industrial mathematics - i

1. industrial mathematics - i TIP – FTP – UB

2. function industrial mathematics - I

3. What is function ? • Imagine : playing golf, putting a golfball into the hole. • A function is transforming an input x into an output y = f(x). x f y f : x  y / y=f(x) f : x  y / y=f(x) y=f(x)=x2

4. What is function ? • (Try) Which of the following equations is a function ? (a) y = 1 – x2 (b) y = Functions are rules, (c) y = but not all rules are functions. • Function is a relation between a set of inputs and a set of permissible outputs, with a property that each input is related to exactly one output. • Function is a mapping or equivalent rule which connected each object in a sets (domain), with a unique value of f(x) from another sets (range/codomain).

5. Domain, codomain, range • If f mapped or related x  A to y  B, it is : - said that y is a map from x - written as f : x  y or y = f(x) • Sets y  B which is map from x  A is called range or result area. f (a) = 1 range R = {1, 2, 3, 4] f (b) = 2 f (c) = 3 f (d) = 4

6. Domain, codomain, range • Domain = all the input numbers x that a function can process. • Co-domain = all the numbers in the sets y. • Range = complete collection of numbers y that correspond to the numbers is the domain. • y =  domain is -1 ≤ x ≤ 1 , range is 0 ≤ y ≤ 1 • y = x3 , -2 ≤ x ≤ 3  range is -8 ≤ y ≤ 27

7. examples • Define the domain and range for these equations : (a) y = x3 , -2 ≤ x < 3 (b) y = x4 (c) y = , 0 ≤ x ≤ 6 • Let’s say f : R  R with f(x-1) = x2 + 5x, define : (a) f(x) (b) f(-3)

8. Operations of function • Operations of function can be a sum, substract, multiply, or divide with the rules are : • Example : If F(x) = and G(x) = define : a. F+G(x) b. F-G(x) c. F.G(x) d. F/G(x) e. F5

9. Composite function • Function composition is the combining operations of two functionssequentially resulting to another function (composite function). • Function composition is the application of one function to the results of another. y=f(x) z=g(y)/z=g(f(x)) mapping of x  A to z  C is a composition of f and g written (g o f)(x) = g(f(x))

10. Composite function • Composite function is always associative or not commutative. means  f o g ≠ g o f • Example : f : R  R and g : R  R f(x) = 3x – 1 and g(x) = 2x2 + 5 Define : a. (g o f)(x) and b. (f o g)(x) ! a. (g o f)(x)=g(f(x)) = g(3x – 1) = 2(3x – 1)2 + 5 = 2(9x2 – 6x + 1) + 5 = 18x2 – 12x + 2 + 5 = 18x2 – 12x + 7 b. (f o g)(x) = …..??

11. Composite function • How to define a function from a known function composition ? • Example : Given f(x) = 3x – 1 and (f o g)(x) = x2 + 5, define g(x) ! Answer : (f o g)(x) = x2 + 5 f(g(x)) = x2 + 5 3.g(x) – 1 = x2 + 5 3.g(x) = x2 + 6 g(x) = 1/3(x2 + 6) Try  Given g(x) = 2x2 + 2 and (g o f)(x) = x – 3 , define f(x) !

12. Inverse function • Invers function is a function that undoes another function : If an input x into the function f produces an output y, then putting y into the function g produces the output x  g is an invers function of f. If, f : A  B = f : {(a,b,c,1,2,3)|a,b,c  A and 1,2,3  B} Then f-1: B  A = f : {(1,2,3,a,b,c)|1,2,3  B and a,b,c A} • A function f that has an inverse is called invertible; denoted by f-1. f : x  y or y = f(x) f-1 : y  x or x = f-1(y)  y = f-1(x)

13. Inverse function • Inverse Function, another explanation.

14. InversE function • Example : Determine the inverse function from function f(x) = 2x – 6 y = f(x) = 2x – 6 y = 2x – 6 2x = y + 6 x = ½(y + 6) So, x = f-1(y) = ½ (y + 6)  f-1(x) = ½ (x + 6) • Now determine the inverses from this function !! :

15. Composition and Inverse function • How is the function is a combination of composition and invers function ? Function composition Invers function (reverse way) h = (g o f) h-1= f-1 o g-1 (g o f)-1 = f-1 o g-1 • Example : If f : R  R and g : R  R determined by function f(x) = x + 3 and g(x) = 5x – 2 , define (f o g)-1(x) !!

16. Composition and Inversefunction • Example : If f : R  R and g : R  R determined by function f(x) = x + 3 and g(x) = 5x – 2 , define (f o g)-1(x) !! Solution 1 = Find (f o g)(x) first, then define (f o g)-1(x) (f o g)(x) = f(g(x)) = (5x – 2) + 3 y = 5x + 1 5x = y – 1 x = 1/5(y – 1) = 1/5y – 1/5 So, (f o g)-1(x) = 1/5x – 1/5

17. Composition and Inversefunction • Example : If f : R  R and g : R  R determined by function f(x) = x + 3 and g(x) = 5x – 2 , define (f o g)-1(x) !! Solution 2 = Find f-1(x) and g-1(x) first, then use (f o g)-1(x) = (g-1 o f-1)(x) (f o g)-1(x) = (g-1 o f-1)(x) = g-1(f-1(x)) = 1/5(x – 3) + 2/5 = 1/5x – 3/5 + 2/5 = 1/5x – 1/5

18. Tip application

19. TASK 1. If f(x) = 2x + 1 and g(x) = , determine (g o f)-1(x) ! 2. If f(x) = and g(x) = 2x – 1 , determine (fog)-1(x) ! 3. If , find f-1(1) ! 4. f(x) = 2x – 3 , f-1(-1) = ….. 5. If f(x) = and (f o g)(x) = 2x – 1 , find g(x) ! 6. If f(x) = 2x – 1 for –2 < x < 4 and g(x) = for 3 < x < 5 , find the domain and range of ! 7. If f(x+2) = 2x3– 4x + 3

20. TASK score (1). If f(x) = 2x + 1 and g(x) = , determine (g o f)-1(x) ! (gof)(x) = (10) (gof)-1(x) = (15) OR g-1(x) = (5) f-1(x) = (5) (gof)-1(x) = (f-1o g-1) (x) = (15)

21. TASK score (2). If f(x) = and g(x) = 2x – 1 , determine (fog)-1(x) ! (fog)(x) = (10) (fog)-1(x) = (15) OR f-1(x) = (5) g-1(x) = (5) (fog)-1(x) = (g-1 o f-1) = (15)

22. TASK score (3). If , find f-1(1) ! (10)  (5) (4). f(x) = 2x – 3 , f-1(-1) = ….. f-1(x) = (10)  f-1(-1) = = 1 (5)

23. TASK score (5). If f(x) = and (f o g)(x) = 2x – 1 , find g(x) ! (fog)(x) = f(g(x)) = 2x – 1  (5) g(x) = (5) (6) (5) (7) (5)

24. Thank you Industrial mathematics -1