"Basic Properties of Functions" PPT courseware on the concepts and properties of functions (the concept of parity of functions in Lesson 3)

"Basic Properties of Functions" PPT courseware on the concepts and properties of functions (the concept of parity of functions in Lesson 3)

Part One: Learning Objectives

1. Understand the definitions of odd functions and even functions.

2. Understand the characteristics of the graphs of odd and even functions.

3. Master the method of judging the parity of functions.

core competencies

1. Use the characteristics of odd (even) functions to cultivate intuitive imagination.

2. Cultivate logical reasoning skills with the help of the method of judging whether functions are odd or even.

Basic properties of functions PPT, part 2: independent preview and exploration of new knowledge

parity of function

parity even function odd function

Condition: Let the domain of function f(x) be I. If ∀x∈I, both have -x∈I

Conclusion f(-x)=f(x) f(-x)=-f(x)

Image Features Symmetry about ____ Symmetry about ____

Thinking: What are the characteristics of the domain of a function with parity and evenness?

Tip: The domain is symmetrical about the origin.

First try

1. Which of the following functions is an even function ()

A． y=x

B． y=2x2-3

C. y=1x

D. y＝x2，x∈[0,1]

2. The function represented by the following graph has parity and evenness ()

3. Function y=f(x), x∈[-1, a](a>-1) is an odd function, then a is equal to ()

A． -1B． 0

C. 1 D． unconfirmed

4. If f(x) is an even function on R, and f(2)=3, then f(-2)=________.

Basic properties of functions PPT, the third part: cooperative exploration to improve literacy

Judgment of function parity

[Example 1] Determine the parity of the following functions:

(1)f(x)=x3+x;

(2)f(x)=1-x2+x2-1;

(3)f(x)=2x2+2xx+1;

(4)f(x)＝x－1，x<0，0，x＝0，x＋1，x>0.

[Solution] (1) The domain of the function is R and is symmetric about the origin.

And f(-x)=(-x)3+(-x)=-(x3+x)=-f(x),

Therefore the function f(x) is an odd function.

(2) From 1-x2≥0, x2-1≥0, x2=1, that is, x=±1.

Therefore, the domain of the function is {-1,1}, which is symmetric about the origin.

And f(1)=f(-1)=-f(-1)=0, so f(x) is both an odd function and an even function.

(3) The domain of function f(x) is (-∞, -1) ∪ (-1, +∞),

It is not symmetric about the origin, so f(x) is neither an odd function nor an even function.

regular method

Two methods to determine the parity of a function

(1)Definition method:

(2)Image method:

Track training

1. Among the following functions, which one is even? (fill in serial number)

①f(x)＝x3; ②f(x)＝|x|＋1; ③f(x)＝1x2;

④f (x) = x + 1x; ⑤ f (x) = x2, x∈-1,2]．

②③　[For ①, f(-x)=-x3=-f(x), it is an odd function;

For ②, f(-x)=|-x|+1=|x|+1, it is an even function;

For ③, the domain is {x|x≠0}, which is symmetrical about the origin, f(-x)=1-x2=1x2=f(x), then it is an even function;

For ④, the domain is {x|x≠0}, which is symmetrical about the origin, f(-x)=-x-1x=-f(x), then it is an odd function;

For ⑤, the domain is [-1,2], it is not symmetrical about the origin, and it does not have parity or evenness, so it is a non-odd and non-even function. ]

Graph problem of odd and even functions

[Example 2] It is known that the domain of the odd function f(x) is [-5,5], and the image on the interval [0,5] is as shown in the figure.

(1) Draw the image on the interval [-5,0];

(2) Write the set of values ​​of x that make f(x)<0.

[ Solution

From the image of y=f(x) on [0,5], we can know its image on [-5,0], as shown in the figure.

(2) From the image, we know that the value set of x that makes the function value y<0 is (-2, 0) ∪ (2, 5).

regular method

Solving problems using the graphics of odd and even functions skillfully

1 Basis: The graph of odd function ⇔ is symmetrical about the origin, and the graph of even function ⇔ is symmetrical about the y-axis.

2 Solution: Based on the symmetry of the graphs of odd and even functions, problems such as finding function values ​​or drawing graphs of odd and even functions can be solved.

PPT on the basic properties of functions, part 4: Achieving standards in class and solidifying the bases

1. Thinking and analysis

(1) The function f(x)=x2, x∈[0,+∞) is an even function. ()

(2) For the function y=f(x), if there is x such that f(-x)=-f(x), then the function y=f(x) must be an odd function. ()

(3) There is no function that is both an odd function and an even function. ()

(4) If the domain of a function is symmetric about the origin, then the function is either an odd function or an even function. ()

2. Function f(x)＝|x|＋1 is ()

A． odd function

B． even function

C. It is both an odd function and an even function

D. non-odd and non-even functions

3. It is known that the function f(x)=ax2+2x is an odd function, then the real number a=______.

4. It is known that the function y=f(x) is an even function defined on R, and when x≤0, f(x)=x2+2x. The graph of the function f(x) on the left side of the y-axis is now drawn, as the picture shows.

(1) Please complete the graph of the complete function y=f(x);

(2) Write the increasing interval of function y=f(x) based on the image;

(3) Based on the image, write the set of x values ​​that make f(x)<0.

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For more information about the PPT courseware "The Concept and Properties of Functions, the Concept of Parity of Functions, and the Basic Properties of Functions", please click the "Concepts and Properties of Functions" ppt "The Concept of Parity of Functions" ppt "Basic Properties of Functions" ppt tag.

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