Western Normal University Edition First Grade Mathematics Volume 1
People's Education Press First Grade Mathematics Volume 1
Beijing Normal University Edition Seventh Grade Mathematics Volume 1
People's Education Press Third Grade Mathematics Volume 1
People's Education Press Second Grade Mathematics Volume 1
Beijing Normal University Edition Fifth Grade Mathematics Volume 1
Hebei Education Edition Third Grade Mathematics Volume 1
Hebei Education Edition Seventh Grade Mathematics Volume 2
Beijing Normal University Edition Seventh Grade Mathematics Volume 2
Qingdao Edition Seventh Grade Mathematics Volume 1
People's Education High School Mathematics Edition B Compulsory Course 2
Beijing Normal University Edition Eighth Grade Mathematics Volume 1
Qingdao Edition Seventh Grade Mathematics Volume 2
Qingdao Edition Second Grade Mathematics Volume 2
Jiangsu Education Edition Fourth Grade Mathematics Volume 1
People's Education Press First Grade Mathematics Volume 2
Category | Format | Size |
---|---|---|
People's Education High School Mathematics Edition A Compulsory Course 1 | pptx | 6 MB |
Description
"End of Chapter Integration" Concept and Properties of Functions PPT
Part One: Topic 1: Finding the Range of a Function
Example 1 Find the range of the following functions:
(1)y=(5x"-" 1)/(4x+2);(2)y=(x^2 "-" 4x+3)/(2x^2 "-" x"-" 1);
(3)y=(2x^2+4x"-" 7)/(x^2+2x+3);(4)y=2x-√(x"-" 1).
Solution: (1) (Solved with the help of the characteristics of the inverse proportional function)
y=(5x"-" 1)/(4x+2)=(5/4 "(" 4x+2")-" 1"-" 5/2)/(4x+2)=(5/4 " (" 4x+2")-" 7/2)/(4x+2)=5/4-7/(2"(" 4x+2")" ).
∵7/(2"(" 4x+2")" )≠0,∴y≠5/4.
So the value range of the function is {y"∈" R├|y≠5/4┤}.
(2)∵y=(x^2 "-" 4x+3)/(2x^2 "-" x"-" 1)=("(" x"-" 1")(" x"-" 3 ")" )/("(" x"-" 1")(" 2x+1")" )=(x"-" 3)/(2x+1)(x≠1),
And (x"-" 3)/(2x+1)=(1/2 "(" 2x+1")-" 7/2)/(2x+1)=1/2-7/(2"( " 2x+1")" ).
When x=1, the original formula y=(1"-" 3)/(2×1+1)=-2/3.
The value range of the ∴ function is {y"∈" R├|y≠1/2 "and" y≠"-" 3/2┤}.
(3) (Convert into a quadratic equation about x, and then use the discriminant to evaluate the domain)
It is known that the functional formula can be transformed into: yx2+2yx+3y=2x2+4x-7.
(y-2)x2+2(y-2)x+3y+7=0,
When y≠2, consider the above equation as a quadratic equation about x.
∵x∈R,∴Δ≥0, that is, [2(y-2)]2-4(y-2)(3y+7)≥0.
The solution is -9/2≤y<2.
When y=2, 3×2+7≠0, ∴y≠2.
The value range of the ∴ function is -9/2,2.
(4) Let √(x"-" 1)=t, then t≥0, x=t2+1.
∴y=2(t2+1)-t=2t2-t+2=2 t-1/4 2+15/8.
∵t≥0,∴y≥15/8. The value range of ∴ function y=2x-√(x"-" 1) is 15/8,+∞.
Integrate PPT at the end of the chapter, the second part of the content: Topic 2: Using the monotonicity of the function to find the maximum value of the function
Example 2 Let a be a real number, function f(x)=x2+x-a+1, x∈R.
(1) Discuss the parity of function f(x);
(2) Find the minimum value of f(x).
Solution: (1) When a=0, function f(-x)=(-x)2+|-x|+1=f(x), at this time f(x) is an even function .
When a≠0, f(a)=a2+1,f(-a)=a2+2|a|+1,f(-a)≠f(a),f(-a) ≠-f(a).
At this time, the function f(x) is neither an odd function nor an even function.
(2)①When x≤a, function f(x)=x2-x+a+1= x-1/2 2+a+3/4.
If a≤1/2, then the function f(x) decreases monotonically on (-∞,a], so the minimum value of the function f(x) on (-∞,a] is f(a )=a2+1.
If a>1/2, then the minimum value of function f(x) on (-∞,a] is f 1/2 =3/4+a, and f 1/2 ②When x≥a, function f(x)=x2+x-a+1= x+1/2 2-a+3/4. If a≤-1/2, then the minimum value of function f(x) on [a,+∞) is f -1/2 =3/4-a, and f -1/2 ≤f( a). If a>-1/2, then the function f(x) increases monotonically on [a,+∞), so the minimum value of the function f(x) on [a,+∞) is f (a)=a2+1. To sum up, when a≤-1/2, the minimum value of function f(x) is 3/4-a; When -1/2 When a>1/2, the minimum value of function f(x) is a+3/4. Methods and techniques The basic idea of solving problems with parameters is to classify and discuss. The key is to determine the standard of discussion, which requires no repetition or omission. The standard for discussion of parity in this question is zero and non-zero parameters, which correspond to even functions and non-odd functions respectively. Even functions; the discussion standards for maximum and minimum values are relatively complicated and can be seen as two types of standards. One is the zero point of the absolute value (zero point knowledge will be learned in Chapter 4), and the other is the symmetry axis of the parabola and the corresponding interval. Position usually requires the help of the image of the function. Integrate PPT at the end of the chapter, the third part: Topic 3 Application of parity of functions Example 3 If the odd function y=f(x) is a decreasing function defined on [-1,1], and f(1-a)+f(1-a2)>0, find the Ranges. Solution: According to the properties of odd functions, -f(1-a2)=f(a2-1), that is, f(1-a)+f(1-a2)>0 is equivalent to f(1-a)>f (a2-1), And because f(x) is a decreasing function defined on [-1,1], So {■("-" 1≤1"-" a≤1"," @"-" 1≤a^2 "-" 1≤1"," @1"-" a Methods and techniques: Use the property that f(x) is an odd function and a decreasing function, remove f, and equivalently transform the inequality set of a. Variation Training 3 If f(x) is an even function defined on the real number set R and is an increasing function on the interval (-∞,0), and f(2a2+a+1) Solution: Method 1: ∀x1,x2∈(0,+∞), and x1 Because f(x) is an increasing function on the interval (-∞,0), So f(-x1)>f(-x2). And because f(x) is an even function, f(x1)>f(x2), So f(x) is a decreasing function on (0,+∞), Because 2a2+a+1=2 a2+1/2a +1=2 a+1/4 2+7/8,3a2-2a+1=3 a-1/3 2+2/3, So 2a2+a+1 and 3a2-2a+1 are two positive numbers, Therefore, f(2a2+a+1) Method 2: Same as Method 1, it is judged that 2a2+a+1 and 3a2-2a+1 are two positive numbers, then -(2a2+a+1)<0 and -(3a2-2a+1)<0.
File Info
Update Time: 2024-09-07
This template belongs to Mathematics courseware People's Education High School Mathematics Edition A Compulsory Course 1 industry PPT template
"End of Chapter Integration" Concept and Properties of Functions PPT Simple campus recruitment activity planning plan summary enterprise and institution recruitment publicity lecture PPT template is a general PPT template for business post competition provided by the manuscript PPT, simple campus recruitment activity planning plan summary enterprise and institution recruitment promotion Lecture PPT template, you can edit and modify the text and pictures in the source file by downloading the source file. If you want more exquisite business PPT templates, you can come to grid resource. Doug resource PPT, massive PPT template slide material download, we only make high-quality PPT templates!
Tips: If you open the template and feel that it is not suitable for all your needs, you can search for related content "End of Chapter Integration" Concept and Properties of Functions PPT is enough.
How to use the Windows system template
Directly decompress the file and use it with office or wps
How to use the Mac system template
Directly decompress the file and use it Office or wps can be used
Related reading
For more detailed PPT-related tutorials and font tutorials, you can view: Click to see
How to create a high-quality technological sense PPT? 4 ways to share the bottom of the box
Notice
Do not download in WeChat, Zhihu, QQ, built-in browsers, please use mobile browsers to download! If you are a mobile phone user, please download it on your computer!
1. The manuscript PPT is only for study and reference, please delete it 24 hours after downloading.
2. If the resource involves your legitimate rights and interests, delete it immediately.
3. Contact information: service@daogebangong.com
"End of Chapter Integration" Concept and Properties of Functions PPT, due to usage restrictions, it is only for personal study and reference use. For commercial use, please go to the relevant official website for authorization.
(Personal non-commercial use refers to the use of this font to complete the display of personal works, including but not limited to the design of personal papers, resumes, etc.)