Module 7: Rational Functions and Function Operations

 

This photo shows heavy traffic.

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In Try This 3 you may have noticed that it is possible to use the graphs of functions to interpret a composition of those functions. One method of using graphs of functions to interpret the composite function is to work through each function separately.

 

For example, to find the driving time d, first determine the congestion C (which is number of cars on the road) at a specific time t (which is the number of hours after midnight). Then use the value of congestion to determine the driving time. Using function notation, the function d = D(C(t)) can be found by starting at a t-value and using c = C(t) to determine a c-value. The c-value is then used to determine the d-value using d = D(c).



In this diagram, the function c equals C at t can be used to show that a t-value of 14 leads to a c-value of 4 decimal 5. The function d equals D at c is then used to show that a c-value of 4 decimal 5 gives a d-value of 1 decimal 1.

Here, a t-value of 14 h after midnight is used to determine a c-value of 4.5 number of cars.
This c-value is then used to determine a d-value of 1.1 h to drive across town.

 

If a graph of the composite function is available, it is possible to determine the information from the previous graph in one step.

 

This graph shows that a t-value gives a d-value using the function d equals D at C at t.

This relationship can be used to determine the graph of f(g(x)) using the graphs of f(x) and g(x) by determining the points one at a time. However, an easier way to determine a graph is to use equations if they are available. The equations of f(x) and g(x) can then be used to determine an equation for f(g(x)), which can then be plotted, as you did earlier in the lesson.



textbook

If you would like to see another example that uses the graphs of functions and their composition, read “Example 3” on pages 503 and 504 of the textbook.

 

To look at another example of how a problem can be solved using a composition of functions, read “Example 5” on pages 505 and 506 of the textbook. Note how the two formulas are composed before the value of V is substituted in to be solved in one step instead of two.