Analysis of Functions

 

Functions are traditionally represented on a two dimensional Cartesian axis. The most familiar version is to have the horizontal axis labeled as x and the vertical axis labeled as a function of x. This paper will analyze some simple functions along with a few manipulations thereof. The following three functions are to be considered:
f(x) = 2x + 5
g(x) = x2 – 3
h(x) = (1/3) * (7 – x).

For the first part, the goal is to determine (f – h) (4). The first step is to find the general form for the difference of the two functions f(x) and h(x):
2x + 5 – ((1/3) * (7 – x))
(1/3) * (6x + 15) – (1/3) * (7 – x)                Rewrite f(x)
(1/3) * (6x + 15 – 7 + x)                                       Factor out (1/3)
(1/3) * (7x + 8)                                           Collect terms
Now substitute 4 in for x:
(1/3) * (28 + 8) = (1/3) * 36 = 12.

For the composition functions, the entire inner function is inserted into the outer function. Consider the following two examples:

f(g(x)) = f(x2 – 3)
= 2(x2 – 3) + 5
= 2x2 – 1

h(g(x)) = h(x2 – 3)
= (1/3) * (7 – (x2 – 3))
= (1/3) * (10 – x2).

Graphing a function is fairly straight forward. For each value of x on the horizontal axis, match it with the function value on the vertical axis. In order to transform the function to the right by 6 units, it is necessary to replace x with x – 6. In order to transform the function downward by seven units, it is necessary to subtract 7 from the value of the function. This gives:
g(x) = x2 – 3
gT(x) = (x – 6)2 – 3 – 7
= (x – 6)2 – 10.
The graph on the following page shows the original function in blue and the transformed function in red.

To find the inverse of a function, simply interchange the function expression and the variable x. For the functions f(x) and h(x) we find:
f(x) = 2x + 5                                      Original function f(x)
x = 2 * f’(x) + 5                       Interchange as described
x – 5 = 2 * f’(x)                       Subtract 5 from both sides
f’(x) = (1/2) * (x – 5)               Divide both sides by 2

h(x) = (1/3) * (7 – x)                Original function h(x)
x = (1/3) * (7 – h’(x))              Interchange as described
3x = 7 – h’(x)                          Multiply both sides by 3
h’(x) = 7 – 3x                          Isolate h’(x)

Presented here are some fairly direct applications for simple functions. While complexity can increase as the functions become more intricate, the process remains the same. Each step above can be a powerful tool for analyzing functions under different circumstances.

 

 

References

Annenberg Learner (2012).  Math in Daily Life, retrieved on December Jan 2,

2014, from: http://www.learner.org/interactives/dailymath/index.html

 

Ashford University (2012). Student Guide MAT222: Introduction to Algebra.

Retrieved on Jan. 2, 2014, from: http://classroom.ecollege.com/re

 

Dugopolski, M. (2012). Elementary and intermediate Algebra (4th Ed.). New

York, NY: McGraw-Hill.