Lesson 1: Functions as ModelsA function can be a simple mathemical model or a piece of larger model. Recallthat a functon is just a rule or law f, that expresses the dependency of a variablequantity, y, on another variable quantity x.Example 1: The cost of a pound of orange juice for three consecutuve weekis given by the table below:The Price of Orange Juice Week Week 1 Week 2 Week 3Cost 200 215 230What will be the cost of a pound of orange juice be in Week 4?Solution: The actual cost of a pound of orange juice in Week 4 will be de-termined by a number of factors, such as orange juice production, distribution,sales, etc. These factors are the natural law governing orange juice cost.

Therecent cost of orange juice can be model as:x= the number of weeks since Week 1P(x) = the cost of a pound of orange juice at time x, in pesosThe table above have shown us the details: P(0) = 200, P(1) = 215, and P(2) =230. Then summarizing this information with the function will show us: P(x)= 200 + (15)x.Using the model, we can deduce that P(3) = 200 +(15)(3) = 245 pesos. We cannow predict that the cost of a pound of orange juice in Week 4 will be 245 pesos.

Note that this may or may not be accurate. The model between the relation-ship of orange juice and it’s cost is based entirely on an observation of previouspatterns.Example 2.1Lesson 2: Evaluating FunctionsTo evaluate a function1.

Substitute the given value in the function of x.2.Replace all the variable xwith the value of the function.3.

Then compute and simplify the given function.Example 1: Given the function: f(x ) = 2 x+ 1, nd f(6).Substitute 6 in place holder x,f(6) = 2 x+ 1Replace all the variable of xwith 6,f(6) = 2(6) + 1Then compute function. f(6) = 12 + 1f (6) = 13Therefore, f(6) = 13. It can also write in ordered pair (6,13).

Example 2: Given the function f(x ) = x2+ 2 x+ 4 when x= 4. Substitute-4 in the place holder x,f( 4) = x2+ 2 x+ 4Replace the all the variables with 6, f( 4) = ( 4) 2+ 2( 4) + 4f ( 4) = (16) + ( 8) + 4f ( 4) = 12Therefore, f( 4) = 12 or simply as ( 4;12) :Example 3: Given g(x ) = x2+ 2 x- 1. Find g(2y).

Answer in terms of y.g(2 y) = x2+ 2 x 1g (2 y) = (2 y)2+ 2(2 y) 1g (2 y) = 4 y2+ 4 y 1Therefore, 4( y)2+ 4 y 1:2Example 4: Givenf(x ) = 2 x2+ 4 x- 12, nd f(2 x+ 4).Solution:f(2 x+ 4) = 2 x2+ 4 x 12= 2(2 x+ 4) 2+ 4(2 x+ 4) 12= 2(2 x+ 4)(2 x+ 4) + 4(2 x+ 4) 12= 2(4 x2+ 16 x+ 16) + 4(2 x+ 4) 12= (8 x2+ 32 x+ 32) + (8 x+ 16) 12Combine like terms f(2 x+ 4) = 8 x2+ (32 x+ 8 x) + (32 + 16 12)= 8 x2+ 40 x+ 36= 2(2 x2+ 10 x+ 9)Therefore, f(2 x+ 4) = 2(2 x2+ 10 x+ 9).Example 5: Given f(x ) = x2-x – 4.

If f(m ) = 8, compute the value of mSolution: Make the function f(x ) equivalent to f(m )x 2 x 4 = 8x 2 x 12 = 0( x 4)( x+ 3) = 0x 4 + 0; x+ 3 = 0x = 4; x= 3Therefore, the value of a can be either 4 or -3.3Exercises:Evaluate the functionsgiven:1. p(x ) = 2x + 1, nd p(-2)2. p(x ) = 4 x, nd p(-4)3.

g(n ) = 3 n2+ 6, nd g(8)4. g(x ) = x3+ 4 x, nd g(5)5. f(n ) = n3+ 3 n2, nd f(-5)6. w(a ) = a2+ 5 a, nd w(7)7. p(a ) = a3- 4 a, nd p(-6)8.

f(n ) = 4 3n+ 8 5, ndf(-1)9. f(x) = -1 + 1 4x;nd f(3 4)10. h(n) = n3+ 6 n, nd h(4)4Answers in Exercises:1. 52.

-163. 1984. 1455. -506. 847.

-1928. 4 159. – 13 1610. 885Lesson 3: Operations on FunctionsLet h(x) and g(x) be functions, and the operations on these two functions isshown below: Adding two functions as:(h+g)(x) = h(x)+g(x) Subtracting two functions as:(h-g)(x) = h(x) – g(x) Multiplying two functions as:(h g)(x) = h(x) g(c) Dividing two functions as:( h g)(x) = h(x ) g(x ) ; whereg(x ) 6= 0Example 1:Let f(x) = 4x + 5 and g(x) = 3x.

Find (f+g)(x), (f-g)(x), (f g)(x), and ( f g)(x). (f+g)(x) = (4x+5) + (3x) = 7x+5 (f-g)(x) = (4x+5) – (3x) = x+5 (f g)(x) = (4x+5) (3x) = 12 x2+5x (f g)(x) = 4x +5 3xExample 2:Let f(x)= 3x+2 and g(x)= 5x-1. Find (f+g)(x), (f-g)(x), (f g)(x), and ( f g)(x).

(f+g)(x) = (3x+2) + (5x-1) = 8x+1 (f-g)(x) = (3x+2) – (5x-1) = -2x+3 (f g) = (3x+2) (5x-1) = 15 x2+7x -2 (f g)(x) = 3x +2 5x 1Example 3:Let v(x) = x3and w(x) = 3 x2+5x. Find (v+w)(x), (v-w)(x), (v w)(x), and( v w)(x). (v+w)(x) = ( x3) + (3 x2+5x) = x3+ 3 x2+5x (v-w)(x) = ( x3) (3×2+5x) = x3 3x 2-5x (v w) = ( x3) (3×2+5x) = 3 x5+ 5 x4 (v w)(x) = ( x3 3x 2+5 x) = xx 2 x(3 x+5) = x2 3x +5Example 4:Let f(x) = 4 x3+ 2 x2+4x + 1 and g(x) = 3 x5+ 4 x2+8x-12. Find (f+g)(x),(f-g)(x), (f g)(x), and ( f g)(x).

6(f+g)(x) = (4 x3+ 2 x2+4x+1) + (3 x5+ 4 x2+8x-12) = 3 x5+ 4 x3+ 6 x2+12x-11 (f-g)(x) = (4 x3+ 2 x2+4x+1) – (3 x5+ 4 x2+8x-12) = 3x 5+ 4 x3 2x 2-4x+13 (f g)(x) = (4 x3+ 2 x2+4x+1) (3 x5+ 4 x2+8x-12)= 12 x8+ 6 x7+ 12 x6+ 19 x5+ 40 x4 16×3+ 12 x2 40x 12 (f g)(x) = (4×3+2 x2+4 x+1) (3×5+4 x2+8 x 12)Example 5:Let h(x) = 1 and g(x) = x4 x3+ x2-1. Find (h+g)(x), (h-g)(x), (h g)(x),and ( h g)(x). (h+g)(x) = (1) + ( x4 x3+ x2-1) = x4 x3+ x2 (h-g)(x) = (1) – ( x4 x3+ x2-1) = x4 x3+ x2+2 (h g)(x) = (1) (x 4 x3+ x2-1) = x4 x3+ x2-1 (h g)(x) = 1 x4 x3+ x2 17Exercises:1. If h(x) = 7x+3 and g(x) = 2 x2+1. Find (f+g)(x)2. If f(x) = x5-18 and g(x) = x2- 6x + 9, what is the vaue of (g-h)(x)?3.

If t(x) = 25 x5and s(x) = 55 x8, what is the value of ( t s)(x)?4. If v(x) = x3and w(x) = x2+ 4, solve (v w)(x)?5. If f(x) = 4x + 11 and g(x) = 5x + 9, nd (f+g)(x).6.

If f(z) = 7z – 4 and g(z) = z-2, nd (f-g)(x).7. If f(x) =8 x2-20 and g(x) =-4, nd( f g)(x).8. If f(x) = 2x+2 and g(x) = 9 x2, what is the value of (f g)(x)?9.

If f(x) = 7 x2+ 8x -3 and g(x) = 7x, solve for (f g)(x)?10. If f(x) = 35 x8- 45x and g(x) = 5x, what is the value of ( f g)(x).8Answers to Operations on Functions Exercises:1. 2 x2+7x +42.

x5 x2+ 6x – 273. 5x 11×34. x5+ 4 x35. 9x +206.

6z -27. 2x 2+ 58. 18 x3+ 18 x29. 49 x3+ 56 x2- 2110.7 x7- 99