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CAmodule1discussion.docx

 Module 1: Functions

Discussion Instructions:

Please review  the class discussion directions . They tell you how to prepare and reflect before responding to the prompt below. They also include detailed guidelines on how you must react to classmates and then reply to the responses you receive. Please pay close attention to the response directions. You must post, respond, and reply as directed to earn your discussion points .

Discussion writing prompt

1.

1.  Introduce yourself to your instructor and classmates. Make sure to include the following:

· Indicate the academic program in which you are enrolled.

· Explain why you decided to enroll in this program.

2. Describe how you might apply functions, dependent and independent variables, relations, domain, range, composite, inverse, and linear functions in real-world applications. Identify the field and describe how it applies in the field.

 

If you are new to Canvas, follow  these directionsLinks to an external site.  for participating in the discussion and to review the  academic expectations  for your submission.

Submission reminders:

· Submit your initial discussion post by 11:59 pm ET on Wednesday.  Respond to two of your classmates' discussion posts, and reply to the responses you received by 11:59 pm ET on Sunday.

· Contribute a minimum of 250 words to the initial post.

· Follow established  netiquetteLinks to an external site. guidelines when participating in forums.

 

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CAmodule1lecture.docx

Module 1:  Functions

Functions  

Consider the equation: y = 2x + 3, some ordered pairs that satisfy this equation are: (-2,-1), (-1,1), (0,3).

Dependent and Independent Variables :

In this equation: y = 2x +3, the value of  y depends on the value of  x, then “y” is the dependent variable and “x” is the independent variable.

Relation, Domain, and Range:

relation is any set of ordered pairs of the form (x, y). The of x-coordinates is called the  domain of the relation. The set of y-coordinates is called the  range of the relation

Function:

Is a relation in which each element of the domain corresponds to exactly one element in the range.

 

Example: Give the domain and range, then determine whether the relation is a function.

A. {(1,4), (2,3), (3,5), (-1,3), (0,6)}

B. {(-1,3), (4,2), (3,1), (2,6), (3,5)}

Solution:

A. The domain is {1,2,3,-1,0} and the range is {4,3,5,6}. The number 3 was included only once, even though appears in both pairs (2,3) and (-1,3). Since no x-value corresponds to more than one y-value in the range, this relation is a  function.

B. The domain is {-1,4,3,2} and the range is {3,2,1,6,5}. Since the ordered pairs (3,1) and (3,5) have the same first coordinate (x) and a different second coordinate, each value does not correspond to exactly one value in the range; therefore, this relation is  not a function.

From this example, an alternate definition of  function is a set of ordered pairs in which no first coordinate is repeated.

 

Composite Function

Given the function: f[g(x)] = (f o  g)(x)

Function  f[g(x)]   is called a  composition of  f with  or the  composite function of  with  g

Example:

Given f(x) = x – 3 and g(x) = x + 7, find:

a) (f o  g)(x)

b) (f o  g)(2)

c) (g o  f)(x)

d) (g o  f)(2)

Solution:

a) (f o  g)(x) 

since f(x) = x + 4, then (f o  g)(x) = f[g(x)] = (x + 7) – 3 = x + 4

(f o  g)(x) = x + 4

b) (f o  g)(2) = x + 4 = 2 + 4 = 6

(f o  g)(2) = 6

c) (g o  f)(x)

since g(x) = x + 7, then (g o  f)(x) = g[f(x)] = (x – 3) + 7 = x + 4

(g o  f)(x) = x + 4

d) (g o  f)(2) = x + 4 = 2 +4 = 6

(g o  f)(2) = 6

Inverse Function

One-to-one functions consist of ordered pairs in which each x-coordinate corresponds to exactly one y-coordinate and each y-coordinate corresponds to exactly one x-coordinate. Such a relationship allows creating  new functions called the  inverse function. Only one-to-one functions have inverses.

Examples:

Function f(x): {(0,1), (1,4), (-3,-2)}

The inverse function will be: f−1(x)= {(1,0), (4,1), (-2,-3)

Function f(x) = 4x + 2, this function also can be expressed as: 

y = 4x + 2

Interchange x and y:

x = 4y +2, solving for y:

x – 2 = 4y

x−24=y

The inverse will be: f−1(x) = y

x−24=y

Linear Function or Linear Equation

A linear function is a function of the form f(x) = mx + b or y = mx + b

· The graph of any linear equation is a straight line.

· The domain of any linear function is all real numbers.

· If m ≠ 0, then the range of any linear function is all real numbers.

· m, is the slope of the line (is a measure of the steepness of a line)

· b, is the y-intercept

Linear Equations (mathsisfun.com)Links to an external site.

Example: 

Find the linear equation that passes through the points (2,3) and (1,4) steps: 1) find the slope (m), m = y2−y1x2−x1

let (2, 3) be (x1,y1) and (1, 4) be (x2,y2)

 m=y2−y1x2−x1=4−31−2=1−1=−1

 

2) find the point-slope form of the equation of the line

y–y1=m(x–x1), replacing the values of m, x1 , and y1 [ordered pair (2,3)]

y – 3 = - 1 (x – 2)

y – 3 = -x + 2

y = -x + 2 + 3

y = - x + 5

References

Angel, A., & Rundle, D. (2018).  Intermediate algebra for college students. (10th edition). Pearson.