Lesson 1: Functions as ModelsA function happen to be a simple mathemical model or a piece of larger model.Recall that a functon is just a rule or law f, that expresses the dependency of avariable y, on another variable x.Example 1: The graph of a function f is shown below:(a) Find the values f(1) and f(5).(b) What is the domain and range of f ? Solution:(a) We see from the graph that the point (1,3) lies on the graph of f, so thevalue of f at 1 is f(1) = 3. While x = 5, the graph lies about 0.7 unit below thex- axis. Therefore, we estimate that f(5) = -0.

7.(b) Notice that f(x) is dened when 0 x 7, so the domain of f is theclosed interval 0,7. See that f takes on all values from the interval -2 to 4, sothe range of f is-2y 4 = -2,4Example 2: Sketch the graph and nd the domain and range of the func-tions:(a) f(x) = 2x- 1(b) g(x) = x2Solution:(a)The equation of graph is y = 2x-1, and recognize this as being the equa-tion of a line with slope 2 and y-intercept -1.

1Recall: The slope- intercept form of the equation of a line y = mx + b.This enables us to sketch the graph below. The expression is dened for allreal numbers, so the domain of is the set of all real numbers, which we denoteby R. The graph shows that the range is also R. (b) Since g(2) = 22= 4 and g(-1) = ( 1) 2= 1, we could plot the points (2,4)and (-1,1), together with a few other points on the graph, and join them to pro-duce the graph below.

The equation of the graph is y = x2, which represents aparabola. The domain of g is R. The range of g consists of all values of g(x),and that is the all numbers of the form x2. But x2 0 for all numbers x andany positive number of y is a square.

Therefore, the range of g is y| y 0 =0, 1). This can also be seen from the gure below. Example 3:2Lesson 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 1:3Example 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 -3.4Exercises: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)5Answers in Exercises:1. 52. -163. 1984. 1455. -506.

847. -1928. 4 159. – 13 1610. 886Lesson 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).7(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 18Exercises: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).

9Answers 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- 910