- 1. Functions Lesson 1 www.MathAcademy.sg c⃝ 2015 Math Academy www.MathAcademy.sg 1
- 2. Set notations [a, b] = {x ∈ R : a ≤ x ≤ b} (a, b] = {x ∈ R : a < x ≤ b} (a, b) = {x ∈ R : a < x < b} (a, ∞) = {x ∈ R : x > a} R = set of real numbers R+ = set of positive real numbers c⃝ 2015 Math Academy www.MathAcademy.sg 2
- 3. Rule, domain and range X f (X) 1 2 3 3 4 5 f Domain Range A function f , is deﬁned by its rule and domain. f : x → x + 2 rule , x ∈ [0, ∞). domain f (x) = x + 2, x ≥ 0. Remark: When we state a function, we must always state both its rule and domain. The domain Df , is the set of all possible x values. The range Rf , is the set of all possible y values.c⃝ 2015 Math Academy www.MathAcademy.sg 3
- 4. Rule, domain and range X f (X) 1 2 3 3 4 5 f Domain Range A function f , is deﬁned by its rule and domain. f : x → x + 2 rule , x ∈ [0, ∞). domain f (x) = x + 2, x ≥ 0. Remark: When we state a function, we must always state both its rule and domain. The domain Df , is the set of all possible x values. The range Rf , is the set of all possible y values.c⃝ 2015 Math Academy www.MathAcademy.sg 4
- 5. Finding Range Sketch the graph to ﬁnd Rf . Rf is the range of y-values that the graph takes. c⃝ 2015 Math Academy www.MathAcademy.sg 5
- 6. Example (1) Find the range of the following: (a) f : x → x2 − 1, x ∈ [−1, 2], (b) g : x → ex + 1, x ∈ R. Solution: (a) −1 1 2 −1 3 x y ∴ Rf = [−1, 3]. (b) 1 x y ∴ Rg = (1, ∞). c⃝ 2015 Math Academy www.MathAcademy.sg 6
- 7. Horizontal line test A function is said to be 1 − 1 if for every y ∈ Rf , there is only ONE x such that f (x) = y. −1 1 1 x y It is not 1-1 since the line y = 1 cuts the graph twice. x y It is 1-1 since every horizontal line cuts the graph at most once. Horizontal Line Test f is NOT a 1 − 1 function if there is a horizontal line that cuts the graph at MORE THAN ONE point. f IS a 1 − 1 function if any horizontal line y = a, where a ∈ Rf , cuts the graph AT ONLY ONE point. c⃝ 2015 Math Academy www.MathAcademy.sg 7
- 8. Horizontal line test A function is said to be 1 − 1 if for every y ∈ Rf , there is only ONE x such that f (x) = y. −1 1 1 x y It is not 1-1 since the line y = 1 cuts the graph twice. x y It is 1-1 since every horizontal line cuts the graph at most once. Horizontal Line Test f is NOT a 1 − 1 function if there is a horizontal line that cuts the graph at MORE THAN ONE point. f IS a 1 − 1 function if any horizontal line y = a, where a ∈ Rf , cuts the graph AT ONLY ONE point. c⃝ 2015 Math Academy www.MathAcademy.sg 8
- 9. Diﬀerentiation test for 1-1 f is 1 − 1 if f ′ (x) > 0 for ALL x in the domain (strictly increasing functions) or f ′ (x) < 0 for ALL x in the domain (strictly decreasing functions) c⃝ 2015 Math Academy www.MathAcademy.sg 9
- 10. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (a) With the aid of a diagram, explain why f is not 1-1. [2] Show not 1-1 through graph Sketch the graph, give the equation of a SPECIFIC horizontal line that cuts the graph in at least 2 points. (a) −3 1 3 y = f (x) x y The line y = 1 cuts the graph of y = f (x) more than once, therefore f is not 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 10
- 11. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (a) With the aid of a diagram, explain why f is not 1-1. [2] Show not 1-1 through graph Sketch the graph, give the equation of a SPECIFIC horizontal line that cuts the graph in at least 2 points. (a) −3 1 3 y = f (x) x y The line y = 1 cuts the graph of y = f (x) more than once, therefore f is not 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 11
- 12. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (b) If the domain of f is restricted to the set {x ∈ R : x ≥ k}, state with a reason the least value of k for which the function is 1-1. [2] (b) −3 1 3 y = f (x) x y Show 1-1 through graph Sketch the graph, explain that any horizontal line cuts the graph at only 1 point. Least value of k is 1. When x ≥ 1, any horizontal line y = a for a ∈ Rf , cuts the graph of y = f (x) at only 1 point, hence it is 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 12
- 13. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (b) If the domain of f is restricted to the set {x ∈ R : x ≥ k}, state with a reason the least value of k for which the function is 1-1. [2] (b) −3 1 3 y = f (x) x y Show 1-1 through graph Sketch the graph, explain that any horizontal line cuts the graph at only 1 point. Least value of k is 1. When x ≥ 1, any horizontal line y = a for a ∈ Rf , cuts the graph of y = f (x) at only 1 point, hence it is 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 13
- 14. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (b) If the domain of f is restricted to the set {x ∈ R : x ≥ k}, state with a reason the least value of k for which the function is 1-1. [2] (b) −3 1 3 y = f (x) x y Show 1-1 through graph Sketch the graph, explain that any horizontal line cuts the graph at only 1 point. Least value of k is 1. When x ≥ 1, any horizontal line y = a for a ∈ Rf , cuts the graph of y = f (x) at only 1 point, hence it is 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 14
- 15. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (c) By considering the derivative of f (x), prove that f is a one-one function for the domain you have found in (b). [2] Show 1-1 through diﬀerentiation Show that f ′ (x) is either > 0 or < 0 for ∀x ∈ Df . block: This method cannot be used to show that a function is not 1-1. (c) For x ≥ 1, f (x) = x2 + 2x − 3. f ′ (x) = 2x + 2 ≥ 2(1) + 2 since x ≥ 1 = 4 > 0 Since f ′ (x) > 0 for x ≥ 1, f is a strictly increasing function, and hence it is 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 15
- 16. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (c) By considering the derivative of f (x), prove that f is a one-one function for the domain you have found in (b). [2] Show 1-1 through diﬀerentiation Show that f ′ (x) is either > 0 or < 0 for ∀x ∈ Df . block: This method cannot be used to show that a function is not 1-1. (c) For x ≥ 1, f (x) = x2 + 2x − 3. f ′ (x) = 2x + 2 ≥ 2(1) + 2 since x ≥ 1 = 4 > 0 Since f ′ (x) > 0 for x ≥ 1, f is a strictly increasing function, and hence it is 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 16
- 17. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (c) By considering the derivative of f (x), prove that f is a one-one function for the domain you have found in (b). [2] Show 1-1 through diﬀerentiation Show that f ′ (x) is either > 0 or < 0 for ∀x ∈ Df . block: This method cannot be used to show that a function is not 1-1. (c) For x ≥ 1, f (x) = x2 + 2x − 3. f ′ (x) = 2x + 2 ≥ 2(1) + 2 since x ≥ 1 = 4 > 0 Since f ′ (x) > 0 for x ≥ 1, f is a strictly increasing function, and hence it is 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 17
- 18. Example (2) The function f is deﬁned by f : x → | − x2 − 2x + 3|, x ∈ R. (c) By considering the derivative of f (x), prove that f is a one-one function for the domain you have found in (b). [2] Show 1-1 through diﬀerentiation Show that f ′ (x) is either > 0 or < 0 for ∀x ∈ Df . block: This method cannot be used to show that a function is not 1-1. (c) For x ≥ 1, f (x) = x2 + 2x − 3. f ′ (x) = 2x + 2 ≥ 2(1) + 2 since x ≥ 1 = 4 > 0 Since f ′ (x) > 0 for x ≥ 1, f is a strictly increasing function, and hence it is 1-1. c⃝ 2015 Math Academy www.MathAcademy.sg 18
- 19. Inverse functions Df Rf x y f f (x) = y f −1 f −1 (y) = x Properties of inverse function 1. For f −1 to exist, f must be a 1-1 function. 2. Df −1 = Rf . 3. Rf −1 = Df . 4. (f −1 )−1 = f . c⃝ 2015 Math Academy www.MathAcademy.sg 19
- 20. Inverse functions Geometrical relationship between a function and its inverse (i) The graph of f −1 is the reﬂection of the graph f about the line y = x. (ii) (a, b) lies on f ⇔ (b, a) lies on f −1 . (iii) x = k is an asymptote of f ⇔ y = k is an asymptote of f −1 Remark: The notation f −1 stands for the inverse function of f . It is not the same as 1 f . c⃝ 2015 Math Academy www.MathAcademy.sg 20
- 21. Inverse functions Geometrical relationship between a function and its inverse (i) The graph of f −1 is the reﬂection of the graph f about the line y = x. (ii) (a, b) lies on f ⇔ (b, a) lies on f −1 . (iii) x = k is an asymptote of f ⇔ y = k is an asymptote of f −1 Remark: The notation f −1 stands for the inverse function of f . It is not the same as 1 f . c⃝ 2015 Math Academy www.MathAcademy.sg 21
- 22. Inverse functions Geometrical relationship between a function and its inverse (i) The graph of f −1 is the reﬂection of the graph f about the line y = x. (ii) (a, b) lies on f ⇔ (b, a) lies on f −1 . (iii) x = k is an asymptote of f ⇔ y = k is an asymptote of f −1 Remark: The notation f −1 stands for the inverse function of f . It is not the same as 1 f . c⃝ 2015 Math Academy www.MathAcademy.sg 22
- 23. Inverse functions Geometrical relationship between a function and its inverse (i) The graph of f −1 is the reﬂection of the graph f about the line y = x. (ii) (a, b) lies on f ⇔ (b, a) lies on f −1 . (iii) x = k is an asymptote of f ⇔ y = k is an asymptote of f −1 Remark: The notation f −1 stands for the inverse function of f . It is not the same as 1 f . c⃝ 2015 Math Academy www.MathAcademy.sg 23
- 24. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (i) Sketch the graph of y = f (x). Your sketch should indicate the position of the graph in relation to the origin. (i) 4 1 y = f (x) x y c⃝ 2015 Math Academy www.MathAcademy.sg 24
- 25. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (i) Sketch the graph of y = f (x). Your sketch should indicate the position of the graph in relation to the origin. (i) 4 1 y = f (x) x y c⃝ 2015 Math Academy www.MathAcademy.sg 25
- 26. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (ii) Show that the inverse function f −1 exists and ﬁnd f −1 (x) in similar form. Showing inverse exists To show f −1 exists, we only need to show that f is 1-1. (ii) Every horizontal line y = a for a > 1 cuts the graph of y = f (x) at only 1 point, hence it is 1-1 and f −1 exists. y = x2 − 8x + 17 = x2 − 8x + (−4)2 − (−4)2 + 17 = (x − 4)2 + 1 x − 4 = ± √ y − 1 x = ± √ y − 1 + 4 x = √ y − 1 + 4 or − √ y − 1 + 4 [rej since x > 4] ∴ f −1 (x) = √ x − 1 + 4, x > 1. Note: You must state the domain of f −1 !c⃝ 2015 Math Academy www.MathAcademy.sg 26
- 27. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (ii) Show that the inverse function f −1 exists and ﬁnd f −1 (x) in similar form. Showing inverse exists To show f −1 exists, we only need to show that f is 1-1. (ii) Every horizontal line y = a for a > 1 cuts the graph of y = f (x) at only 1 point, hence it is 1-1 and f −1 exists. y = x2 − 8x + 17 = x2 − 8x + (−4)2 − (−4)2 + 17 = (x − 4)2 + 1 x − 4 = ± √ y − 1 x = ± √ y − 1 + 4 x = √ y − 1 + 4 or − √ y − 1 + 4 [rej since x > 4] ∴ f −1 (x) = √ x − 1 + 4, x > 1. Note: You must state the domain of f −1 !c⃝ 2015 Math Academy www.MathAcademy.sg 27
- 28. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (ii) Show that the inverse function f −1 exists and ﬁnd f −1 (x) in similar form. Showing inverse exists To show f −1 exists, we only need to show that f is 1-1. (ii) Every horizontal line y = a for a > 1 cuts the graph of y = f (x) at only 1 point, hence it is 1-1 and f −1 exists. y = x2 − 8x + 17 = x2 − 8x + (−4)2 − (−4)2 + 17 = (x − 4)2 + 1 x − 4 = ± √ y − 1 x = ± √ y − 1 + 4 x = √ y − 1 + 4 or − √ y − 1 + 4 [rej since x > 4] ∴ f −1 (x) = √ x − 1 + 4, x > 1. Note: You must state the domain of f −1 !c⃝ 2015 Math Academy www.MathAcademy.sg 28
- 29. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (ii) Show that the inverse function f −1 exists and ﬁnd f −1 (x) in similar form. Showing inverse exists To show f −1 exists, we only need to show that f is 1-1. (ii) Every horizontal line y = a for a > 1 cuts the graph of y = f (x) at only 1 point, hence it is 1-1 and f −1 exists. y = x2 − 8x + 17 = x2 − 8x + (−4)2 − (−4)2 + 17 = (x − 4)2 + 1 x − 4 = ± √ y − 1 x = ± √ y − 1 + 4 x = √ y − 1 + 4 or − √ y − 1 + 4 [rej since x > 4] ∴ f −1 (x) = √ x − 1 + 4, x > 1. Note: You must state the domain of f −1 !c⃝ 2015 Math Academy www.MathAcademy.sg 29
- 30. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (ii) Show that the inverse function f −1 exists and ﬁnd f −1 (x) in similar form. Showing inverse exists To show f −1 exists, we only need to show that f is 1-1. (ii) Every horizontal line y = a for a > 1 cuts the graph of y = f (x) at only 1 point, hence it is 1-1 and f −1 exists. y = x2 − 8x + 17 = x2 − 8x + (−4)2 − (−4)2 + 17 = (x − 4)2 + 1 x − 4 = ± √ y − 1 x = ± √ y − 1 + 4 x = √ y − 1 + 4 or − √ y − 1 + 4 [rej since x > 4] ∴ f −1 (x) = √ x − 1 + 4, x > 1. Note: You must state the domain of f −1 !c⃝ 2015 Math Academy www.MathAcademy.sg 30
- 31. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (iii) On the same diagram as in part (i), sketch the graph of y = f −1 . (iii) (i) The graph of f −1 is the reﬂection of the graph f about the line y = x. (ii) (a, b) lies on f ⇔ (b, a) lies on f −1. (iii) x = k is an asymptote of f ⇔ y = k is an asymptote of f −1 1 4 1 4 y = f (x) y = f −1 (x) y = x x y c⃝ 2015 Math Academy www.MathAcademy.sg 31
- 32. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (iv) Write down the equation of the line in which the graph of y = f (x) must be reﬂected in order to obtain the graph of y = f −1 , and hence ﬁnd the exact solution of the equation f (x) = f −1 (x). (iv) It must be reﬂected along the line y = x. Since f and f −1 intersect at the line y = x, ﬁnding exact solution of the equation f (x) = f −1 (x) is equivalent to ﬁnding f (x) = x x2 − 8x + 17 = x x2 − 9x + 17 = 0 x = 9 ± √ 92 − 4(1)(17) 2 = 9 + √ 13 2 or 9 − √ 13 2 (rej as it is not in Df ) c⃝ 2015 Math Academy www.MathAcademy.sg 32
- 33. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (iv) Write down the equation of the line in which the graph of y = f (x) must be reﬂected in order to obtain the graph of y = f −1 , and hence ﬁnd the exact solution of the equation f (x) = f −1 (x). (iv) It must be reﬂected along the line y = x. Since f and f −1 intersect at the line y = x, ﬁnding exact solution of the equation f (x) = f −1 (x) is equivalent to ﬁnding f (x) = x x2 − 8x + 17 = x x2 − 9x + 17 = 0 x = 9 ± √ 92 − 4(1)(17) 2 = 9 + √ 13 2 or 9 − √ 13 2 (rej as it is not in Df ) c⃝ 2015 Math Academy www.MathAcademy.sg 33
- 34. Example (3) The function f is deﬁned by f : x → x2 − 8x + 17 for x > 4. (iv) Write down the equation of the line in which the graph of y = f (x) must be reﬂected in order to obtain the graph of y = f −1 , and hence ﬁnd the exact solution of the equation f (x) = f −1 (x). (iv) It must be reﬂected along the line y = x. Since f and f −1 intersect at the line y = x, ﬁnding exact solution of the equation f (x) = f −1 (x) is equivalent to ﬁnding f (x) = x x2 − 8x + 17 = x x2 − 9x + 17 = 0 x = 9 ± √ 92 − 4(1)(17) 2 = 9 + √ 13 2 or 9 − √ 13 2 (rej as it is not in Df ) c⃝ 2015 Math Academy www.MathAcademy.sg 34
- 35. Example (4) Function g is deﬁned by g : x → x 2 − 3x for x ∈ R, If the domain of g is restricted to the set {x ∈ R : x ≥ a}, ﬁnd the least value of a for which g−1 exists. Hence, ﬁnd g−1 and state its domain. [4] [a = 1.5, g−1 (x) = 3 2 + √ x + 9 4 , x ≥ − 9 4 ] 0 3 (1.5, -2.25) y = g(x) From the graph, g is 1-1 for x ≥ 1.5. Hence, g−1 exists when a = 1.5 y = x 2 − 3x = (x − 1.5) 2 − 2.25 y + 2.25 = (x − 1.5) 2 x − 1.5 = √ y + 2.25 or − √ y + 2.25 x = 1.5 + √ y + 2.25 or 1.5 − √ y + 2.25( rej since x ≥ 1.5) ∴ g−1 x = 1.5 + √ x + 2.25, x ≥ −2.25c⃝ 2015 Math Academy www.MathAcademy.sg 35
- 36. Example (4) Function g is deﬁned by g : x → x 2 − 3x for x ∈ R, If the domain of g is restricted to the set {x ∈ R : x ≥ a}, ﬁnd the least value of a for which g−1 exists. Hence, ﬁnd g−1 and state its domain. [4] [a = 1.5, g−1 (x) = 3 2 + √ x + 9 4 , x ≥ − 9 4 ] 0 3 (1.5, -2.25) y = g(x) From the graph, g is 1-1 for x ≥ 1.5. Hence, g−1 exists when a = 1.5 y = x 2 − 3x = (x − 1.5) 2 − 2.25 y + 2.25 = (x − 1.5) 2 x − 1.5 = √ y + 2.25 or − √ y + 2.25 x = 1.5 + √ y + 2.25 or 1.5 − √ y + 2.25( rej since x ≥ 1.5) ∴ g−1 x = 1.5 + √ x + 2.25, x ≥ −2.25c⃝ 2015 Math Academy www.MathAcademy.sg 36