The concepts of monotonicity and parity of functions

Parity
1. Definition
In general, for the function f (x)
(1) If f (- x) = - f (x) exists for any X in the function domain, then f (x) is called an odd function
(2) If f (- x) = f (x) for any X in the function domain, then f (x) is called an even function
(3) If f (- x) = - f (x) and f (- x) = f (x) hold simultaneously for any X in the domain of function definition, then f (x) is both odd and even, which is called odd and even function
(4) If f (- x) = - f (x) and f (- x) = f (x) can not hold for any X in the domain of function definition, then f (x) is neither odd nor even, which is called non odd non even function
Note: 1. The odd and even properties are the whole properties of a function
② The domain of odd and even functions must be symmetric about the origin. If the domain of a function is not symmetric about the origin, then the function must not be odd (or even)
(analysis: to judge the parity of a function, the first step is to check whether its domain of definition is symmetrical about the origin, and then strictly follow the definition of parity, simplify, sort out, and compare with F (x) to get a conclusion.)
③ The basis of judging or proving whether a function has parity is the definition
2. The characteristics of odd even function image
Theorem the graph of odd function is centrosymmetric with respect to origin, and the graph of even function is centrosymmetric with respect to y axis or axis
If f (x) is an odd function, the image of F (x) is symmetric about the origin
Point (x, y) → (- x, - y)
If an odd function increases monotonically in a certain interval, it also increases monotonically in its symmetric interval
If an even function increases monotonically in a certain interval, it decreases monotonically in its symmetric interval
Monotonicity:
In general, Let f (x) be defined as I:
If for the values X1 and X2 of any two independent variables belonging to an interval in I, when X1 and X2 have f (x1) < f (x2), then f (x) is an increasing function in this interval
If for the values X1 and X2 of any two independent variables belonging to an interval in I, if x1f (x2), then f (x) is a decreasing function in this interval
If the function y = f (x) is an increasing or decreasing function in a certain interval, then it is said that y = f (x) has (strict) monotonicity in this interval. This interval is called the monotone interval of y = f (x). In the monotone interval, the image of increasing function is ascending, and that of decreasing function is descending
Note: (1) the monotonicity of function is also called the increase and decrease of function;
(2) Monotonicity of function is a local concept for a certain interval;
(3) The method steps to determine the monotonicity of function in a certain interval are as follows:
a. Let x1, X2 ∈ a given interval, and x1

The concept of mathematical function in senior one
Are the two functions f (x) = √ (X & # 178; - 1) and G (x) = √ (x + 1) ·√ (x-1) equal

Unequal
Two functions should be equal to meet two conditions: 1, the same analytic function, 2, the same domain
The analytic expressions of these two functions are the same (can be changed into the same), but we need to find the domain of definition
The first: x ^ 2-1 ≥ 0, X ≥ 1 or X ≤ - 1
The second: x + 1 ≥ 0 and X-1 ≥ 0, the solution is x ≥ 1,
So the definition is different from, so the two functions are different

On the non differentiable points of functions
What are the non differentiable points of function f (x)? Besides the discontinuous points and some continuous but non differentiable points of function f (x), do they also include the meaningless points in the corresponding derivative?
I'm talking about meaningless points just like you said, but there's an example
f(x) = (x - 4)( x + 1)^(2/3)
f'(x) = 5(x - 1)/[3(x + 1)^(1/3)]
Where x = - 1 is the non differentiable point, but this seems to be meaningful in the original function. When it comes to the derivative function, it is meaningless and non differentiable. What's the matter?

In short, the point where the derivative does not exist
I don't know what kind of "pointless in derivative" you mean
Is this the case
For example, f (x) = LNX, f '(x) = 1 / X. then x = 0 is the meaningless point in F' (x)
In fact, f (x) = LNX itself has no definition at x = 0. Of course, we can't find a derivative at this point
In the derivative function, x = - 1 is meaningless. To know whether it can be derived at - 1, we have to use the definition
1. First of all, it is easy to obtain that the function is continuous at x = - 1, which is a necessary condition for derivability
2. Find the left derivative
f'((-1)-)=lim{x->(-1)-}[f(x)-f(-1)]/[x-(-1)]
=lim{x->(-1)-}[(x - 4)( x + 1)^(2/3)]/(x+1)
=lim{x-

What are differentiable functions and non differentiable functions? What are the conditions?

Let y = f (x) be a function of one variable. If y has a derivative y '= f' (x) at x = x [0], then y is said to be differentiable at x = x [0]
Conditions: 1) if f (x) is continuous at x0, then when a tends to 0, [f (x + a) - f (x)] / A has a limit, then f (x) is said to be differentiable at x0
(2) If M and f (m) are derivable at any point on the interval (a, b), then f (x) is said to be derivable on (a, b)
Discontinuous functions are certainly not differentiable
The other is that although the function is continuous, the left derivative and the right derivative at a certain point are not equal. For the problems of the left derivative and the right derivative, please refer to the University's "mathematical analysis"

1. The image of the function f (x) and its derivative f '(x) defined on R are continuous curves, and for the real number a, B (A0, f' (b) f (b);
(3) Existence X. belongs to [a, b], f (X.) f (a); (4) existence X. belongs to [a, b], f (a) - f (b) > F '(X.) (a-b)
The number of correct conclusions is () a.1 B.2 C.3 D.4
1. The image of the function f (x) and its derivative f '(x) defined on R are continuous curves, and for the real number a, B (A0, f' (b) f (b);
(3) There is X. It belongs to [a, b], f (x). ）> = f (a); (4) there exists X. It belongs to [a, b], f (a) - f (b) > F '(x). )(a-b)
The number of correct conclusions is () a.1 B.2 C.3 D.4

The image of function f (x) and its derivative f '(x) defined on R are continuous curves, and for real numbers a and B (a < b), f' (a) > 0, f '(b) < 0, which means that there is x0 in the interval (a, B), so that f' (x0) = 0,
So the function f (x) has a maximum point in the interval (a, b). At the same time, it shows that the function has at least one increasing interval and one decreasing interval in the interval [a, b]
From the above analysis, we can see that the function f (x) does not necessarily have zero point in the interval [a, b], so ① is not correct;
Because the function has the maximum point in the interval (a, b), and the abscissa of the maximum point in the same subtraction interval with the real number B is the existence of an x0, so ② is correct;
If f (b) > F (a), x0 = a, then ③ is incorrect;
When f (a) > F (b), and x0 is the abscissa of the maximum point, the conclusion is correct
So choose B

Function image if the function image is above the X axis, then the function value must be zero_ On the contrary, if the function image is below the x-axis,
If the function image is above the X axis, then the function value must be zero_ On the contrary, if the function image is below the x-axis, then the function value must be zero_ .
If the function image is on the left side of the y-axis, then the value of the independent variable is certain_ On the contrary, if the function image is on the right side of the y-axis, then the value of the independent variable must be constant___ .

If the function image is above the X axis, then the function value must be zero_ Greater than zero__ On the contrary, if the function image is below the x-axis, then the function value must be zero_ Less than zero__ .
If the function image is on the left side of the y-axis, then the value of the independent variable is certain_ Less than zero__ On the contrary, if the function image is on the right side of the y-axis, then the value of the independent variable must be constant_ Greater than zero__ .

Draw the image y = - 0.5x-1, according to the image: 1. The intersection coordinate of function image and X axis 2. The function image is above X axis, the value of X
Range 3. The value range of X when the function image is below X

1,（-2,0）
2,X-2

For a linear function y = 3x + 2, when x (), the function image is above the X axis

For the linear function y = 3x + 2, when x (> 2 / 3), the function image is above the x-axis

Given that the image of the function y = X2 - (m-2) x + m passes through a point (- 1,15), let its image and X-axis intersect at points a and B, point C is on the image, and s △ ABC = 1, the coordinates of point C are obtained

Since the image of the function y = X2 - (m-2) x + m passes through the point (- 1,15), there is 1 + (m-2) + M = 15, and the solution is m = 8; therefore, the analytical formula of the parabola is y = x2-6x + 8, ｜ a (2,0), B (4,0) (let a point be on the left side of B point), so AB = 2, and s △ ABC = 12ab · | YC | = 1, and the solution is | YC | = 1

The function f (x) = Tan ω x (ω ＞ 0) has two adjacent tangent lines y = π 4. If the length of the line is π 4, then the value of F (π 4) is ()
A. 0B. 1C. -1D. π4

The length of the line segment obtained from the two adjacent tangent lines y = π 4 of the function image is π 4. The period of the function f (x) is π 4. From π ω = π 4, we get ω = 4. So we choose a

The concepts of monotonicity and parity of functions