1. Dr. Omar Al Jadaan Assistant Professor – Computer Science & Mathematics General Education Department Mathematics

2. COMBINATIONS OF FUNCTIONS AND INVERSE FUNCTIONS Combination of functions: If f(x) and g(x) are two functions then f(x) + g(x), f(x) – g(x), f(g(x)) and f(x)/g(x) are combination of two functions. Domain: The domain is the intersection of the domain of f(x) and g(x). In other words their domain is where the domain of f(x) overlaps the domain of g(x). Note that for f(x)/g(x) if g(x) ≠ 0.

3. Examples Find f+g(x), f – g(x), f(g(x)) and f/g(x) and their domain if f(x) = x 2 - 4 and g(x) = x + 2. Solution: f(x) + g(x) = x 2 -4 + x + 2 = x 2 + x -2 f(x) – g(x) = x 2 -4 –x -2 = x 2 -x – 6 f(g(x)) = f(x+2) = (x+2) 2 – 4 f/g(x) = (x 2 -4)/(x + 2) = (x -2)

4. Graph of example Graph of function f(x) = x 2 - 4 and g(x)= x+2 Domain: - ∞ ≤ x ≤ +∞ x y 0 2 -2 -4

5. Example Find f ○ g(x), g ○ f(x), f ○ g(-1), f ○ g(0) and g ○ f(1) for f(x) = x 2 - 1 and g(x)= x + 1. Solution: f ○ g(x) = f(g(x)) = f(x+1) =(x+1) 2 - 1 g ○ f(x) = g(f(x)) = g(x 2 - 1) = x 2 – 1 + 1 = x 2 f ○ g(-1) = 0 – 1 = -1 f ○ g(0) = 1 – 1 = 0 g ○ f(1) =1

6. Example Refer to the figure, the white line graph is of f(x) and the color is the graph of g(x). Find f ○ g(-1), f ○ g(3), f ○ g(5) and g ○ f(0). Solution: f ○ g(-1)=f(-2)=0 f ○ g(3)=f(-2)=0 f ○ g(5)=f(0)=-1 g ○ f(0)=g(-1)=-2 x y

7. Skill practice 1. f(x) = 3x 2 +x and g(x) = x – 4 a) Find f + g(x), f – g(x), fg(x), and f/g(x). b) Find f ○ g(x) and g ○ f(x). c) Find f ○ g(1) and g ○ f(0). 2. find f ○ g(x), and g ○ f(x). f(x) = (2x -3)/(x+4) and g(x) = x/(x-1)

8. Skill Practice 3. Refer to the figure, the white line graph is of f(x) and the color is the graph of g(x). Find f ○ g(1), f ○ g(4) and g ○ f(-2). x y

9. Skill Practice Solution: 1. a) f + g(x)= 3x 2 + 2x -4 f – g(x)= 3x 2 + 4 fg(x)= 3x 3 - 11x 2 -4x f/g(x)= (3x 2 + x)/(x-4) b) f ○ g(x)= 3x 2 – 23x + 44 g ○ f(x)= 3x 2 + x – 4 c) f ○ g(1)=f(-3) =24 g ○ f(0)=g(0)=-4

10. Skill Practice Solution: 2. f ○ g(x)= (-x+3)/(5x-4) g ○ f(x)= (2x-3)/(x-7) 3. f ○ g(1) =f(4)=1 f ○ g(4) =f(0)=-2 g ○ f(-2)=g(1)=4

11. INVERSE FUNCTIONS The inverse of a function f(x) is f -1 (x). We use functional decomposition to show that two functions are inverse of each other. In other words, f(x) is inverse of g(x) if f ○ g(x) = x and g ○ f(x) = x.

12. Example The graph of the function f(x) is given below. Find the f -1 (x). x y

13. Example Solution: Make a table of values for f(x) and switch the x and y columns for f -1 (x). To get the table for f -1 (x), switch x and y value. xy= f -1 (x) -3-5 0-3 10 31 55 xy=f(x) -5-3 0 01 13 55

14. Example Solution: the color graph is the graph of f -1 (x). It is a reflection of the graph of f(x) across the line y=x x y

15. Example Inverse function test: A function has an inverse if its graph passes test – if any horizontal line touches the graph in more than one place, then the function will not have inverse. Here it has an inverse function because any horizontal line touches the graph in one place only. x y

16. Skill Practice 1. Show that the following functions are inverses: a) f(x) = ½ x + 7 and g(x) = 2x – 14. b) f(x) = (x+2)/(x-3) and g(x) = (3x+2)/(x-1)

17. Skill practice 2. Does the given graph have an inverse function? If yes then the draw its graph. x y

18. Skill practice 3. Does the graph have an inverse function? If yes, find its inverse. x y 0 2 -2 -4

19. Skill practice 1. Solution: a) f(g(x)) = f(2x_14) = x g(f(x) = g( ½ x +7) = x b) f(g(x)) = f((3x+2)/(x-1)) = x g(f(x))= g((x+2)/(x-3))=x

20. Skill practice 2. Solution: Yes. The color graph is an inverse function. x y

21. Skill practice 3. Solution: No, because it Fails horizontal function test. It passes Through two points. x y 0 2 -2 -4

22. Graph of Exponential and Logarithmic functions Exponential function: It is given in the form f(x) = a x where a is any positive number except 1. The graph of the function f(x) = 2 x is given by:

23. Example The graph of f(x) = 2 x : x y 0 1

24. Logarithmic function: Rewrite the log 3 9 = 2. The base of the logarithm is the base of the exponent, so 3 will be raised to a power of 2. So it gives the following: log 3 9 = 2 rewritten as an exponent is 3 2 = 9 In general y = log a x means x = a y

25. Example The graph of f(x) = log 2 x : x y 0 1

26. Skill practice 1. rewrite the logarithmic equations as exponential equations. 1. log 4 16=2 2. log 100 10 = ½ 2. rewrite the exponential equations as logarithmic equations. 1. 3 4 = 812. 5 3 = 125 3. draw the graph of the following functions: 1. y = 3 x 2. x = log 3 y

27. Skill practice 4. Suppose a bacteria culture contains 2500 bacteria at 1:00 and at 1:30 there are 6000. what is the hourly growth rate? 5. solve the following equations: a) Log (x+2) = 2b) 2 ln (x) – ln(x+1)=0 c) log(x) (x+1) =1 d) ln (x+1)/x = 3

28. Skill practice 1. Solution: 1. 4 2 = 16 2. 100 1/2 = 10 2. Solution: 1. log 3 81= 4 2. log 5 125 = 3

29. Skill practice Solution 3. The graph of f(x)=y= 3 x The graph of log 3 y=x x y 0 1

30. Skill practice 4. Solution: Initially t = 0 hours, the population is 2500 it means you substitute t =0 in the growth formula n(t) = n 0 e rt n(0) = n 0 e 0 2500 = n 0 Half hour later, the population of bacteria is 6000, so t = 0.5. 6000 = 2500 e 0.5r 2.4 = e 0.5r ln 2.4 = 0.5r r = (ln2.4)/0.5 = 1.75 The bacteria are increasing at the rate of 175% per hour.

31. Trigonometric function Trigonometry has been used for over 2 thousand years to solve many real world problems, among them surveying, navigation and problems in engineering and medical sciences. The unit circle is the basis of analytic trigonometry. Angles have two sides, the initial side and the terminal side. A positive angle rotates counterclockwise, A negative angle rotates clockwise,.

32. Angles on the unit circle are measured in radians. 2 π radians = 360 0. Initial side Terminal side p(x,y) 1 θ There are six Trig functions: Sinθ=y, cosθ=x, tanθ=y/x, secθ=1/x, cscθ=1/y, cotθ=x/y

33. Examples 1. Find all six trigonometric functions for θ. a) The terminal point for θ is (24/25, 7/25). b) θ = π /3. a) sin θ = 7/25, cos θ = 24/25, csc θ =25/7 sec θ = 25/24, tan θ = 7/24, and cot θ =24/7 b) cos π /3 = ½ sin π /3 = (3) 1/2 /2 sec π /3 = 2 csc = 2/ (3) 1/2 tan π /3 = (3) 1/2 cot π /3 = 1/ (3) 1/2

34. Skill practice 1. Find all six trigonometric functions for θ. a) the terminal points for θ is (3/5,4/5). b) θ = π /6

35. Skill practice Solution: 1. a) sin θ = 4/5, cos θ = 3/5, csc θ =5/4 sec θ = 5/3, tan θ = 4/3, and cot θ =3/4 b) π /6 = 30 0. So the value of sin 30 0 = 1/2, cos 30 0 = (3) 1/2 /2, csc 30 0 =2 sec 30 0 = 2/ (3) 1/2, tan 30 0 = 1/ (3) 1/2, and cot 30 0 = (3) 1/2

36. Graph of trigonometric functions The graph of trigonometric function is a record of each cycle around the circle. For the function f(x)=sin x, x is the angle and f(x) is the y- coordinate of the terminal point determined by the angle x.

37. Graph of sin x function For x values select the interval -2 π to 2 π. 1 2π2π -2π x y π-π-π

38. Graph of cos x function For x values select the interval -2 π to 2 π. 1 2π2π -2π x y π-π-π

39. Graph of tan x function For x values select the interval -2 π to 2 π, if x≠ π /2, - π /2, 3 π /2, -3 π /2 1 2π2π -2π x y π -π-π

40. Examples 1. Find all six trigonometric ratios for θ. 2. A person is standing 300 feet from the base of a five story building. He estimates that the angle of elevation to the top of the building is 63 0. Approximately how tall is the building? 3 5 4 θ

41. Examples 3. Solve the triangle. 3 c a 30

42. Examples Solution: 1. sin θ = 3/5, cos θ = 4/5, tan θ = 3/4, sec θ =5/4, csc θ = 5/3 and cot θ = 4/3. 2. We need to find b in the following triagle: tan 63 0 = b/300 b = 300 * 1.96 = 588 So the building is 588 feet tall. 300 b 63

43. Example Solution: 3. sin 30 0 = 3/c, so ½ = 3/c, c = 6. a 2 + 3 2 = 6 2 a 2 = 36 -9 = 27 a = (27) 1/2 Another angle = 90 -30 = 60 0.

44. Skill practice 1. Find all trigonometric ratios for θ. 2. A plane is flying at an altitude of 5000 feet. The angle of elevation to the plane from a car traveling on a highway is about 38.7 0. How far apart are the plane and car? θ 1 3 2

45. Skill practice 3. Solve the triangle. 60 b 4 a

46. Inverse trigonometric function Only one-to-one function can have inverses, and the trigonometric functions are certainly not one-to-one. We can limit their domain and force them to be one-to-one. Limiting the sin function to the interval x = – π /2 to π /2 and the range is [- 1, 1]. If we limit the cos function to the interval from x = 0 to x = π, then we will have another one-t-one function. We express inverse of sin function as f(x)= arc sin(x) or sin -1.

47. Inverse trigonometric Graph of inverse trigonometric functions. π/2 sin xcos xtan x π/2 1 1 π/2 -π/2 π

48. Example arc sin 2 1/2 /2 = π /4 arc tan 3 1/2 = π /3 arc cos (-1) = π