function on the range of 2 to 5! Then the maximum value is obtained at x=2 and fmax=f(2)=0

The minimum value of the function is taken at x=5, and fmin=f(5)=-15

Since a is the endpoint of the closed interval, the function f(x)=x can be equal to f(a)=a. Did you notice that this is a power function of the form y=x, and a 1, and a is an odd number, so this is an odd function. y increases with x, but here y decreases over the interval (0,a), does this mean that a is actually a negative number?

We can see that when x takes 0 and 1, it satisfies the power function y=x, but x actually can't take 0, because 0 belongs to the open interval, so the answer comes out, a= 1.

I typed so many words, I hope it can be adopted, dear hehe.

Derivative of f(x) gives f'(x) = 3x 2-1 when f'(x)=3x 2-1<=0, f(x) is monotonically decreasing to get x to belong to [-root number 3 3, root number 3 3].

So a = root number 3 3

First of all, we must know the cubic variance formula a 3-b 3 = (a-b) * (a 2 + ab + b 2).

Beginning to prove, if x1 x2 is taken, then f(x2)-f(x1)=x2 3-x-x1 3+x1=(x2-x1)*(x2 +x1*x2+x1 2)-(x2-x1)=(x2-x1)*(x2 2+x2*x1+x1 2-1) is divided by equality to know that when x2=x1, that is, x 2+x 2+x 2=3x 2=1, there is a turn, at this time x=the root number of thirds is three, so a = the root number of thirds is three.

Special values generally occur at special moments, and special values are equal.

You can use the definition method, compulsory 1 has, see it yourself, you can also ask for guidance, this method is simpler, in elective 2-2, see it yourself.

Take x1>x2>0, so f(x1)-f(x2)=x1-x2+a x1-a x2=x1-x2+a(x2-x1) x1*x2=(x1-x2)(x1*x2-a) x1*x2

So when x1*x2>a, x1*x2>x2*x2 is obviously true when x2 > root number a, so when x is greater than or equal to the root number a, and when it is greater than 0 and less than the root number a, for x<0, the same can be obtained (or obtained from the property of the odd function) in (- root number a), is the increasing function In (- root number a,0) is the subtraction function.

That's an absolute perfect score.

This is the checkmark function.

On (- root a), (root a, + is the increment function.

On (-root, 0), (0, root, a) is the subtraction function.

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

When x>0, f(x)>1 i.e. f(1)≠0 f(0)=1 let x>y>0, then.

f(x)-f(y)=

Do you want to copy the wrong questions? Example: f(x+y)=f(x)f(y)whether f(x+y)=f(x)-f(y) [the middle is subtracted and not multiplied].

2 Conclusion: f(x) = -2x, so there is a maximum value of 6 on [-3,3) and no minimum value.

1) When x is an integer, f(x)=-2x.

First, f(0)=0 is obtained from f(0+0)=2f(0).

From f(0)=f(a-a)=f(a)+f(-a), f is an odd function.

For the positive integer n there is f(na) = f(a) + f(a) +f(a)=nf(a)，f(-na)=-f(na)=-nf(a)。

So for all integers there is f(x)=-xf(-1)=-2x.

2) When x is a rational number, f(x) = -2x.

Let x=p q, p and q are non-zero integers, then qf(x)=f(qx)=f(p)=-2p, so f(x)=-2p q=-2x.

The cauchy method can only prove the rational number, and the next step is to extend to the irrational number with a condition such as "f(x)<0 at x>0".

3) When x is an irrational number, f(x)=-2x.

Suppose there is an irrational number t such that f(t)!=-2t, you may wish to set f(t)<-2t (otherwise -t can be investigated), order.

u=-(f(t)+2t) 2>0, then any rational number w between (t,t+u) satisfies 0-2t-2u=f(t).

Push out f(w-t)=f(w)-f(t)>0, contradictory.

1.Let x1 x2, so x1-x2 0, f(x1)=f[(x1-x2)+x2]= f(x1-x2)xf(x2), according to the known when x>0, f(x)>1, so f(x1-x2) >1 (because x1-x2 0).

So f(x1) f(x2), f(x) are increments on r.

1.Take x=y=0, then f(x+y)=f(0)=f 2(0), f(0)=0 or 1, let f(0)=0, when x>0, f(x+0)=f(x)f(0)=0 contradicts the condition, so f(o)=1;

Let x>0, then f(x+(-x))=f(x)f(-x)=1, so f(-x)=1 f(x)>0; So on r f(x)>0;

Let y be any positive number, then the condition f(x+y) f(x)=f(y)>1, so f(x+y)>f(x); From the arbitrariness of x+y>x, i.e., y, we know that on r, f is an increasing function.

2.f(0)=f(0)+f(0), f(0)=0; f(x-x)=f(x)+f(-x)=0,f(x)=-f(-x);That is, f(x) is an odd function, and only x>0 can be discussed;

For a>b>0, there is f(a)-f(b)=f(a)+f(-b)=f(a-b)<0, that is, f decreases on x>0, so f decreases on r;

f(2)=f(1+1)=2f(1)=-4,f(3)=f(2+1)=f(2)+f(1)=-6;f(-3)=-f(3)=6;

So there is a maximum value of 6 on [-3,3) and no minimum value.

Solution: y=-x 50+162x-21000-1 50[x -50*162x]-21000-1 50[x -8100x+4050 ]+4050 50-21000

1 50 (x-4050) +328050-21000-1 50 (x-4050) +307050 function opening downward, x (0, 4050), monotonically decreasing.

When x=4050, the monthly income of the leasing company is the largest, with a maximum of 307,050 yuan.

The rent is 4,050 yuan, and the monthly income is the largest, which is 963,150 yuan.

That's how it should ......