Composition and Inverse
Jessica Cantu
MAT 222 Week5 Final Assignment
John Gomillion
June 13, 2014

Composition and Inverse

Functions can be very useful because they give us the option to change up an expression by using different values. These values and functions can help a person or a company be successful and wise. Functions also extend independent (x) and dependent (y) variables by graphing in the coordinate plane. This is very helpful because it creates 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) Following the rules of composition, each function can be calculated separately and then subtracted. f(4) = 2(4) + 5 The x is replaced with the 4 from the first problem. f(4) = 8 + 5 Order of operations is used to evaluate the function. f(4) = 13 h(4) = (7 – 4)/3 Here 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 substituting the values and subtracting. The next problem will have two pairs of the functions that will be composed into each other. What is going to be done here is to replace the x in the f function with the g function.
(f ° g)(x) = f(g(x))
(f ° g)(x) = f(x2 – 3) f is now going to work on the rule of g. And 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 – 1 The final results.
Now we will compose the following: (h° g)(x) = h(g(x)) which will be solved 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) Through distributive property, the rule of h is applied to g. 3
(h° g)(x) = 10 – x2 The final results 3

The next task is to transform g(x) so the graph is moved 6 units to the right and 7 units downward 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 downward means 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 – 10 The final task 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:
Subtract 5 to both sides. Multiply both sides by 7. x – 5 = 2y 3x = 7 – y
Last solving step:
Divide both sides of the problem on the left by 2; and subtract 7 from both sides of the problem on the right. x – 5 = y 3x – 7 = -y 2
Here are the inverse functions: f -1(x) = x – 5 h-1(x) = - (3x – 7) 2

In conclusion, there are many instances in the real world, where math can come in handy. It can save a person and a business a lot of time and money. People seem to think that math is so hard and not even important in the real world, but they could not be more wrong. If one takes the time to learn and understand how equations and functions work then one can learn to appreciate the value of it.

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