Composition of functions

Colorado State University Global Campus Wk 4 Math Discussion

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Mathematics

Colorado State University Global Campus

Question Description

I’m working on a algebra discussion question and need guidance to help me study.

 

Step-by-step solved example for the Composition of functions

Here are some Discussion assignment clarifications, with a solved example, for this week’s topic:
Since most of the people have questions about the composition of functions, we have to work together on this assignment. Please review section 3.4 for the composition of functions’ definitions and examples, as this subject is challenging, but interesting, and useful, and requires some very good analytical and attention to detail skills.

Note 1:

Somewhere, in the fine lines, I stated that the temperature conversion is just a starting point for our composition of functions’ analysis. The only topic that we discuss here is about the process of composition of functions which is not an easy subject. Please note that a conversion formula (C to F or F to C) is considered the function for our practice exercise.

Please see below my solved example, with detailed explanations and comments. I will choose the month of October, where the average temperature in my case is m = 63, (M or m=month).

Find: F(Fahrenheit) = F(M) = F(Oct) = F(63) = 63

C(Celsius) = C(M) = C(Oct) = C(63) = 5/9(M-32) = 5/9(63-32) = 5/9 x 31= 155/9 = 17.2

1. Calculate (C ᵒ F)(M) for the month of your choice (October), and show formula and steps:

(C ᵒ F)(M) = (C ᵒ F)(63) = C(F(63)) = C(63) = 155/9 (it is OK to repeat the steps)

2. Discuss the meaning of the function (C ∘ F)(M) = C(F(M))

A “Function Composition” is applying one function to the results of another function. In this operation the function C is applied to the result of applying the function F to M. (Please do not limit the discussion to the topic of temperature conversion only as we can have any letters/variables for our practice exercises).

Note 2:

Since we can have any letters/variables for our functions it is not important what a temperature conversion is but how do we solve a composition of functions, which can become complicated fast. In other words, how do we de-compose (break up) a function into a composition of other functions? How do we make things simpler in order to find a solution?

3. How does the composition of functions in parts 1 and 2 compare to (F ∘ C)(M)? Are they the same? or are they inverse?

a). Composition of Celsius and Fahrenheit (result in Celsius):

(C ᵒ F)(M) = (C ᵒ F)(63) = C(F(63)) = C(63) = 5/9(63-32) = 5/9×31 = 155/9

b). Composition of Fahrenheit and Celsius (result in Fahrenheit):

(F ᵒ C)(M) = (F ᵒ C)(63) = F(C(63)) = F(155/9) = 9/5C+32 = 9/5×155/9+32 = 31+32 = 63

As shown above the two functions are inverse.

Note 3:

We use the Celsius conversion formula to solve for variable F (Fahrenheit):

C = 5/9(F-32) Celsius (from Fahrenheit)

F = 9/5C + 32 Fahrenheit (from Celsius)

For C = 155/9 (easier to simplify if we keep the results as a fraction) we have:

F = 9/5×155/9 +32 = 31+32 = 63

Note 4: Final Observations

In a business setting you have a maximum of 3 or 4 minutes to present your work. This is why your comments must be concise and precise. Also, you have to be able to say a lot with very little. In terms of analytical and attention to detail skills, I think this assignment is one of the best, and an eye opener, since Algebra is the second most unforgiving type of mathematics (second only to Statistics).

I end my comment with a Composition of functions’ Summary:

  • A “Function Composition” is applying one function to the results of another.
  • (g º f)(x) = g(f(x)), first apply f( ), then apply g( )
  • Some functions can be de-composed into two (or more) simpler functions.==============================================================================================================================================================================QUESTION
  • Consider the following two functions:
    • F(M): The average temperature (F) in Fahrenheit during month (M) of the year.
    Month (M) F (M)
    January 42
    February 43
    March 52
    April 64
    May 72
    June 83
    July 86
    August 84
    September 73
    October 65
    November 55
    December 44
    • C(F): The conversion formula to calculate the temperature in Celsius (C) based on the temperature in Fahrenheit (F).

    C(F)=59(F32)CF=59F-32Your task for this discussion is as follows:

    1. Calculate (C ∘ F) for the month of your choice.
    2. Discuss the meaning of the function (C ∘ F)(M).
    3. Now think about the composition, (F ∘ C). Does this composition make sense? If so, how does it compare to (C ∘ F)? If not, why not?