M.2a.01. dai cuong ve ham so
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This document discusses functions and their properties. It provides examples of different types of functions, including quadratic, absolute value, and piecewise defined functions. It examines the domains and ranges of functions, and how the graph of a function relates to its equation. Examples are given of determining the value of functions for specific inputs, and how the graph or equation of a function changes when the variable is replaced by its opposite.
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(𝐓𝐋𝐄 𝟏𝟎𝟎) (𝐋𝐞𝐬𝐬𝐨𝐧 𝟏)-𝐏𝐫𝐞𝐥𝐢𝐦𝐬
𝐃𝐢𝐬𝐜𝐮𝐬𝐬 𝐭𝐡𝐞 𝐄𝐏𝐏 𝐂𝐮𝐫𝐫𝐢𝐜𝐮𝐥𝐮𝐦 𝐢𝐧 𝐭𝐡𝐞 𝐏𝐡𝐢𝐥𝐢𝐩𝐩𝐢𝐧𝐞𝐬:
- Understand the goals and objectives of the Edukasyong Pantahanan at Pangkabuhayan (EPP) curriculum, recognizing its importance in fostering practical life skills and values among students. Students will also be able to identify the key components and subjects covered, such as agriculture, home economics, industrial arts, and information and communication technology.
𝐄𝐱𝐩𝐥𝐚𝐢𝐧 𝐭𝐡𝐞 𝐍𝐚𝐭𝐮𝐫𝐞 𝐚𝐧𝐝 𝐒𝐜𝐨𝐩𝐞 𝐨𝐟 𝐚𝐧 𝐄𝐧𝐭𝐫𝐞𝐩𝐫𝐞𝐧𝐞𝐮𝐫:
-Define entrepreneurship, distinguishing it from general business activities by emphasizing its focus on innovation, risk-taking, and value creation. Students will describe the characteristics and traits of successful entrepreneurs, including their roles and responsibilities, and discuss the broader economic and social impacts of entrepreneurial activities on both local and global scales.
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M.2a.01. dai cuong ve ham so
- 1. 1 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog 1 . – D – – hay – x y = f(x) VD1: = HD 3 f(x) 2 x 1 |x| 2 x 1 0 x 1 x 1 x 1 vµ x 2 |x| 2 0 |x| 2 x 2 TOPPER. Chú ý 0 0 1 f(x) f(x) f(x) g(x) g(x) 0
- 2. 2 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog VD2: = HD x 2 f(x) 2x 3 3 x 3 2x 3 0 x 3 x 32 23 x 0 x 3 VD3: = HD – Ta có x2 – 2x + 3 = (x – 1)2 2 1 f(x) x 2x 3 |x| 1 VD4: = HD 0 x m – 1 2x f(x) x m 1 VD5: = HD y x m 1 2x m x m 1 x m 1 0 m 2x m 0 x 2 m 1 0 m 0m 0 2 m – 1 (0 ; 2) m 1 0 m 1 m 1 2 m 3
- 3. 3 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog 2 (b) Tính f(–2), f(–1), , f(1). 3 2 2x 1 khi x < 1 f(x) 1 x khi 1 x 1 2 f( ) 2 y x m 2x m 1 1 (a) (b) (c) (d) 2x 1 y |x| 1 x y |x| 2 x 1 y x 1 y 2x 1 2 x
- 4. 4 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog 2 M(x0 ; y0) (G) x0 D và y0 = f(x0) x y x y O TOPPER. Chú ý 0 ; y0 y0 = f(x0) VD6: 2 ? = HD 2 1 = 02 O VD7 0 ; y0 0 ; y0 y = x2 = 0 ; y0 2 – mx + 2 + m khi ta có: hay2 0 0 0 y x mx 2 m 2 0 0 0 y x 2 m(1 x ) 0 0 2 00 0 1 x 0 x 1 y 3y x 2 TOPPER. Chú ý y y = c O c x 1) = f(x2) x1, x2 K x K (c là
- 5. 5 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog 4 (C): A(–1 ; 1), B(–1 ; –1), C(1 ; –1), D(1 ; 0) 5 0 ; y0 mx 1 y x m
- 6. 6 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog 3 x1, x2 K, x1 > x2 f(x1) > f(x2) x1, x2 K, x1 > x2 f(x1) < f(x2) x1, x2 K, x1 x2, x1, x2 K, x1 x2, 2 1 2 1 f(x ) f(x ) 0 x x 2 1 2 1 f(x ) f(x ) 0 x x VD8 trên = HD 1 và x2 1 x2 ta có: . 2 1 2 1 2 1 2 1 f(x ) f(x ) (2x 3) (2x 3) 2 0 x x x x
- 7. 7 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog 6 y = f(x) = x2 và( ; 1) (1 ; ). 7 và 2 x 1 ( ; 1) (1 ; ). 8 (a) và (b) 3 y x ( ; 0) (0 ; ). y x 1 [1 ; ). ax y f(x) . x 2 (2; ). (2; ).
- 8. 8 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog D thì –x D: VD9: 3 – 4x = HD –x và f(–x) = 2(–x3) – 4(–x) = –2x3 + 4x = –f(x). VD10: = HD [–2 ; 2] thì –x [–2 ; 2] và y f(x) 2 x 2 x 2 x 0 2 x 2 2 x 0 f( x) 2 x 2 x f(x) TOPPER. Chú ý 4 x D thì –x D và f(–x) = f(x) x D thì –x D và f(–x) = – f(x).
- 9. 9 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog A (a) f(x) = –2x + 5 (b) f(x) = –x3 + 2x (c) (d) 3 f(x) x 2 2 f(x) x 2|x| B 3 y f(x) 2x|x | C 3 + (m2 – 1)x2 VD11: = 0 hay x –1 . y x 1 y x 1 [ 1; ) [ 1; ).
- 10. 10 www.tuhoc.edu.vn htttp://tuhoc.edu.vn/blog 1 (a) (b) {–2 ; 2} (c) x [–1 ; ) và x 1 (d) x 1 [ ; 2] 2 2 ; 1] (b) f(–2) = –5; f(–1) = 0; ; f(1) = 0 2 2 f( ) 2 2 3 0 x m x m 0 m 1 2x m 1 0 x 2 m 0 m 1 0 2 4 5 0 ; y0 khi ta có: hay 0 m x0y0 + 1 = m(x0 + y0) x0 khi: mx 1 x m 0 0 0 mx 1 y x m 0 0 0 0 x y my mx 1 0 0 0 0 x y 0 x y 1 0 m 1, m 1 7 và( ; 1) (1; ) 8 [1; ) A B (b) a < 0 (2; ) C m = 1