Composition and Inverse

Functions provide an opportunity for manipulating expressions using different values.
These values can help business owners, data analysts, and even the consumer compare rates and data. Functions also extend independent (x) and dependent (y) variables by graphing in the coordinate plane and creating a visual demonstration of the relationship.
The following functions will be used in the required problems. f(x) = 2x+5 g(x) = x2+3 h(x) = (7-x)/3
The first task is to compute (f – h)(4).
(f – h)(4) = f(4) – h(4) Because of rules of composition, each function may be calculated separately and then subtracted. f(4) = 2(4) + 5 The ‘x’ was replaced with the 4 from the problem. f(4) = 8 + 5 Order of operations was used to evaluate the function. f(4) = 13 h(4) = (7 - 4)/3 The same process is used for h(4) and f(4). h(4) = 3/3 h(4) = 1 (f – h)(4) = 13 – 1 (f – h)(4) = 12 This is the solution after plugging the values in and subtracting.
Next, two pairs of the functions will be composed into each other. One option to find the solution for the function, g(x), is to calculate it and then substitute for the ‘x’ value in the f(x).
The option used here is to replace the ‘x’ in the ‘f’ function with the ‘g’ function. This means the rule of ‘f’ will work on ‘g’. This means the rule of ‘f’ will work on the problem: (f°g)(x) = f(g(x))

(f° g)(x) = f(g(x)) (f° g)(x) = f(x2 + 3) ‘f’ is now going to work on the rule of ‘g’. ‘G’ replaces the ‘x’.
(f° g)(x) = 2(x2 + 3) + 5 The rule of ‘f’ is applied to ‘g’.
(f° g)(x) = 2x2 +6 +5 Simplifying by using distributive property and order of operations.
(f° g)(x) = 2x2 +11 The final results
Now we will compose the following: (h°g)(x) = h(g(x)) in the same manner as the previous problem.
(h°g)(x) = h(g(x)) The rule of ‘h’ will work on ‘g’. (h°g)(x) = h(x2 + 3) Substitute the ‘g’ function in for the ‘x’. (h°g)(x) = [7 – (x2 + 3)]/3 Through distributive property, the rule of ‘h’ is applied to ‘g’.

(h°g)(x) = (4 - x2 )/3 The final results.
The next task is to transform g(x) so the graph is placed 6 units to the right and 7 units down from where it would be right now. Six units to the right means a -6 would be included with ‘x’ to be squared. Seven units down means, we need to put -7 outside of the squaring. The new function will look like this after using order of operations to simplify: g(x) = x2 + 3 G(x) = (x-6)2 + 3 -7
G(x) = (x-6)2 – 4
The final requirement is to find the inverse of two functions, f and h. To find the inverse the function are written with ‘y’ instead of the function name, then the places of ‘x’ and ‘y’ will be switched, and solve for ‘y’ again. Here are the functions: f(x) = 2x+5 h(x) = (7-x)/3
Here we replace f(x) and h(x) with ‘y’: y = 2x+5 y = (7-x)/3
Here we switch the ‘y’ and the ‘x’: x = 2y+5 x = (7-y)/3
Now we solve for ‘y’: Add -5 to both sides. Multiply both sides by 3. x-5 = 2y 3x = 7-y
One more solving step: Divide both sides by 2 for the left problem. Subtract 7 and multiply with (-1) from both sides for each problem
(x-5)/2 = y -3x+7 = y
Presenting the inverse functions: f-1(x) = (x-5)/2 h-1(x) = 7 – 3x
A composite function represents, in one function, the results of an entire chain of dependent functions. The composition of functions is always associative. That is, if ‘f’, ‘g’, and ‘h’ are three functions with suitably chosen domains and co domains, then f ∘ (g ∘ h) = (f ∘ g) ∘ h, where the parentheses serve to indicate that composition is to be performed first for the parenthesized functions.
The graphs of inverses share a unique relationship. They are mirror images of each other across the line y = x. Therefore, you can verify that functions are inverses by graphing them

References Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY:
McGraw-Hill Publishing.