MAT 111 Functions Lecture Notes PDF

Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 1 Functions_Part 1 Dr. Sameh Basha ...

Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 1 Functions_Part 1 Dr. Sameh Basha  Standard Sets of Numbers Natural numbers: ℕ = 1,2,3, … , Integer numbers: ℤ = … , −3, −2, −1,0,1,2,3, … 𝑝 Rational numbers: ℚ = { :𝑝 ∈ ℤ 𝑎𝑛𝑑 𝑞 ∈ ℕ} 𝑞 Real numbers: ℝ 2  Notes: Every natural number n in ℕ also will be integer number in ℤ Natural numbers: ℕ = 1,2,3, … , Integer numbers: ℤ = … , −3, −2, −1,0,1,2,3, … Notice that every integer n is also a rational number, since we can write it as the quotient of two integers: n = n/1. 3 4 Note There exist some number not in ℚ and exist in the real line ℝ (i.e., 𝑝 can not represent in the form ) 𝑞 as 𝜋, 2, 5, … Which are called irrational numbers (ℝ\ℚ). ℝ = ℚ ∪ (ℝ\ℚ) 5 Equations and Inequalities Equations and inequalities are both mathematical sentences formed by relating two expressions to each other. In an equation, the two expressions are deemed equal which is shown by the symbol =. Ex: x=y. Ex: 3𝑥 + 5 = 11 In an inequality, the two expressions are not necessarily equal which is indicated by the symbols: >, y, x, 𝒃 (ii) if 𝑎 > 𝑏 and b > 𝑐 then 𝐚 > 𝒄 (iii) if 𝑎 > 𝑏 and 𝑐 ∈ ℝ then 𝒂 + 𝒄 > 𝒃 + 𝒄 (iv) if 𝑎 > 𝑏 and 𝑐 ∈ ℝ then 𝒂 − 𝒄 > 𝒃 − 𝒄 (v) if 𝑎 > 𝑏 and 𝑐 > 0 then 𝒂𝒄 > 𝒃𝒄 (vi) if 𝑎 > 𝑏 and 𝑐 < 0 then 𝒂𝒄 < 𝒃𝒄 (vii) if 𝑎 > 𝑏 >0 then 𝟏 < 𝟏 𝒂 𝒃 9 Absolute Value Functions The distance from number 3 to zero is: 3 The distance from number -3 to zero is: 3 The absolute value of a number is the distance from the number to zero. 𝒙, 𝒙 ≥ 𝟎 𝒙 =ቊ −𝒙, 𝒙 < 𝟎 Ex: 𝟑 =𝟑, −𝟑 = 𝟑, −𝟕 = 𝟕 10 Note: 𝒙𝟐 = ±𝒙 𝒙, 𝒙 ≥ 𝟎 𝒙 =ቊ −𝒙, 𝒙 < 𝟎 𝒙𝟐 = 𝒙 11  𝟑 = 𝟑 , −𝟑 = 𝟑 𝟐 𝟐 (𝟑) = 𝟗 , (−𝟑) = 𝟗 12  Inequalities Properties If a,b,c, and d are real numbers then: (viii) if 𝑎2 ≥ 𝑏 2 then 𝑎 ≥ 𝑏 (ix) if 𝑎 > 𝑏 >0 then 𝒂𝟐 > 𝒃𝟐 (x) if 0 > 𝑎 > 𝑏 then 𝒂𝟐 < 𝒃𝟐 (xi) if 𝑎 > 𝑏 >0 then 𝒂 > 𝒃 (xii) if 0 > 𝑎 > 𝑏 then 𝒂 < 𝒃 13 Absolute Value Functions The distance from number 3 to zero is: 3 The distance from number -3 to zero is: 3 The absolute value of a number is the distance from the number to zero. 𝒙, 𝒙 ≥ 𝟎 𝒙 =ቊ −𝒙, 𝒙 < 𝟎 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈, 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 ≥ 𝟎 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 = ቊ −(𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈), 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 < 𝟎 14  𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈, 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 ≥ 𝟎 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 = ቊ −(𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈), 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 < 𝟎 Ex: 𝟐𝒙 − 𝟑, 𝟐𝒙 − 𝟑 ≥ 𝟎 𝟐𝒙 − 𝟑 = ቊ −(𝟐𝒙 − 𝟑), 𝟐𝒙 − 𝟑 < 𝟎 𝟐𝒙 − 𝟑, 𝟐𝒙 ≥ 𝟑 =ቊ −(𝟐𝒙 − 𝟑), 𝟐𝒙 < 𝟑 𝟑 𝟑 𝟐𝒙 − 𝟑, 𝒙≥ 𝟐𝒙 − 𝟑, 𝒙≥ = 𝟐= 𝟐 𝟑 𝟑 𝟑 − 𝟐𝒙, 𝒙< −(𝟐𝒙 − 𝟑), 𝒙< 𝟐 𝟐 15 Theorem Let 𝑎 > 0, then 𝒙 ≤𝒂 ⟹ −𝒂≤𝒙≤𝒂 𝒙 > 𝒂 ⟹ 𝒙 > 𝒂 𝑶𝑹 𝒙 < −𝒂 16  Ex: Solve the following inequalities: 2𝑥 − 3 < 5 Sol: 2𝑥 − 3 < 5 ⇒ −5 < 2𝑥 − 3 < 5 ⇒ −5 + 3 < 2𝑥 < 5 + 3 ⇒ −2 < 2𝑥 < 8 −2 8 ⇒ 7 Sol: 2𝑥 + 3 > 7 ⇒ 2𝑥 + 3 > 7 OR 2𝑥 + 3 < −7 ⇒ 2𝑥 > 4 OR 2𝑥 < −10 ⇒𝑥>2 OR 𝑥 < −5 S.s= −∞, −5 ∪ 2, ∞ =ℝ − −5,2 20 Theorem Let 𝑎 > 0, then 𝒙 ≤𝒂 ⟹ −𝒂 ≤𝒙≤𝒂 𝒙 > 𝒂 ⟹ 𝒙 > 𝒂 𝑶𝑹 𝒙 < −𝒂 Note: 𝒙𝟐 = 𝒙 (𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈)𝟐 = 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 21 Example: Solve the following inequalities: 1- 2𝑥 − 1 < 5 2- 5𝑥 + 2 ≥ 3 3- 2𝑥 + 3 ≤ 4 4- 3 𝑥 2 − 1 < 7 22 Function 23 Function Overview: Functions are fundamental to the study of calculus. In this chapter we review what functions are and how they are visualized as graphs, how they are combined and transformed, and ways they can be classified. Functions are a tool for describing the real world in mathematical terms. A function can be represented by an equation (algebraically), a graph, a numerical table, or a verbal description. 24 Definition 25 Function A function is a rule that assign to each element 𝑥 in a set 𝐷 exactly one element 𝑦 in a set 𝐸. 26 Function A function is a rule that assign to each element 𝑥 in a set 𝐷 exactly one element 𝑦 in a set 𝐸. 27 Function D E D E 1 1 1 1 4 2 2 4 5 3 3 5 8 4 4 9 Not Function Not Function D E D E 1 1 1 1 2 4 4 2 3 5 5 4 4 7 Function Function 28 Function A function is a rule that assign to each element 𝑥 in a set 𝐷 exactly one element 𝑦 in a set 𝐸. 29 Function D E Domain f 1 1 Co-domain f = Df 2 4 3 5 4 8 30 Function D E Domain f 1 1 Co-domain f = Df 4 2 = 𝟏, 𝟒, 𝟓, 𝟖, 𝟏𝟎 5 3 = 𝟏, 𝟐, 𝟑 , 𝟒 8 4 10 Range f = Rf = 𝟏 , 𝟒, 𝟓 , 𝟖 31  Function D E D E Domain f = Df = 𝟏, 𝟐, 𝟑, 𝟒 Domain f = Df = 𝟏, 𝟐, 𝟑, 𝟒 1 1 1 1 4 2 4 2 Co-domain f = 𝟏, 𝟒, 𝟓, 𝟖 Co-domain f = 𝟏, 𝟒, 𝟓, 𝟖, 𝟏𝟐 5 3 5 3 8 4 8 4 Range f = Rf = 𝟏, 𝟒, 𝟓, 𝟖 Range f = Rf = 𝟏, 𝟒, 𝟓, 𝟖 12 Function Function D E D E Domain f = Df = 𝟏, 𝟐, 𝟑, 𝟒 1 1 1 1 4 Co-domain f = 2 2 𝟏, 𝟒, 𝟓 4 5 3 3 5 8 4 4 Range f = Rf = 𝟏, 𝟒, 𝟓 9 Function Not Function 32 Function Co-domain f = 𝒇 𝒙 = 𝒙𝟐 + 𝟐 𝟏, 𝟑, 𝟒, 𝟓, 𝟔, 𝟏𝟏, 𝟏𝟖, 𝟐𝟎 1 3 Domain f = Df = 𝟏, 𝟐, 𝟑, 𝟒 4 Range f = Rf = 𝟑, 𝟔, 𝟏𝟏, 𝟏𝟖 1 5 2 6 3 11 4 18 20 33 Function 𝟏 Generally, 𝒇 𝒙 = Domain f = Df=ℝ − {𝟐} 𝒙−𝟐 −𝟏 Domain f = Df = 𝟎, 𝟏, 𝟑, 𝟒 0 𝟐 1 3 -1 4 1 2 𝟏 𝟐 34 Domain function Generally, Domain the function 𝑓(𝑥) is the set of all possible values of the input 𝑥 that acceptable by the function 𝑓(𝑥). (i.e., that can be processed by the function 𝑓(𝑥) ). 𝐷𝑓 = ℝ − {𝑝𝑜𝑖𝑛𝑡𝑠 𝑡ℎ𝑎𝑡 𝑚𝑎𝑘𝑒 𝑎 𝑝𝑟𝑜𝑏𝑙𝑒𝑚 (𝑁𝑜𝑡 𝑎𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒 𝑏𝑦 𝑓(𝑥))} Points that make the Points that make the denominator = zero value inside the square root is negative 35 Function 𝟏 𝒇 𝒙 = 𝒙−𝟐 Domain f = Df=ℝ − {𝟐} 36 Example: Find 𝐷𝑓 1- 𝑓 𝑥 = 4 − 𝑥 2 2- 𝑓 𝑥 = 𝑥 2 − 9 1 3- f(x)= 𝑥−4 1 4- 𝑓 𝑥 = 𝑥 2 +2 5- 𝑓 𝑥 = 𝑥2 + 2 37 Example: Find 𝐷𝑓 1- 𝑓 𝑥 = 4 − 𝑥 2 Answer: The function 𝑓(𝑥) to be defined should: 4 − 𝑥2 ≥ 0 ⟹ 4 ≥ 𝑥2 ⟹ 𝑥 2≤ 4 ⟹ 𝑥 ≤2 ⟹ −2 ≤ 𝑥 ≤ 2 𝐷𝑓 = −2,2 38 Example: Find 𝐷𝑓 2- 𝑓 𝑥 = 𝑥 2 − 9 Answer: The function 𝑓(𝑥) to be defined should: 𝑥2 − 9 ≥ 0 ⇒ 𝑥 2≥ 9 ⇒ 𝑥 ≥3 ⇒ 𝑥 ≥ 3 𝑂𝑅 𝑥 ≤ −3 𝐷𝑓 = −∞, −3 ∪ 3, ∞ = ℝ − −3,3 39 Example: Find 𝐷𝑓 1 3- f(x)= Draft: 𝑥−4 Answer: 𝒙−𝟒=𝟎 ⇒𝒙=𝟒 𝐷𝑓 = ℝ − 4 40 Example: Find 𝐷𝑓 1 4- 𝑓 𝑥 = Draft: 𝑥 2 +2 𝒙𝟐 + 𝟐 = 𝟎 Answer: ⇒ 𝒙𝟐 = −𝟐 ∵ 𝑥 2 + 2 ≠ 0, ∀𝑥 ∈ ℝ Impossible ∴ 𝐷𝑓 = ℝ 41 Example: Find 𝐷𝑓 5- 𝑓 𝑥 = 𝑥 2 + 2 Answer: 𝐷𝑓 = ℝ, Since it is a polynomial 42 Example: Find Df and Rf: 1- 𝑓 𝑥 = 9 − 𝑥 2 2- 𝑓 𝑥 = 𝑥 2 − 4 3- f(x)=1 − 3 9 − 𝑥 2 4- f(x)=1 − 3 2𝑥 2 − 1 5- 𝑓 𝑥 = 4− 𝑥 1 6- 𝑓 𝑥 = 𝑥−4 1 7- 𝑓 𝑥 = 4+𝑥 2 43 Example: Find Df and Rf: 1- 𝑓 𝑥 = 9 − 𝑥 2 To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 Answer: ⟹ −3 ≤ 𝑥 ≤ 3 The function 𝑓(𝑥) to be defined ⟹ 0 ≤ 𝑥2 ≤ 9 should: ⟹ 0 ≥ −𝑥 2 ≥ −9 9 − 𝑥2 ≥ 0 ⟹ −9 ≤ −𝑥 2 ≤ 0 ⟹ 9 ≥ 𝑥2 ⟹ 0 ≤ 9 − 𝑥2 ≤ 9 ⟹ 𝑥 2≤ 9 ⟹ 0 ≤ 9 − 𝑥2 ≤ 3 ⟹ 𝑥 ≤3 ⟹ 0 ≤ 𝑓(𝑥) ≤ 3 ⟹ −3 ≤ 𝑥 ≤ 3 ⟹ 𝑅𝑓 = 0,3 ⟹ 𝐷𝑓 = −3,3 44 Note: If −𝟑 ≤ 𝒙 ≤ 𝟑 ⟹ 𝟗 ≤ 𝒙𝟐 ≤ 𝟗 If −𝟑 ≤ 𝒙 ≤ 𝟑 ⟹ 𝟎 ≤ 𝒙𝟐 ≤ 𝟗 45 Note: If −𝟑 ≤ 𝒙 ≤ 𝟑 ⟹𝟑≤ 𝒙 ≤𝟑 If −𝟑 ≤ 𝒙 ≤ 𝟑 ⟹𝟎≤ 𝒙 ≤𝟑 46 Example: Find Df and Rf: 1- 𝑓 𝑥 = 9 − 𝑥 2 𝐷𝑓 = −3,3 𝑅𝑓 = 0,3 47 Example: Find Df and Rf: 2- 𝑓 𝑥 = 𝑥 2 − 4 To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 Answer: ⟹ 𝑥 ∈ ℝ − −2,2 The function 𝑓(𝑥) to be defined ⟹ 𝑥 ≥2 should: ⟹ 𝑥2 ≥ 4 𝑥2 − 4 ≥ 0 ⟹ 𝑥2 − 4 ≥ 0 ⟹ 𝑥2 ≥ 4 ⟹ 𝑥2 − 4 ≥ 0 ⟹ 𝑥 ≥2 ⟹ 𝑓(𝑥) ≥ 0 ⟹ 𝑥 ≥ 2 𝑂𝑅 𝑥 ≤ −2 ⟹ 𝑅𝑓 = 0, ∞ ⟹ 𝐷𝑓 = −∞, −2 ∪ 2, ∞ ⟹ 𝐷𝑓 = ℝ − −2,2 48 Example: Find Df and Rf: 2- 𝑓 𝑥 = 𝑥 2 − 4 𝐷𝑓 = −∞, −2 ∪ 2, ∞ ⟹ 𝐷𝑓 = ℝ − −2,2 𝑅𝑓 = 0, ∞ 49 Example: Find Df and Rf: To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 3- f(x)=1 − 3 9 − 𝑥 2 ⟹ −3 ≤ 𝑥 ≤ 3 Answer: ⟹ 0 ≤ 𝑥2 ≤ 9 ⟹ 0 ≥ −𝑥 2 ≥ −9 The function 𝑓(𝑥) to be defined should: ⟹ −9 ≤ −𝑥 2 ≤ 0 ⟹ 0 ≤ 9 − 𝑥2 ≤ 9 9 − 𝑥2 ≥ 0 ⟹ 0 ≤ 9 − 𝑥2 ≤ 3 ⟹ 9 ≥ 𝑥2 ⟹ 0 ≤ 3 9 − 𝑥2 ≤ 9 ⟹ 𝑥2 ≤ 9 ⟹ 0 ≥ −3 9 − 𝑥 2 ≥ −9 ⟹ −9 ≤ −3 9 − 𝑥 2 ≤ 0 ⟹ 𝑥 ≤3 ⟹ −8 ≤ 1 − 3 9 − 𝑥 2 ≤ 1 ⟹ −3 ≤ 𝑥 ≤ 3 ⟹ −8 ≤ 𝑓(𝑥) ≤ 1 ⟹ 𝐷𝑓 = −3,3 ⟹ 𝑅𝑓 = −8,1 50 Example: Find Df and Rf: 3- f(x)=1 − 3 9 − 𝑥 2 𝐷𝑓 = −3,3 𝑅𝑓 = −8,1 51 Example: Find Df and Rf: To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 4- f(x)=1 − 3 2𝑥 2 −1 1 1 Answer: ⟹𝑥 ∈ℝ− − , 2 2 The function 𝑓(𝑥) to be defined should: ⟹ 𝑥 ≥ 1 2 2𝑥 2 − 1 ≥ 0 1 ⟹ 𝑥2 ≥ ⟹ 2𝑥 2 ≥1 2 1 ⟹ 2𝑥 2 ≥ 1 ⟹ 𝑥2≥ 2 1 ⟹ 2𝑥 2 − 1 ≥ 0 ⟹ 𝑥 ≥ 2 ⟹ 2𝑥 2 − 1 ≥ 0 1 1 ⟹ 𝑥≥ 𝑂𝑅 𝑥 ≤− ⟹ 2𝑥 2 − 1 ≥ 0 2 2 1 1 ⟹ 𝐷𝑓 = −∞, − ∪ ,∞ ⟹ −3 2𝑥 2 − 1 ≤ 0 2 2 1 1 ⟹ 1 − 3 2𝑥 2 − 1 ≤ 1 ⟹ 𝐷𝑓 = ℝ − − , 2 2 ⟹ 𝑓(𝑥) ≤ 1 ⟹ 𝑅𝑓 = −∞, 1 52 Example: Find Df and Rf: 4- f(x)=1 − 3 2𝑥 2 − 1 1 1 𝐷𝑓 = ℝ − − , 2 2 𝑅𝑓 = −∞, 1 53 Example: Find Df and Rf: 5- 𝑓 𝑥 = 4 − 𝑥 To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 Answer: ⟹ −4 ≤ 𝑥 ≤ 4 The function 𝑓(𝑥) to be defined ⟹0≤ 𝑥 ≤4 should: ⟹ 0 ≥ − 𝑥 ≥ −4 ⟹4− 𝑥 ≥0 ⟹ −4 ≤ − 𝑥 ≤ 0 ⟹4≥ 𝑥 ⟹0≤4− 𝑥 ≤4 ⟹ 𝑥 ≤4 ⟹0≤ 4− 𝑥 ≤2 ⟹ −4 ≤ 𝑥 ≤ 4 ⟹ 0 ≤ 𝑓(𝑥) ≤ 2 ⟹ 𝐷𝑓 = −4,4 ⟹ 𝑅𝑓 = 0,2 54 Example: Find Df and Rf: 5- 𝑓 𝑥 = 4 − 𝑥 𝐷𝑓 = −4,4 𝑅𝑓 = 0,2 55 Example: Find Df and Rf: 1 6- 𝑓 𝑥 = To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 𝑥−4 Answer: Note: ⟹𝑥 ∈ℝ− 4 ⟹ 𝐷𝑓 = ℝ − 4 𝑥−4=0 ⟹ 𝑥 ∈ −∞, 4 ∪ 4, ∞ 𝑥=4 𝑥 ∈ −∞, 4 𝑥 ∈ 4, ∞ ⟹ 𝐷𝑓 = −∞, 4 ∪ 4, ∞ ⟹ −∞ < 𝑥 < 4 ⟹4 𝒂 ⟹ 𝒙 > 𝒂 𝑶𝑹 𝒙 < −𝒂 Note: 𝒙𝟐 = 𝒙 (𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈)𝟐 = 𝒂𝒏𝒚𝒕𝒉𝒊𝒏𝒈 60  The graph of a function is a curve in the xy-plane. But the question arises: Which curves in the xy-plane are graphs of functions? 61 The Vertical Line Test A curve in the xy-plane is the graph of a function of x If and only if No vertical line intersects the curve more than once. 62 The Vertical Line Test A curve in the xy-plane is the graph of a function of x if and only if no vertical line intersects the curve more than once. 63 Four Ways to Represent a Function There are four possible ways to represent a function: verbally (by a description in words) numerically (by a table of values) visually (by a graph) algebraically (by an explicit formula) 64 Four Ways to Represent a Function A. Verbally: The cost C of mailing an envelope depends on its weight w. Although there is no simple formula that connects w and C, the post offie has a rule for determining C when w is known. Numerically: The human population of the world P depends on the time t. The table gives estimates of the world population P(t) at time t, for certain years. For instance, P(1950)= 2,560,000,000 Graphically: The vertical acceleration a of the ground as measured by a seismograph during an earthquake is a function of the elapsed time t. Algebrically: The area A of a circle depends on the radius 2 r of the circle. The rule that connects r and A is given by the equation 𝐴 = 𝜋𝑟. 65 piecewise Defined Functions The functions in the following four examples are defied by different formulas in different parts of their domains. Such functions are called piecewise defied functions. 66 Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 2_Part 1 Functions_Part 2 Dr. Sameh Basha  Symmetry A subset 𝑨 ⊆ ℝ is said to be symmetric around the origin if: ∀ 𝒙: 𝒙 ∈ 𝑨 ⟹ −𝒙 ∈ 𝑨 For every number 𝒙 in 𝑨 anumber −𝒙 is also in 𝑨. Ex: 𝑨 = −𝟓, 𝟓 Symmetric around the origin. B= −𝟏,𝟐 Not Symmetric around the origin. 2 ∈ 𝐵 𝑏𝑢𝑡 − 2 ∉ 𝐵 C= −𝟑, 𝟑 Not Symmetric around the origin. −3 ∈ 𝐶 𝑏𝑢𝑡 − −3 = 3 ∉ 𝐶 2  Function Not Symmetric Symmetric Symmetric Symmetric with respect to y- Neither even around the axis. nor odd origin. Even function Odd function 3  Symmetric functions Even function Odd function The function 𝒇: 𝑨 ⟶ 𝑩 is said to be even function The function 𝒇: 𝑨 ⟶ 𝑩 is said to be odd function if: if: (1) The domain of 𝒇(𝒙) (= 𝑫𝒇=A) is Symmetric (1) The domain of 𝒇(𝒙) (= 𝑫𝒇=A) is Symmetric around the origin. around the origin. (2) 𝒇 −𝒙 = 𝒇 𝒙 ; ∀𝒙 ∈ 𝑫𝒇 (2) 𝒇 −𝒙 = −𝒇 𝒙 ; ∀𝒙 ∈ 𝑫𝒇 The domain of the even function is Symmetric The domain of the odd function is Symmetric around the origin. around the origin. The graph of the even function is symmetric with The graph of the odd function is symmetric with respect to the y-axis. respect to the origin. 4  Symmetric function Not Symmetric function Even function Odd functions 5 Example: Determine whether the following functions even, odd, or neither nor. 1- 𝑓 𝑥 = 𝑥2 2- 𝑓 𝑥 = 𝑥3 3- f(x): −2,5 ⟶ ℝ , 𝑓 𝑥 = 𝑥 2 4- 𝑓 𝑥 = 𝑥2 + 𝑥3 1 5-𝑓 𝑥 = (𝑥+1)2 3 6-𝑓 𝑥 = 2 − 4−𝑥 2 7- 𝑓 𝑥 = 𝑥 𝑥2 − 5 6 Example: Determine whether the following functions even, odd, or neither nor. 1- 𝑓 𝑥 = 𝑥 2 Answer: (i) 𝐷𝑓 = ℝ which is symmetric around the origin (ii) 𝑓 −𝑥 = (−𝑥)2 = 𝑥 2 = 𝑓 𝑥 ∴ 𝑓 −𝑥 = 𝑓 𝑥 ; ∀𝑥 ∈ ℝ 𝑓 𝑥 = 𝑥 2 is even function. 7 Example: Determine whether the following functions even, odd, or neither nor. 2- 𝑓 𝑥 = 𝑥 3 Answer: (i) 𝐷𝑓 = ℝ which is symmetric around the origin (ii) 𝑓 −𝑥 = (−𝑥)3 = −𝑥 3 = −𝑓 𝑥 ∴ 𝑓 −𝑥 = −𝑓 𝑥 ; ∀𝑥 ∈ ℝ 𝑓 𝑥 = 𝑥 3 is odd function. 8 Example: Determine whether the following functions even, odd, or neither nor. 3-f(x): −2,5 ⟶ ℝ , 𝑓 𝑥 = 𝑥 2 Answer: (i) 𝐷𝑓 = −2,5 which is not symmetric around the origin. 𝑖𝑖 𝑓 𝑥 = 𝑥 2 is neither even nor odd function on −2,5. 9  Note: f(x): 𝐴 ⟶ 𝐵 Domain f Co-domain f 𝐷𝑓 = 𝐴 10 Example: Determine whether the following functions even, odd, or neither nor. 4- 𝑓 𝑥 = 𝑥 2 + 𝑥 3 Answer: (i) 𝐷𝑓 = ℝ which is symmetric around the origin (ii) 𝑓 −𝑥 = (−𝑥)2 +(−𝑥)3 = 𝑥 2 − 𝑥 3 = −(−𝑥 2 + 𝑥 3 ) ∴ 𝑓 −𝑥 ≠ 𝑓 𝑥 ; ∀𝑥 ∈ ℝ ∴ 𝑓 −𝑥 ≠ −𝑓 𝑥 ; ∀𝑥 ∈ ℝ 𝑓 𝑥 = 𝑥 2 + 𝑥 3 is neither even nor odd function. 11 Example: Determine whether the following functions even, odd, or neither nor. 1 5-𝑓 𝑥 = Draft: (𝑥+1)2 (𝑥 + 1)2 = 𝟎 Answer: ⇒𝑥+1=𝟎 ⇒ 𝑥 = −𝟏 (i) 𝐷𝑓 = ℝ − −1 which is not symmetric around the origin. 1 𝑓 𝑥 = is neither even nor odd function. (𝑥+1)2 12 increasing and Decreasing Functions The graph shown in this Figure rises from A to B, falls from B to C, and rises again from C to D. The function f is said to be increasing on the interval 𝑎, 𝑏 , decreasing on 𝑏, 𝑐 , and increasing again on 𝑐, 𝑑. Notice that if 𝑥1 and 𝑥2 are any two numbers between a and b with 𝑥1 < 𝑥2 , then f(𝑥1 ) < 𝑓(𝑥2 ). 13 Definition 14  3 𝑓 𝑥 = 𝑥 is increasing on ℝ. 15  2 𝑓 𝑥 = 𝑥 is decreasing on the interval −∞, 0 and increasing on the interval 0, ∞. 16 Ex: Find the intervals of increasing and decreasing of the following functions: 1 − 𝑓 𝑥 = 𝑥2 + 2 2-𝑓 𝑥 = 9 − 𝑥 2 3- 𝑓 𝑥 = 𝑥 2 − 4 4- 𝑓 𝑥 = 4− 𝑥 17 Ex: Find the intervals of increasing and decreasing of the following functions: 1 − 𝑓 𝑥 = 𝑥2 + 2 Answer: 𝐷𝑓 = ℝ Note: ∀𝑥1 , 𝑥2 ∈ ℝ: 𝑥1 < 𝑥2 if 𝒂 > 𝒃 then 𝒂𝟐 ‫𝟐𝒃 ؟ ؟؟‬ ⟹ 𝑥12 < 𝑥22 Note: if 𝒂 > 𝒃 >0 then 𝒂𝟐 > 𝒃𝟐 ⟹ 𝑥12 +2 < 𝑥22 + 2 if 0 > 𝒂 > 𝒃 then 𝒂𝟐 < 𝒃𝟐 ⟹ 𝑓(𝑥1 ) < 𝑓(𝑥2 ) ⟹ 𝑓 𝑥 is increasing on ℝ 18 Ex: Find the intervals of increasing and decreasing of the following functions: 1 − 𝑓 𝑥 = 𝑥2 + 2 Answer: 𝐷𝑓 = ℝ = −∞, 𝟎 ∪ 𝟎, ∞ ∀𝒙𝟏 , 𝒙𝟐 ∈ −∞, 𝟎 : 𝒙𝟏 < 𝒙𝟐 ∀𝒙𝟏 , 𝒙𝟐 ∈ 𝟎, ∞ : 𝒙𝟏 < 𝒙𝟐 ⟹ 𝒙𝟐𝟏 > 𝒙𝟐𝟐 ⟹ 𝒙𝟐𝟏 < 𝒙𝟐𝟐 ⟹ 𝒙𝟐𝟏 +𝟐 > 𝒙𝟐𝟐 + 𝟐 ⟹ 𝒙𝟐𝟏 +𝟐 < 𝒙𝟐𝟐 + 𝟐 ⟹ 𝒇(𝒙𝟏 ) > 𝒇(𝒙𝟐 ) ⟹ 𝒇(𝒙𝟏 ) < 𝒇(𝒙𝟐 ) ⟹ 𝒇 𝒙 is decreasing on −∞, 0 ⟹ 𝒇 𝒙 is increasing on 0, ∞ 19  𝑓 𝑥 = 𝑥2 + 2 ⟹ 𝒇 𝒙 is decreasing on −∞, 0 ⟹ 𝒇 𝒙 is increasing on 0, ∞ 20 Ex: Find the intervals of increasing and decreasing of the following functions: 2 − 𝑓 𝑥 = 9 − 𝑥2 Answer: The function 𝑓(𝑥) to be defined should: 9 − 𝑥 2 ≥ 0 ⟹ 9 ≥ 𝑥 2 ⟹ 𝑥 2 ≤ 9 ⟹ 𝑥 ≤ 3 ⟹ −3 ≤ 𝑥 ≤ 3 𝐷𝑓 = −3,3 = −𝟑, 𝟎 ∪ 𝟎, 𝟑 ∀𝒙𝟏 , 𝒙𝟐 ∈ −𝟑, 𝟎 : 𝒙𝟏 < 𝒙𝟐 ∀𝒙𝟏 , 𝒙𝟐 ∈ 𝟎, 𝟑 : 𝒙𝟏 < 𝒙𝟐 ⟹ 𝒙𝟐𝟏 > 𝒙𝟐𝟐 ⟹ 𝒙𝟐𝟏 < 𝒙𝟐𝟐 ⟹ −𝒙𝟐𝟏 < −𝒙𝟐𝟐 ⟹ −𝒙𝟐𝟏 > −𝒙𝟐𝟐 ⟹ 𝟗 − 𝒙𝟐𝟏 < 𝟗 − 𝒙𝟐𝟐 ⟹ 𝟗 − 𝒙𝟐𝟏 > 𝟗 − 𝒙𝟐𝟐 ⟹ 𝟗 − 𝒙𝟐𝟏 < 𝟗 − 𝒙𝟐𝟐 ⟹ 𝟗 − 𝒙𝟐𝟏 > 𝟗 − 𝒙𝟐𝟐 ⟹ 𝒇(𝒙𝟏 ) < 𝒇(𝒙𝟐 ) ⟹ 𝒇(𝒙𝟏 ) > 𝒇(𝒙𝟐 ) ⟹ 𝒇 𝒙 is increasing on −𝟑, 𝟎 ⟹ 𝒇 𝒙 is decreasing on 𝟎, 𝟑 21  𝑓 𝑥 = 9 − 𝑥2 ⟹ 𝒇 𝒙 is increasing on −𝟑, 𝟎 ⟹ 𝒇 𝒙 is decreasing on 𝟎, 𝟑 22 Ex: Find the intervals of increasing and decreasing of the following functions: 3- 𝑓 𝑥 = 𝑥 2 − 4 Answer: The function 𝑓(𝑥) to be defined should: 𝑥 2 − 4 ≥ 0 ⟹ 𝑥 2 ≥ 4 ⟹ 𝑥 ≥ 2 ⟹ 𝑥 ≥ 2 𝑂𝑅 𝑥 ≤ −2 𝐷𝑓 = −∞, −2 ∪ 2, ∞ ∀𝒙𝟏 , 𝒙𝟐 ∈ −∞, −𝟐 : 𝒙𝟏 < 𝒙𝟐 ∀𝒙𝟏 , 𝒙𝟐 ∈ 𝟐, ∞ : 𝒙𝟏 < 𝒙𝟐 ⟹ 𝒙𝟐𝟏 > 𝒙𝟐𝟐 ⟹ 𝒙𝟐𝟏 < 𝒙𝟐𝟐 ⟹ 𝒙𝟐𝟏 −𝟒 > 𝒙𝟐𝟐 − 𝟒 ⟹ 𝒙𝟐𝟏 −𝟒 < 𝒙𝟐𝟐 − 𝟒 ⟹ 𝒙𝟐𝟏 − 𝟒 > 𝒙𝟐𝟐 − 𝟒 ⟹ 𝒙𝟐𝟏 − 𝟒 < 𝒙𝟐𝟐 − 𝟒 ⟹ 𝒇(𝒙𝟏 ) > 𝒇(𝒙𝟐 ) ⟹ 𝒇(𝒙𝟏 ) < 𝒇(𝒙𝟐 ) ⟹ 𝒇 𝒙 is decreasing on −∞, −𝟐 ⟹ 𝒇 𝒙 is increasing on 𝟐, ∞ 23  𝑓 𝑥 = 𝑥2 − 4 ⟹ 𝒇 𝒙 is decreasing on −∞, −𝟐 ⟹ 𝒇 𝒙 is increasing on 𝟐, ∞ 24 Ex: Find the intervals of increasing and decreasing of the following functions: 4- 𝑓 𝑥 = 4 − 𝑥 Answer: The function 𝑓(𝑥) to be defined should: 4 − 𝑥 ≥ 0 ⟹ 4 ≥ 𝑥 ⟹ 𝑥 ≤ 4 ⟹ −4 ≤ 𝑥 ≤ 4 ⟹ 𝐷𝑓 = −4,4 ∀𝒙𝟏 , 𝒙𝟐 ∈ −𝟒, 𝟎 : 𝒙𝟏 < 𝒙𝟐 ∀𝒙𝟏 , 𝒙𝟐 ∈ 𝟎, 𝟒 : 𝒙𝟏 < 𝒙𝟐 ⟹ 𝒙𝟏 < 𝒙𝟐 ⟹ 𝒙𝟏 > 𝒙𝟐 ⟹ − 𝒙𝟏 < − 𝒙 𝟐 ⟹ − 𝒙𝟏 > − 𝒙𝟐 ⟹ 𝟒 − 𝒙𝟏 < 𝟒 − 𝒙𝟐 ⟹ 𝟒 − 𝒙𝟏 > 𝟒 − 𝒙𝟐 ⟹ 𝟒 − 𝒙𝟏 < 𝟒 − 𝒙𝟐 ⟹ 𝟒 − 𝒙𝟏 > 𝟒 − 𝒙𝟐 ⟹ 𝒇(𝒙𝟏 ) < 𝒇(𝒙𝟐 ) ⟹ 𝒇(𝒙𝟏 ) > 𝒇(𝒙𝟐 ) ⟹ 𝒇 𝒙 is increasing on −𝟒, 𝟎 ⟹ 𝒇 𝒙 is decreasingon 𝟎, 𝟒 25  𝑓 𝑥 = 4− 𝑥 ⟹ 𝒇 𝒙 is increasing on −𝟒, 𝟎 ⟹ 𝒇 𝒙 is decreasingon 𝟎, 𝟒 26 Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 2_Part 2 Functions_Part 3 Dr. Sameh Basha  Bounded functions Definition: Let set 𝒮 ⊆ ℝ, we said that: A real number 𝐿 is an upper bound to 𝒮 if: 𝑥 ≤ 𝐿; ∀𝑥 ∈ 𝒮 A real number 𝑙 is a lower bound to 𝒮 if: 𝑥 ≥ 𝑙; ∀𝑥 ∈ 𝒮 Definition: A set 𝒮 ⊆ ℝ considered to be bounded from above if it has an upper bound. A set 𝒮 ⊆ ℝ considered to be bounded from below if it has a lower bound. We say that 𝒮 ⊆ ℝ is bounded if it is bounded from above and bounded from below. Dr. Sameh Basha Example: Let 𝑆1 = 0,3 Is 𝑆1 Bounded or not? Solution: 𝑆1 = 0,3 𝑆1has an upper bounds 𝟑, 𝟒, 𝟓, … 𝑆1has a lower bounds 𝟎, −𝟏, −𝟐, … 𝑆1bounded from above and bounded from below 𝑆1 is Bounded Dr. Sameh Basha Example: Let 𝑆2 = 0,3 Is𝑆2Bounded or not? Solution: 𝑆2 = 0,3 𝑆2 has an upper bounds 𝟑, 𝟒, 𝟓, … 𝑆2 has a lower bounds 𝟎, −𝟏, −𝟐, … 𝑆2 bounded from above and bounded from below 𝑆2 is Bounded Dr. Sameh Basha Example: Let 𝑆3 = −∞, 3 Is 𝑆3 Bounded or not? Solution: 𝑆3 = −∞, 3 𝑆3 has an upper bounds 𝟑, 𝟒, 𝟓, … 𝑆3 does not have a lower bound. 𝑆3 bounded from above only. 𝑆3 is Not Bounded Dr. Sameh Basha Example: Let 𝑆4 = −3, ∞ Is 𝑆4 Bounded or not? Solution: 𝑆4 = −3, ∞ 𝑆4 does not have an upper bounds. 𝑆4 has a lower bounds. −𝟑, −𝟒, −𝟓, … 𝑆4 bounded from below only. 𝑆4 is Not Bounded Dr. Sameh Basha  Bounded function Range f=Rf Rf= −∞,𝒃 Rf= 𝒂,∞ Rf= 𝒂,𝒃 Rf= −∞,∞ Rf= −∞,𝒃 Rf= 𝒂,∞ Rf= 𝒂,𝒃 Bounded from Not Bounded above only Bounded from )Not Bounded( below only Bounded )Not Bounded( Dr. Sameh Basha  Bounded function Range f=Rf Bounded Bounded Not from below from above Bounded Bounded only only Dr. Sameh Basha Example: Is the following function bounded or not bounded? 1- 𝑓 𝑥 = 𝑥 2 − 9 2- 𝑓 𝑥 = 1 − 3 9 − 𝑥 2 3-𝑓 𝑥 = 3 + 𝑥 − 2 3-𝑓 𝑥 = 2 − 𝑥 − 1 Dr. Sameh Basha Example: Is the following function bounded or not bounded? 2- 𝑓 𝑥 = 𝑥 2 − 9 To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 Answer: ⟹ 𝑥 ∈ ℝ − −3,3 The function 𝑓(𝑥) to be defined ⟹ 𝑥 ≥3 should: ⟹ 𝑥2 ≥ 9 𝑥2 − 9 ≥ 0 ⟹ 𝑥2 − 9 ≥ 0 ⟹ 𝑥2 ≥ 9 ⟹ 𝑥2 − 9 ≥ 0 ⟹ 𝑥 ≥3 ⟹ 𝑓(𝑥) ≥ 0 ⟹ 𝑥 ≥ 3 𝑂𝑅 𝑥 ≤ −3 ⟹ 𝑅𝑓 = 0, ∞ ⟹ 𝐷𝑓 = −∞, −3 ∪ 3, ∞ f(x) is Bounded from below only ⟹ 𝐷𝑓 = ℝ − −3,3 f(x) is not Bounded Dr. Sameh Basha Example: Is the following function bounded or not bounded? 2- f(x)=1 − 3 9 − 𝑥 2 To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 ⟹ −3 ≤ 𝑥 ≤ 3 Answer: ⟹ 0 ≤ 𝑥2 ≤ 9 The function 𝑓(𝑥) to be ⟹ 0 ≥ −𝑥 2 ≥ −9 defined should: ⟹ −9 ≤ −𝑥 2 ≤ 0 9 − 𝑥2 ≥ 0 ⟹ 0 ≤ 9 − 𝑥2 ≤ 9 ⟹ 0 ≤ 9 − 𝑥2 ≤ 3 ⟹ 9 ≥ 𝑥2 ⟹ 0 ≤ 3 9 − 𝑥2 ≤ 9 ⟹ 𝑥2 ≤ 9 ⟹ 0 ≥ −3 9 − 𝑥 2 ≥ −9 ⟹ 𝑥 ≤3 ⟹ −9 ≤ −3 9 − 𝑥 2 ≤ 0 ⟹ −8 ≤ 1 − 3 9 − 𝑥 2 ≤ 1 ⟹ −3 ≤ 𝑥 ≤ 3 ⟹ −8 ≤ 𝑓(𝑥) ≤ 1 ⟹ 𝐷𝑓 = −3,3 ⟹ 𝑅𝑓 = −8,1 f(x) is Bounded Dr. Sameh Basha Example: Is the following function bounded or not bounded? 2- 𝑓 𝑥 = 3 + 𝑥 − 2 To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 Answer: ⟹ 𝑥 ∈ 2, ∞ The function 𝑓(𝑥) to be defined ⟹𝑥≥2 should: ⟹𝑥−2≥0 𝑥−2≥0 ⟹ 𝑥−2≥0 ⟹𝑥≥2 ⟹ 𝑓(𝑥) ≥ 0 ⟹ 𝐷𝑓 = 2, ∞ ⟹ 𝑅𝑓 = 0, ∞ f(x) is Bounded from below only f(x) is not Bounded Dr. Sameh Basha Example: Is the following function bounded or not bounded? 2- 𝑓 𝑥 = 2 − 𝑥 − 1 To find 𝑅𝑓 : Let 𝑥 ∈ 𝐷𝑓 Answer: ⟹ 𝑥 ∈ 1, ∞ The function 𝑓(𝑥) to be defined ⟹𝑥≥1 should: ⟹𝑥−1≥0 𝑥−1≥0 ⟹ 𝑥−1≥0 ⟹𝑥≥1 ⟹− 𝑥−1≤0 ⟹ 𝐷𝑓 = 1, ∞ ⟹2− 𝑥−1≤2 ⟹ 𝑓(𝑥) ≤ 2 ⟹ 𝑅𝑓 = −∞, 2 f(x) is Bounded from above only f(x) is not Bounded Dr. Sameh Basha Mathematical Models: A Catalog of Essential Function Recorded Lecture on Canvas: Mathematical Models A Catalog of Essential Functions_Lecture 3_Part3_Fall'22-23.mp4 Dr. Sameh Basha Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 3 New Functions from Old Functions Dr. Sameh Basha  Basic functions 𝒇 𝒙 = 𝒙 𝒇 𝒙 = −𝒙 𝒇 𝒙 =𝒙 𝒇 𝒙 = 𝒙𝟐 𝒇 𝒙 = −𝒙𝟐 𝒇 𝒙 = 𝒙𝟑 Dr. Sameh Basha  Basic functions 𝒇 𝒙 = 𝟑 𝒙 𝟏 𝒇 𝒙 = 𝒇 𝒙 = 𝒙 𝒙 𝒇 𝒙 = 𝒄𝒐𝒔 𝒙 𝒇 𝒙 = 𝒔𝒊𝒏 𝒙 Dr. Sameh Basha 𝒇 𝒙 = 𝒕𝒂𝒏 𝒙  New Functions from Old Functions We start with the basic functions and obtain new functions by shifting, stretching, and reflecting their graphs. We also show how to combine pairs of functions by the standard arithmetic operations and by composition. Dr. Sameh Basha  New Functions from Old Functions Transformations Combinations Compositions Shifting Stretching Reflecting Dr. Sameh Basha  New Functions from Old Functions Transformations Combinations Compositions Shifting Stretching Reflecting Dr. Sameh Basha Vertical and Horizontal Shifts Suppose 𝒄 > 𝟎, 𝒚=𝒇 𝒙 +𝒄 Shift the graph of 𝒚 = 𝒇 𝒙 a distance c units upward. 𝒚=𝒇 𝒙 −𝒄 Shift the graph of 𝒚 = 𝒇 𝒙 a distance c units downward. 𝒚=𝒇 𝒙+𝒄 Shift the graph of 𝒚 = 𝒇 𝒙 a distance c units to the left 𝒚=𝒇 𝒙−𝒄 Shift the graph of 𝒚 = 𝒇 𝒙 a distance c units to the right Dr. Sameh Basha Dr. Sameh Basha 𝒚 = 𝒙𝟐 𝒚 = 𝒙𝟐 + 𝟐 Dr. Sameh Basha 𝒚 = 𝒙𝟐 𝒚 = 𝒙𝟐 − 𝟐 Dr. Sameh Basha  𝒚 = 𝒙𝟐 𝒚 = (𝒙 + 𝟐)𝟐 Dr. Sameh Basha  𝒚 = 𝒙𝟐 𝒚 = (𝒙 − 𝟐)𝟐 Dr. Sameh Basha  𝒚 = 𝒙𝟐 𝒚= 𝒙𝟐 +𝟐 𝒚 = 𝒙𝟐 − 𝟐 𝒚 = (𝒙 + 𝟐)𝟐 𝒚 = (𝒙 − 𝟐)𝟐 Dr. Sameh Basha  New Functions from Old Functions Transformations Combinations Compositions Shifting Stretching Reflecting Dr. Sameh Basha Vertical and Horizontal Stretching Suppose 𝒄 > 𝟏, 𝒚 = 𝒄𝒇 𝒙 stretch the graph of 𝒚 = 𝒇 𝒙 vertically by a factor of c units 𝟏 Shrink the graph of 𝒚 = 𝒇 𝒙 vertically by a factor of c units 𝒚 = ( )(𝒇 𝒙 ) 𝒄 𝒚 = 𝒇 𝒄𝒙 Shrink the graph of 𝒚 = 𝒇 𝒙 horizontally by a factor of c units 𝟏 stretch the graph of 𝒚 = 𝒇 𝒙 horizontally by a factor of c units 𝒚=𝒇 𝒙 𝒄 Dr. Sameh Basha  New Functions from Old Functions Transformations Combinations Compositions Shifting Stretching Reflecting Dr. Sameh Basha Reflecting Suppose 𝒄 > 𝟏, 𝒚 = −𝒇 𝒙 Reflect the graph of 𝒚 = 𝒇 𝒙 about the x-axis 𝒚 = 𝒇 −𝒙 Shrink the graph of 𝒚 = 𝒇 𝒙 about the y-axis Dr. Sameh Basha Dr. Sameh Basha Example: Given the graph of 𝑦 = 𝑥, use the transformation to graph 𝑦 = 𝑥 − 2, 𝑦 = 𝑥 − 2, 𝑦 = − 𝑥 , 𝑦 = 2 𝑥, 𝑦 = −𝑥 Answer: 𝒚= 𝒙 𝒚= 𝒙−𝟐 𝒚= 𝒙−𝟐 Dr. Sameh Basha Example: Given the graph of 𝑦 = 𝑥, use the transformation to graph 𝑦 = 𝑥 − 2, 𝑦 = 𝑥 − 2, 𝑦 = − 𝑥 , 𝑦 = 2 𝑥, 𝑦 = −𝑥 Answer: 𝒚= 𝒙 𝒚=− 𝒙 𝒚 = −𝒙 Dr. Sameh Basha Example: Given the graph of 𝑦 = 𝑥, use the transformation to graph 𝑦 = 𝑥 − 2, 𝑦 = 𝑥 − 2, 𝑦 = − 𝑥 , 𝑦 = 2 𝑥, 𝑦 = −𝑥 Answer: 𝒚= 𝒙 𝒚=𝟐 𝒙 Dr. Sameh Basha Example: Sketch the curve: 𝑦 = 𝑥 2 + 6𝑥 + 10 Answer: 𝒚 = 𝒙𝟐 + 𝟔𝒙 + 𝟏𝟎 by completing the square. 𝒚 = (𝒙 + 𝟑)𝟐 − 𝟑 𝟐 + 𝟏𝟎 𝒚 = (𝒙 + 𝟑)𝟐 −𝟗 + 𝟏𝟎 𝒚 = (𝒙 + 𝟑)𝟐 +𝟏 Dr. Sameh Basha  𝒚 = 𝒙𝟐 𝒚 = (𝒙 + 𝟑)𝟐 𝒚 = (𝒙 + 𝟑)𝟐 +1 Dr. Sameh Basha Note: Completing the square: 𝒚 = 𝟏𝒙𝟐 + 𝒃𝒙 + 𝒄 𝟐 𝟐 𝒃 𝒃 𝒚= 𝒙+ − +𝒄 𝟐 𝟐 𝒚 = 𝒙𝟐 + 𝟖𝒙 + 𝟒 𝟐 𝟐 𝟖 𝟖 𝒚= 𝒙+ − +𝟒 𝟐 𝟐 𝟐 𝟐 = 𝒙+𝟒 𝟐 − 𝟏𝟐 𝒚= 𝒙+𝟒 − 𝟒 +𝟒 Dr. Sameh Basha Note: Completing the square: 𝒚 = 𝟏𝒙𝟐 − 𝒃𝒙 + 𝒄 𝟐 𝟐 𝒃 𝒃 𝒚= 𝒙− − +𝒄 𝟐 𝟐 𝒚 = 𝒙𝟐 − 𝟖𝒙 + 𝟒 𝟐 𝟐 𝟖 𝟖 𝒚= 𝒙− − +𝟒 𝟐 𝟐 𝟐 𝟐 = 𝒙−𝟒 𝟐 − 𝟏𝟐 𝒚= 𝒙−𝟒 − 𝟒 +𝟒 Dr. Sameh Basha Example: The graph of y=f(x) is given. Match each equation with its graph and give reasons for your choices a- 𝒚 = 𝒇(𝒙 − 𝟒) b- 𝒚 = 𝒇 𝒙 + 𝟑 c- 𝟏 𝒚= 𝒇 𝒙 𝟑 d- 𝒚 = −𝒇 𝒙 + 𝟒 e- 𝒚 = 𝟐𝒇 𝒙 + 𝟔 Dr. Sameh Basha Example: The graph of y=f(x) is given. Match each equation with its graph and give reasons for your choices 𝒚 = 𝟐𝒇 𝒙 + 𝟔 𝒚=𝒇 𝒙 +𝟑 a- 𝒚 = 𝒇(𝒙 − 𝟒) b- 𝒚 = 𝒇 𝒙 + 𝟑 𝒚 = 𝒇(𝒙 − 𝟒) c- 𝟏 𝒚= 𝒇 𝒙 𝟑 d- 𝒚 = −𝒇 𝒙 + 𝟒 e- 𝒚 = 𝟐𝒇 𝒙 + 𝟔 𝟏 𝒚= 𝒇 𝒙 𝟑 𝒚 = −𝒇 𝒙 + 𝟒 Dr. Sameh Basha  New Functions from Old Functions Transformations Combinations Compositions Shifting Stretching Reflecting Dr. Sameh Basha Combinations of two functions Two functions f and g can be combined to form new functions: 𝒚 = 𝒇 + 𝒈, 𝒚 = 𝒇 − 𝒈, 𝒚 = 𝒇𝒈, 𝒚 = 𝒇/𝒈 (𝒇 + 𝒈)(𝒙) = 𝒇(𝒙) + 𝒈(𝒙), (𝒇 − 𝒈)(𝒙) = 𝒇(𝒙) − 𝒈(𝒙), (𝒇𝒈)(𝒙) = 𝒇(𝒙) 𝒈(𝒙), 𝒇 𝒇(𝒙) ( )(𝒙) = 𝒈 𝒈(𝒙) Dr. Sameh Basha Combinations of two functions Two functions f and g can be combined to form new functions: 𝒚 = 𝒇 + 𝒈, 𝒚 = 𝒇 − 𝒈, 𝒚 = 𝒇𝒈, 𝒚 = 𝒇/𝒈 I𝒇 𝑫𝒇 = 𝑨, and 𝑫𝒈 = 𝑩, Then: 𝑫𝒇+𝒈 = 𝑨 ∩ 𝑩, 𝑫𝒇−𝒈 = 𝑨 ∩ 𝑩, 𝑫𝒇𝒈 = 𝑨 ∩ 𝑩, 𝑫𝒇/𝒈 = {𝒙 ∈ 𝑨 ∩ 𝑩|𝒈(𝒙) ≠ 𝟎}, Dr. Sameh Basha Example: Let 𝑓 𝑥 = 𝑥 and 𝑔 𝑥 = 2 − 𝑥. Find (𝒇 + 𝒈)(𝒙), (𝒇 − 𝒈)(𝒙), 𝒇 (𝒇𝒈)(𝒙), ( )(𝒙), 𝑫𝒇+𝒈 , 𝑫𝒇−𝒈 , 𝑫𝒇𝒈 , 𝑫𝒇/𝒈. Note: Note: 𝒈 f 𝒙 = 𝒙 𝒈 𝒙 = 𝟐−𝒙 Answer: 𝒙≥𝟎 𝑫𝒇 = 𝟎, ∞ 𝟐−𝒙≥𝟎 𝟐≥𝒙 𝒇(𝒙) = 𝑥, 𝑫𝒇 = 𝟎, ∞ , 𝒙≤𝟐 𝑫𝒈 = −∞, 𝟐 g(𝒙) = 2 − 𝑥, 𝑫𝒈 = −∞, 𝟐 , (𝒇 + 𝒈)(𝒙) = 𝒇(𝒙) + 𝒈(𝒙) = 𝑥 + 2 − 𝑥 𝑫𝒇+𝒈 = 𝟎, ∞ ∩ −∞, 𝟐 = 𝟎, 𝟐 (𝒇 − 𝒈)(𝒙) = 𝒇(𝒙) − 𝒈(𝒙) = 𝑥 − 2 − 𝑥 𝑫𝒇−𝒈 = 𝟎, ∞ ∩ −∞, 𝟐 = 𝟎, 𝟐 (𝒇𝒈)(𝒙) = 𝒇(𝒙) 𝒈(𝒙) = 𝑥 2−𝑥 𝑫𝒇𝒈 = 𝟎, ∞ ∩ −∞, 𝟐 = 𝟎, 𝟐 𝒇 𝒇(𝒙) 𝑥 𝑫𝒇/𝒈 = {𝒙 ∈ 𝑫𝒇 ∩ 𝑫𝒈 |𝒈(𝒙) ≠ 𝟎} = 𝟎, 𝟐 ( )(𝒙) = = 𝒈 𝒈(𝒙) 2−𝑥 Dr. Sameh Basha Example: 𝒇 Let 𝑓 𝑥 = 𝑥2 and 𝑔 𝑥 = 𝑥 − 1. Find ( )(𝒙), 𝑫𝒇/𝒈. 𝒈 Answer: 𝒇(𝒙) = 𝑥 2 , 𝑫𝒇 = ℝ, g(𝒙) = 𝑥 − 1, 𝑫𝒈 = ℝ, 𝒇 𝒇(𝒙) 𝑥2 ( )(𝒙) = = 𝑫𝒇/𝒈 = {𝒙 ∈ 𝑫𝒇 ∩ 𝑫𝒈 |𝒈(𝒙) ≠ 𝟎} = ℝ-{1} 𝒈 𝒈(𝒙) 𝑥−1 Dr. Sameh Basha  New Functions from Old Functions Transformations Combinations Compositions Shifting Stretching Reflecting Dr. Sameh Basha Composition of two functions 2 Suppose that u = 𝑔 𝑥 = 𝑥 + 1 We want to compute y = 𝑓 𝑢 = 𝑢 We will compute this by substitution: y = 𝑓 𝑢 = 𝑢 = 𝑔 𝑥 = 𝑥2 + 1 This procedure is called Composition Dr. Sameh Basha  Composition of two functions Definition: Given two functions 𝒇 and 𝒈, the composite function 𝒇𝒐𝒈 (also called the composition of 𝒇 and 𝒈 )is defined by: 𝒇𝒐𝒈 𝒙 = 𝒇(𝒈 𝒙 ) Definition: Given two functions 𝒇 and 𝒈, the composite function 𝒈𝒐𝒇 (also called the composition of 𝒇 and 𝒈 )is defined by: 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 ) Dr. Sameh Basha  Composition of two functions Definition: Given two functions 𝒇 and 𝒈, the composite function 𝒇𝒐𝒈 (also called the composition of 𝒇 and 𝒈 )is defined by: 𝒇𝒐𝒈 𝒙 = 𝒇(𝒈 𝒙 ) Input: 𝒙 𝒈 𝒈(𝒙) 𝒇 𝒇(𝒈(𝒙)) 𝒇𝒐𝒈 The domain of 𝒇𝒐𝒈 is the set of all 𝒙 in the domain of 𝒈 such that 𝒈(𝒙) is in the domain 𝒇. 𝒇𝒐𝒈 𝒙 is defined whenever both 𝒈(𝒙) and 𝒇(𝒈 𝒙 ) are defined. Dr. Sameh Basha  Composition of two functions Definition: Given two functions 𝒇 and 𝒈, the composite function 𝒇𝒐𝒈 (also called the composition of 𝒇 and 𝒈 )is defined by: 𝒇𝒐𝒈 𝒙 = 𝒇(𝒈 𝒙 ) Input: 𝒙 𝒈 𝒈(𝒙) 𝒇 𝒇(𝒈(𝒙)) 𝒇𝒐𝒈 The domain of 𝒇𝒐𝒈 is the set of all 𝒙 in the domain of 𝒈 such that 𝒈(𝒙) is in the domain 𝒇. 𝑫𝒇𝒐𝒈 = 𝒙 ∈ 𝑫𝒈 : 𝒈(𝒙) ∈ 𝑫𝒇. Dr. Sameh Basha  Composition of two functions Definition: Given two functions 𝒇 and 𝒈, the composite function 𝒈𝒐𝒇 (also called the composition of 𝒇 and 𝒈 )is defined by: 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 ) Input: 𝒙 𝒇 𝒇(𝒙) 𝒈 𝒈(𝒇(𝒙)) 𝒈𝒐𝒇 The domain of 𝒈𝒐𝒇 is the set of all 𝒙 in the domain of 𝒇 such that 𝒇(𝒙) is in the domain 𝒈. 𝒈𝒐𝒇 𝒙 is defined whenever both 𝒇(𝒙) and 𝒈(𝒇 𝒙 ) are defined. Dr. Sameh Basha  Composition of two functions Definition: Given two functions 𝒇 and 𝒈, the composite function 𝒈𝒐𝒇 (also called the composition of 𝒇 and 𝒈 )is defined by: 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 ) Input: 𝒙 𝒇 𝒇(𝒙) 𝒈 𝒈(𝒇(𝒙)) 𝒈𝒐𝒇 The domain of 𝒈𝒐𝒇 is the set of all 𝒙 in the domain of 𝒇 such that 𝒇(𝒙) is in the domain 𝒈. 𝑫𝒈𝒐𝒇 = 𝒙 ∈ 𝑫𝒇 : 𝒇(𝒙) ∈ 𝑫𝒈. Dr. Sameh Basha Example: Let 𝑓 𝑥 = 𝑥 2 and 𝑔 𝑥 = 𝑥 − 3. Find the composite functions 𝒇𝒐𝒈 and 𝒈𝒐𝒇. Answer: 𝒇(𝒙) = 𝑥 2 , g(𝒙) = 𝑥 − 3 𝒇𝒐𝒈 𝒙 = 𝒇(𝒈 𝒙 )=𝒇 𝒙 − 𝟑 = (𝒙 − 𝟑)𝟐 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 𝒙𝟐 = 𝒙𝟐 − 𝟑 Dr. Sameh Basha Example: Let 𝑓 𝑥 = 𝑥 2 and 𝑔 𝑥 = 𝑠𝑖𝑛𝑥. Find the composite functions 𝒇𝒐𝒈 and 𝒈𝒐𝒇. Answer: 𝒇(𝒙) = 𝑥 2 , g(𝒙) = sin 𝑥 𝒇𝒐𝒈 𝒙 = 𝒇(𝒈 𝒙 )=𝒇 𝒔𝒊𝒏 𝒙 = (𝒔𝒊𝒏 𝒙)𝟐 = 𝒔𝒊𝒏𝟐 𝒙 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 𝒙𝟐 = 𝒔𝒊𝒏 𝒙𝟐 = 𝒔𝒊𝒏 𝒙𝟐 Dr. Sameh Basha Example: Let 𝑓 𝑥 = 𝑥 and 𝑔 𝑥 = 2 − 𝑥. Find the following functions: a) 𝒇𝒐𝒈 b) 𝒈𝒐𝒇 c) 𝒇𝒐𝒇 d) 𝒈𝒐𝒈 Answer: 𝒇(𝒙) = 𝑥, g(𝒙) = 2−𝑥 𝒇𝒐𝒈 𝒙 = 𝒇(𝒈 𝒙 )=𝒇 𝟐−𝒙 = 𝟐−𝒙 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 𝒙 = 𝟐− 𝒙 𝒇𝒐𝒇 𝒙 = 𝒇(𝒇 𝒙 )=𝒇 𝒙 = 𝒙 𝒈𝒐𝒈 𝒙 = 𝒈(𝒈 𝒙 )=𝒈 𝟐−𝒙 = 𝟐− 𝟐−𝒙 Dr. Sameh Basha Example:Let 𝒇 𝒙 = 𝒙 and 𝒈 𝒙 = 𝟐 − 𝒙. Find the composite functions 𝒇𝒐𝒈 and 𝒈𝒐𝒇 and their domains. Answer: 𝒇(𝒙) = 𝑥, g(𝒙) = 2−𝑥 Note: 𝑫𝒈 = −∞, 𝟐 , 𝒈 𝒙 = 𝟐−𝒙 𝑫𝒇 = 𝟎, ∞ , 𝟐−𝒙≥𝟎 𝟐≥𝒙 𝒇𝒐𝒈 𝒙 = 𝒇 𝒈 𝒙 =𝒇 𝟐 − 𝒙 = 𝟐−𝒙 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 𝒙 = 𝟐− 𝒙 𝒙≤𝟐 𝑫𝒈 = −∞, 𝟐 𝑫𝒇𝒐𝒈 = 𝒙 ∈ 𝑫𝒈 : 𝒈(𝒙) ∈ 𝑫𝒇. 𝑫𝒈𝒐𝒇 = 𝒙 ∈ 𝑫𝒇 : 𝒇(𝒙) ∈ 𝑫𝒈. = 𝒙 ∈ −∞, 𝟐 : 𝟐 − 𝒙 ∈ 𝟎, ∞ = 𝒙 ∈ 𝟎, ∞ : 𝒙 ∈ −∞, 𝟐 = 𝟎 ≤ 𝒙 < ∞: −∞ < 𝒙 ≤ 𝟐 = −∞ < 𝒙 ≤ 𝟐: 𝟎 ≤ 𝟐 − 𝒙 < ∞ = 𝟎 ≤ 𝒙 < ∞: 𝟎 ≤ 𝒙 ≤ 𝟐 = −∞ < 𝒙 ≤ 𝟐: 𝟎 ≤ 𝟐 − 𝒙 < ∞ = 𝟎 ≤ 𝒙 < ∞: 𝟎 ≤ 𝒙 ≤ 𝟒 = −∞ < 𝒙 ≤ 𝟐: −𝟐 ≤ −𝒙 < ∞ = 𝟎, 𝟒 = −∞ < 𝒙 ≤ 𝟐: −∞ < 𝒙 ≤ 𝟐 = −∞, 𝟐 Dr. Sameh Basha  Example:Let 𝒇 𝒙 = 𝟑𝒙𝟐 − 𝟐 and 𝒈 𝒙 = 𝒙. Find the composite functions 𝒇𝒐𝒈 and 𝒈𝒐𝒇 and their domains. Answer: 𝒇(𝒙) = 𝟑𝒙𝟐 − 𝟐, g(𝒙) = 𝑥 𝑫𝒇 = ℝ, 𝑫𝒈 = 𝟎, ∞ , 𝒇𝒐𝒈 𝒙 = 𝒇 𝒈 𝒙 =𝒇 𝑥 = 𝟑( 𝑥 )𝟐 − 𝟐 = 𝟑𝒙 − 𝟐 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 𝟑𝒙𝟐 − 𝟐 = 𝟑𝒙𝟐 − 𝟐 𝑫𝒈𝒐𝒇 = 𝒙 ∈ ℝ: 𝟑𝒙𝟐 − 𝟐 ∈ 𝟎, ∞ 𝑫𝒇𝒐𝒈 = 𝒙 ∈ 𝑫𝒈 : 𝒈(𝒙) ∈ 𝑫𝒇. = 𝒙 ∈ ℝ: 𝟎 ≤ 𝟑𝒙𝟐 − 𝟐 < ∞ = 𝒙 ∈ 𝟎, ∞ : 𝒙 ∈ ℝ 𝟐 = 𝒙 ∈ ℝ: 𝟐 ≤ 𝟑𝒙𝟐 < ∞ = 𝒙 ∈ ℝ: ≤ 𝒙𝟐 < ∞ = 𝒙 ∈ 𝟎, ∞ : 𝒙 ≥ 𝟎 𝟑 = 𝟎 ≤ 𝒙 < ∞: 𝟎 ≤ 𝒙 < ∞ 𝟐 𝟐 𝟐 = 𝟎 ≤ 𝒙 < ∞: 𝟎 ≤ 𝒙 < ∞ = 𝒙 ∈ ℝ: ≤ 𝒙 < ∞ = 𝒙 ∈ ℝ: 𝒙 ≥ 𝑶𝑹 𝒙 ≤ − 𝟑 𝟑 𝟑 = 𝟎, ∞ 𝟐 𝟐 𝟐 𝟐 = 𝒙 ∈ ℝ: 𝒙 ∈ ℝ − − , =ℝ − − , 𝟑 𝟑 𝟑 𝟑 Dr. Sameh Basha Example:Let 𝒇 𝒙 = 𝒔𝒊𝒏 𝒙 and 𝒈 𝒙 = 𝒙𝟐 + 𝟏. Find the composite functions 𝒇𝒐𝒈 and 𝒈𝒐𝒇 and their domains. Answer: 𝒇(𝒙) = sin 𝑥, g(𝒙) = 𝒙𝟐 + 𝟏 𝑫𝒇 = ℝ, 𝑫𝒈 = ℝ, 𝒇𝒐𝒈 𝒙 = 𝒇 𝒈 𝒙 =𝒇 𝒙𝟐 + 𝟏 = 𝒔𝒊𝒏 (𝒙𝟐 + 𝟏) 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 𝒔𝒊𝒏 𝒙 = 𝒔𝒊𝒏𝟐 𝒙 + 𝟏 𝑫𝒇𝒐𝒈 = 𝒙 ∈ 𝑫𝒈 : 𝒈(𝒙) ∈ 𝑫𝒇. 𝑫𝒈𝒐𝒇 = 𝒙 ∈ 𝑫𝒇 : 𝒇(𝒙) ∈ 𝑫𝒈. = 𝒙 ∈ ℝ: 𝒙𝟐 + 𝟏 ∈ ℝ = 𝒙 ∈ ℝ: sin 𝑥 ∈ ℝ = 𝒙 ∈ ℝ: 𝒙𝟐 + 𝟏 ≥ 𝟏 = 𝒙 ∈ ℝ: −1 ≤ sin 𝑥 ≤ 𝟏 = 𝒙 ∈ ℝ: 𝒙𝟐 ≥ 𝟎 = 𝒙 ∈ ℝ: 𝒙 ∈ ℝ =ℝ = 𝒙 ∈ ℝ: 𝒙 ≥ 𝟎 = 𝒙 ∈ ℝ: 𝒙 ∈ ℝ =ℝ Dr. Sameh Basha  Example:Let 𝒇 𝒙 = 𝒙 + 𝟏 and 𝒈 𝒙 = 𝟒𝒙 − 𝟑. Find the composite functions 𝒇𝒐𝒈 and 𝒈𝒐𝒇 and their domains. Answer: 𝒇(𝒙) = 𝒙 + 𝟏, g(𝒙) = 𝟒𝒙 − 𝟑 𝑫𝒇 = −𝟏, ∞ , 𝑫𝒈 = ℝ, 𝒇𝒐𝒈 𝒙 = 𝒇 𝒈 𝒙 =𝒇 𝟒𝒙 − 𝟑 = 𝟒𝒙 − 𝟑 + 𝟏 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 𝒙 + 𝟏 = 𝟒 𝒙 + 𝟏 -3 = 𝟒𝒙 − 𝟐 𝑫𝒇𝒐𝒈 = 𝒙 ∈ 𝑫𝒈 : 𝒈(𝒙) ∈ 𝑫𝒇. 𝑫𝒈𝒐𝒇 = 𝒙 ∈ 𝑫𝒇 : 𝒇(𝒙) ∈ 𝑫𝒈. = 𝒙 ∈ ℝ: 𝟒𝒙 − 𝟑 ∈ −𝟏, ∞ = 𝒙 ∈ −𝟏, ∞ : 𝒙 + 𝟏 ∈ ℝ = 𝒙 ∈ ℝ: −𝟏 ≤ 𝟒𝒙 − 𝟑 < ∞ = 𝒙 ∈ ℝ: 𝟐 ≤ 𝟒𝒙 < ∞ = 𝒙 ∈ −𝟏, ∞ : 𝒙 + 𝟏 ≥ 𝟎 𝟏 = 𝒙 ∈ −𝟏, ∞ : 𝒙 + 𝟏 ≥ 𝟎 = 𝒙 ∈ ℝ: ≤ 𝒙 < ∞ 𝟐 = 𝒙 ∈ −𝟏, ∞ : 𝒙 ≥ −𝟏 𝟏 = 𝒙 ∈ −𝟏, ∞ : 𝒙 ∈ −𝟏, ∞ = ,∞ = −𝟏, ∞ 𝟐 Dr. Sameh Basha Note: It is possible to take the composition of three or more functions. For instance, The composite function 𝒇𝒐𝒈𝒐𝒉 : 𝒇𝒐𝒈𝒐𝒉 𝒙 = 𝒇(𝒈 𝒉(𝒙) ) 𝒙 Example: Find 𝒇𝒐𝒈𝒐𝒉 if 𝒇 𝒙 = ,𝒈 𝒙 = 𝒙𝟏𝟎 , 𝒉 𝒙 = 𝒙 + 𝟑 𝒙+𝟏 Answer: 𝒇𝒐𝒈𝒐𝒉 𝒙 = 𝒇(𝒈 𝒉(𝒙) ) =𝒇(𝒈 𝒙 + 𝟑 ) =𝒇((𝒙 + 𝟑)𝟏𝟎 ) (𝒙+𝟑)𝟏𝟎 = (𝒙+𝟑)𝟏𝟎 +𝟏 Dr. Sameh Basha  Example:Let 𝑓 𝑥 = 8 − 𝑥 and 𝑔 𝑥 = 1 + 𝑥 2 Find fog, gof, Dfog, and Dgof. 𝑓 𝑥 = 8− 𝑥 Is gof bounded? 8− 𝑥 ≥𝟎 Find the interval of increasing and decreasing for gof 8≥ 𝑥 𝑥 ≤8 Answer: 𝑔 𝑥 = 1 + 𝑥2 −8 ≤ 𝑥 ≤ 8 𝑓 𝑥 = 8− 𝑥 Dg = ℝ Df = −8,8 𝒇𝒐𝒈 𝒙 = 𝒇 𝒈 𝒙 =𝒇 1 + 𝑥2 = 𝟖− 1+ 𝑥2 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 8 − 𝑥 = 𝟏 + ( 8 − 𝑥 )𝟐 = 𝟖 − (1 + 𝑥 2 ) = 𝟖 − 1 − 𝑥 2 = 𝟕 − 𝑥 2 =𝟏+ 8− 𝑥 =𝟗− 𝑥 𝑫𝒇𝒐𝒈 = 𝒙 ∈ 𝑫𝒈 : 𝒈(𝒙) ∈ 𝑫𝒇. 𝑫𝒈𝒐𝒇 = 𝒙 ∈ 𝑫𝒇 : 𝒇(𝒙) ∈ 𝑫𝒈. = 𝒙 ∈ ℝ: 1 + 𝑥 2 ∈ −8,8 = 𝒙 ∈ −8,8 : 8 − 𝑥 ∈ ℝ = 𝒙 ∈ ℝ: −𝟖 ≤ 1 + 𝑥 2 ≤ 𝟖 = 𝒙 ∈ −8,8 : 0 ≤ 8 − 𝑥 < ∞ = 𝒙 ∈ ℝ: 𝟏 ≤ 1 + 𝑥 2 ≤ 𝟖 = 𝒙 ∈ −8,8 : 0 ≤ 8 − 𝑥 < ∞ = 𝒙 ∈ ℝ: 𝟎 ≤ 𝑥 2 ≤ 𝟕 = 𝒙 ∈ −8,8 : −8 ≤ − 𝑥 < ∞ = 𝒙 ∈ ℝ: 𝟎 ≤ 𝒙 ≤ 𝟕 = 𝒙 ∈ −8,8 : −∞ < 𝑥 ≤ 𝟖 Dr. Sameh Basha = 𝒙 ∈ −8,8 : 𝒙 ∈ −8,8 = 𝒙 ∈ ℝ: 𝒙 ∈ − 𝟕, 𝟕 = −8,8 𝑓 𝑥 = 8 − 𝑥 and 𝑔 𝑥 = 1 + 𝑥 2 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 8− 𝑥 = 𝟏 + ( 8 − 𝑥 )𝟐 = 𝟏 + 8 − 𝑥 =𝟗− 𝑥 𝑫𝒈𝒐𝒇 = −8,8 Let 𝒙 ∈ 𝑫𝒈𝒐𝒇 = −8,8 ⇒ −𝟖 ≤ 𝒙 ≤ 𝟖 ⇒𝟎≤ 𝒙 ≤𝟖 ⇒ 𝟎 ≥ − 𝒙 ≥ −𝟖 ⇒ −𝟖 ≤ − 𝒙 ≤ 𝟎 ⇒𝟏≤𝟗− 𝒙 ≤𝟗 ⇒ 𝟏 ≤ 𝒈𝒐𝒇(𝒙) ≤ 𝟗 ⇒ 𝑹𝒈𝒐𝒇 = 𝟏, 𝟗 Dr. Sameh Basha 𝑓 𝑥 = 8 − 𝑥 and 𝑔 𝑥 = 1 + 𝑥 2 𝒈𝒐𝒇 𝒙 = 𝒈(𝒇 𝒙 )=𝒈 8− 𝑥 = 𝟏 + ( 8 − 𝑥 )𝟐 = 𝟏 + 8 − 𝑥 =𝟗− 𝑥 𝑫𝒈𝒐𝒇 = −8,8 = −8,0 ∪ 0,8 ∀𝒙𝟏 , 𝒙𝟐 ∈ −8,0 : 𝒙𝟏 < 𝒙𝟐 ∀𝒙𝟏 , 𝒙𝟐 ∈ 0,8 : 𝒙𝟏 < 𝒙𝟐 ⇒ 𝒙𝟏 > 𝒙𝟏 ⇒ 𝒙𝟏 < 𝒙𝟏 ⇒ − 𝒙𝟏 < − 𝒙𝟐 ⇒ − 𝒙𝟏 > − 𝒙𝟐 ⇒ 𝟗 − 𝒙𝟏 < 𝟗 − 𝒙𝟐 ⇒ 𝟗 − 𝒙𝟏 > 𝟗 − 𝒙𝟐 ⟹ 𝒈𝒐𝒇 (𝒙𝟏 ) < 𝒈𝒐𝒇 (𝒙𝟐 ) ⟹ 𝒈𝒐𝒇 (𝒙𝟏 ) > 𝒈𝒐𝒇 (𝒙𝟐 ) ⟹ 𝒈𝒐𝒇 𝒙 is increasing on −8,0 ⟹ 𝒈𝒐𝒇 𝒙 is decreasing on 0,8 Dr. Sameh Basha Galala University Fall 2024-2025 Faculty of Science Mathematics I (MAT 111) Department of Mathematics Lecture 4 The Limit of a Function_Part 1 Dr. Sameh Basha  Introduction to limits 𝑥−1 Let 𝑓 𝑥 =. 𝑥 2 −1 Suppose we need to find 𝑓 1 𝑥−1 Notice that: 𝑓 𝑥 = not defined when x=1. 𝑥 2 −1 So, let us investigate the behavior of the function for values of x near to 1. near to 1 𝒙𝟏 Dr. Sameh Basha  near to 1 𝒙𝟏 lim 𝟐 = 𝟎. 𝟓 𝒙→𝟏 𝒙 − 𝟏 Dr. Sameh Basha  near to 1 𝒙𝟏 lim 𝟐 = 𝟎. 𝟓 𝒙→𝟏 𝒙 − 𝟏 The closer 𝒙 is to 𝟏 The closer 𝐟(𝒙) is to 𝟎. 𝟓 𝒙−𝟏 → 𝟎. 𝟓 As 𝒙 → 𝟏 𝒙𝟐 −𝟏 Dr. Sameh Basha  Now, Let us investigate the behavior of the function 𝑓 𝑥 = 𝑥 2 − 𝑥 + 2 for values of 𝑥 near to 2. Near to 2 𝒙𝟐 𝒙→𝟐 Dr. Sameh Basha Intuitive Definition of a Limit Dr. Sameh Basha Note Let: 𝑥 2 ; 𝑖𝑓 𝑥 > 0 f 𝑥 =ቊ 5; 𝑖𝑓 𝑥 = 0 𝐷𝑓 = 0, ∞ 𝑓 0 =? ? 𝑓 0 =5 Dr. Sameh Basha  Note: 𝒇(𝒙) is defined 𝒇(𝒙) is defined 𝒇(𝒙) is Not defined at 𝐱 = 𝒂. at 𝐱 = 𝒂. at and and 𝐱 = 𝒂. 𝒇 𝒂 =𝑳 𝒇 𝒂 =𝒃 𝐥𝐢𝐦 𝒇 𝒙 = 𝑳 = 𝒇(𝒂) 𝐥𝐢𝐦 𝒇 𝒙 = 𝑳 𝒙→𝒂 𝒙→𝒂 𝐥𝐢𝐦 𝒇 𝒙 = 𝑳 𝒇 𝒂 =𝒃 𝒙→𝒂 Dr. Sameh Basha Note: Dr. Sameh Basha Note: Dr. Sameh Basha  𝒕𝟐 +𝟗−𝟑 Ex: Estimate the value of 𝒍𝒊𝒎 𝒕→𝟎 𝒕𝟐 As t approaches 0, the values of the function seem to approach 0.1666666... and so we guess that: 𝒕𝟐 + 𝟗 − 𝟑 𝟏 𝒍𝒊𝒎 𝟐 = 𝒕→𝟎 𝒕 𝟔 Dr. Sameh Basha What would have happened if we had taken even smaller values of t? Something strange seems to be happening. If you try these calculations on your own calculator you might get different values, but eventually you will get the value 0 if you make t sufficiently small. 𝟏 Does this mean that the answer is really 0 instead of ? 𝟔 𝟏 No, the value of the limit is 𝟔 The problem is that the calculator gave false values because 𝒕𝟐 + 𝟗 − 𝟑 is very close to 3 when t is small. In fact, when t is sufficiently small, a calculator’s value for 𝒕𝟐 + 𝟗 − 𝟑 is 3.000... to as many digits as the calculator is capable of carrying. rounding errors from the subtraction. Dr. Sameh Basha  One-Sided Limits 𝟎; 𝒊𝒇 𝒕 < 𝟎 𝑯 𝒕 =ቊ 𝟏; 𝒊𝒇 𝒕 ≥ 𝟎 As t approaches 0 from the left, 𝑯 𝒕 approaches 0. As t approaches 0 from the right, 𝑯 𝒕 approaches 1. There is no single number that 𝑯 𝒕 approaches as t approaches 0. Therefore, 𝒍𝒊𝒎 𝑯 𝒕 does not exist. 𝒕→𝟎 As t approaches 0 from the left, 𝑯 𝒕 approaches 0. As t approaches 0 from the right, 𝑯 𝒕 approaches 1. 𝒍𝒊𝒎− 𝑯 𝒕 = 𝟎 𝒍𝒊𝒎+ 𝑯 𝒕 = 𝟏 𝒕→𝟎 𝒕→𝟎 Therefore, 𝒍𝒊𝒎 𝑯 𝒕 does not exist. 𝒕→𝟎 Dr. Sameh Basha One-Sided Limits 𝒕 → 𝟎− indicates that we consider only values of t that are less than 0. 𝒕 → 𝟎+ indicates that we consider only values of t that are greater than 0. Definition: Definition: We write 𝒍𝒊𝒎− 𝒇 𝒙 = 𝑳 We write 𝒍𝒊𝒎+ 𝒇 𝒙 = 𝑳 𝒙→𝟎 𝒙→𝒂 and say the left-hand limit of 𝒇 𝒙 as x approaches a and say the right-hand limit of 𝒇 𝒙 as x approaches a [or the limit of 𝒇 𝒙 as x approaches a from the left] [or the limit of 𝒇 𝒙 as x approaches a from the right] is equal to 𝑳. if we can make the values of 𝒇 𝒙 is equal to 𝑳. if we can make the values of 𝒇 𝒙 arbitrarily close to 𝑳 by taking x to be sufficiently arbitrarily close to 𝑳 by taking x to be sufficiently close to a with x less than a. close to a with x greater than a. Dr. Sameh Basha Definition: Dr. Sameh Basha Ex: The graph of a function t is shown in Figure. Use it to state the values (if they exist) of the following: a) 𝒍𝒊𝒎− 𝒈 𝒙 b) 𝒍𝒊𝒎+ 𝒈 𝒙 c) 𝒍𝒊𝒎 𝒈 𝒙 g) g(5) 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 d) 𝒍𝒊𝒎− 𝒈 𝒙 e) 𝒍𝒊𝒎+ 𝒈 𝒙 f) 𝒍𝒊𝒎 𝒈 𝒙 𝒙→𝟓 𝒙→𝟓 𝒙→𝟓 Answer: a) 𝒍𝒊𝒎− 𝒈 𝒙 = 3 b) 𝒍𝒊𝒎+ 𝒈 𝒙 = 1 c) 𝒍𝒊𝒎 𝒈 𝒙 = Does not Exist 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 g) g(5) = 1 d) 𝒍𝒊𝒎− 𝒈 𝒙 = 2 e) 𝒍𝒊𝒎+ 𝒈 𝒙 = 2 f) 𝒍𝒊𝒎 𝒈 𝒙 = 2 𝒙→𝟓 𝒙→𝟓 𝒙→𝟓 Dr. Sameh Basha Ex: Use the given graph of f to state the value of each quantity, if it exists. If it does not exist. a) 𝒍𝒊𝒎− 𝒇 𝒙 b) 𝒍𝒊𝒎+ 𝒇 𝒙 c) 𝒍𝒊𝒎 𝒇 𝒙 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 d) 𝒇(𝟐) e)𝒍𝒊𝒎 𝒇 𝒙 f)𝒇(𝟒) 𝒙→𝟒 Answer: a) 𝒍𝒊𝒎− 𝒇 𝒙 = 3 b) 𝒍𝒊𝒎+ 𝒇 𝒙 = 1 c) 𝒍𝒊𝒎 𝒇 𝒙 = Does not Exist 𝒙→𝟐 𝒙→𝟐 𝒙→𝟐 d) 𝒇 𝟐 = 3 e)𝒍𝒊𝒎 𝒇 𝒙 = 4 f)𝒇 𝟒 = Does not Exist 𝒙→𝟒 Dr. Sameh Basha Infinite Limits 𝟏 Ex: Find 𝒍𝒊𝒎 𝟐 (if it exists) 𝒙→𝟎 𝒙 Answer: As 𝑥 becomes close to 0, 𝑥 2 also becomes close to 0, 1 and becomes very large. 𝑥2 𝟏 lim 𝟐 = ∞ 𝒙→𝟎 𝒙 Dr. Sameh Basha  lim 𝒇(𝒙) = ∞ 𝒙→𝒂 to indicate that the values of 𝒇 𝒙 tend to become larger and larger (or “increase without bound”) as 𝑥 becomes closer and closer to 𝑎. (𝑓(𝑥) → ∞ as 𝑥 → 𝑎 ) Dr. Sameh Basha Infinite Limits 𝟏 Ex: Find 𝒍𝒊𝒎 − 𝟐 (if it exists) 𝒙→𝟎 𝒙 Answer: As 𝑥 becomes close to 0, 𝑥 2 also becomes close to 0, 1 and - 2 becomes very large negative. 𝑥 𝟏 lim − 𝟐 = −∞ 𝒙→𝟎 𝒙 Dr. Sameh Basha  lim 𝒇 𝒙 = −∞ 𝒙→𝒂 to indicate that the values of 𝒇 𝒙 tend to become larger and larger negative (or “decrease without bound”) as 𝑥 becomes closer and closer to 𝑎. (𝑓 𝑥 → −∞ as 𝑥 → 𝑎 ) Dr. Sameh Basha Similar definitions can be given for the one-sided infinite limits: 𝒍𝒊𝒎 𝒇(𝒙) = ∞ 𝒍𝒊𝒎 𝒇(𝒙) = ∞ 𝒙→𝒂− 𝒙→𝒂+ 𝒍𝒊𝒎 𝒇 𝒙 = −∞ 𝒍𝒊𝒎 𝒇 𝒙 = −∞ 𝒙→𝒂− 𝒙→𝒂+ Note that: 𝒙 → 𝒂− means that we consider only values of x that are less than a, 𝒙 → 𝟐+ means that we consider only values of x that are greater than a Dr. Sameh Basha Vertical asymptote Dr. Sameh Basha The line 𝒙 = 𝒂 is a vertical asymptote line in each of the following: Dr. Sameh Basha For the function A whose graph is shown, state the following. 𝒂) 𝒍𝒊𝒎 𝑨 𝒙 𝒃) 𝒍𝒊𝒎+ 𝑨 𝒙 𝒙→−𝟑 𝒙→𝟐 𝒄) 𝒍𝒊𝒎 𝑨 𝒙 𝒅) 𝒍𝒊𝒎 𝑨 𝒙 𝒙→𝟐− 𝒙→−𝟏 e) The equations of the vertical asymptotes. Answer: 𝒂) 𝒍𝒊𝒎 𝑨 𝒙 = ∞ 𝒃) 𝒍𝒊𝒎+ 𝑨 𝒙 = ∞ 𝒙→−𝟑 𝒙→𝟐 𝒄) 𝒍𝒊𝒎 𝑨 𝒙 = −∞ 𝒅) 𝒍𝒊𝒎 𝑨 𝒙 = −∞ 𝒙→𝟐 − 𝒙→−𝟏 e) The equations of the vertical asymptotes. 𝑥 = −3, 𝑥 = −1, 𝑥 = 2 Dr. Sameh Basha Calculating Limits Using the Limit Laws Dr. Sameh Basha Calculating Limits Using the Limit Laws Dr. Sameh Basha Direct Substitution Property Dr. Sameh Basha Ex: Evaluate the following limits (if exists): 𝟐 𝒙𝟐 −𝟑 1)𝒍𝒊𝒎(𝟐𝒙 − 𝟑𝒙 + 𝟒) 2)𝒍𝒊𝒎 𝟐 𝒙→𝟓 𝒙→𝟐 𝒙 +𝟐 Answer: Answer: 𝒍𝒊𝒎 𝟐𝒙𝟐 − 𝟑𝒙 + 𝟒 𝒙𝟐 −𝟑 𝟐𝟐 −𝟑 𝟒−𝟑 𝟏 𝒙→𝟓 𝒍𝒊𝒎 𝟐 = 𝟐 = = = 𝟐 𝟓 𝟐−𝟑 𝟓 +𝟒 𝒙→𝟐 𝒙 +𝟐 𝟐 +𝟐 𝟒+𝟐 𝟔 = 𝟓𝟎 − 𝟏𝟓 + 𝟒 = 𝟑𝟗 𝒙𝟐 −𝟗 𝒙𝟐 −𝟒 3)𝒍𝒊𝒎 𝟐 4)𝒍𝒊𝒎 𝒙→𝟑 𝒙 +𝟑 𝒙→𝟐 𝒙−𝟐 Answer: Answer: 𝒙𝟐 − 𝟗 𝟑𝟐 − 𝟗 𝟗 − 𝟗 𝒛𝒆𝒓𝒐 𝒙𝟐 −𝟒 𝒙−𝟐 𝒙+𝟐 𝒍𝒊𝒎 𝟐 = 𝟐 = = = 𝒁𝒆𝒓𝒐 𝒍𝒊𝒎 𝟐 =𝒍𝒊𝒎 = 𝒍𝒊𝒎 𝒙 + 𝟐 = 𝟒 𝒙→𝟑 𝒙 + 𝟑 𝟑 +𝟑 𝟗+𝟑 𝟏𝟐 𝒙→𝟐 𝒙 −𝟐 𝒙→𝟐 𝒙−𝟐 𝒙→𝟐 Dr. Sameh Basha Note: Dr. Sameh Basha Theorem 1: Dr. Sameh Basha Theorem 2: Dr. Sameh Basha Theorem 3: Squeeze Theorem (Sandwich Theorem) Dr. Sameh Basha Example: Let 𝑓 𝑥 = ቊ 𝑥 − 4; 𝑥 > 4 8 − 2𝑥; 𝑥 < 4 Evaluate 𝒍𝒊𝒎𝒇(𝒙) (if exists) 𝒙→𝟒 Answer: 𝒍𝒊𝒎 𝒇 𝒙 = 𝒍𝒊𝒎+ 𝑥 − 4 = 4 − 4 = 𝑍𝑒𝑟𝑜 = 𝑍𝑒𝑟𝑜 𝒙→𝟒+ 𝒙→𝟒 𝒍𝒊𝒎− 𝒇 𝒙 = 𝒍𝒊𝒎− 8 − 2𝑥 = 8 − 2 4 = 8 − 8 = 𝑍𝑒𝑟𝑜 𝒙→𝟒 𝒙→𝟒 ∵ 𝒍𝒊𝒎 + 𝒇 𝒙 = 𝒍𝒊𝒎− 𝒇 𝒙 = 𝑍𝑒𝑟𝑜 𝒙→𝟒 𝒙→𝟒 ∴ 𝒍𝒊𝒎 𝒇(𝒙) = 𝒁𝒆𝒓𝒐 𝒙→𝟒 Dr. Sameh Basha Example: Evaluate (if exists) 𝒍𝒊𝒎𝒇(𝒙) where, 𝒙→𝟎 𝑓 𝑥 = 𝑥 Answer: 𝒙; 𝒙≥𝟎 𝒇 𝒙 = 𝒙 =ቊ −𝒙; 𝒙