Function and Inverse Cindy Dean

Topics: Function, Algebra, Mathematics Pages: 5 (495 words) Published: April 24, 2014

Composition and Inverse
Cindy Dean
Intermediate Algebra: MAT 222
February 14, 2014

Composition and Inverse

In this week’s assignment, I will be solving functions with different values and variables. Many companies and businesses, use these methods to either make progress or to change something that will benefit their success. The first function is: (f – h)(4)

f(4) – h(4)I multiplied 4 with each variable.
f(4) = 2(4) + 5The x is replaced with 4.
f(4) = 13I used the order of operation to evaluate this
h(4) = (7 – 3)/3I will repeat the steps that I used in the previous
problem for the variable, h.
h(4) = 3/3I used the order of operation to simplify.
h(4) = 1This is the solution for h.
(f – h)(4) = 13 + 1I combined the values of f and h.
(f – h)(4) = 14This is my solution.
The next problem, I will be finding the solution for g(x). I will replace the x in the f function with the g function. My function will look like this: (f° g)(x) = f(g(x))
(f° g)(x) = f (x^2 – 3)My g function is replacing the x. (f° g)(x) = 2(x^2 – 3) + 5
(f° g)(x) = 2x^2 – 6 + 5I used the order of operation to simplify
my answer.
(fg)(x) = 2x^2 – 1This is my solution.
This problem, I will be composing for (hg)(x).
(h° g)(x) = h(g(x))I will repeat the steps that I have been using. (h° g)(x) = h (x^2 + 5)
(h g)(x) = 5 (x^2 – 3)/-1
(h g)(x) = 2 + x^2/-1This is my final answer.
The next step is to graph the g(x) function and transform it to graph 6 units to the right and 7 units down. g(x) = x^2 – 3This is the original function.
g(x) = (x – 6)^2 – 3 – 7I plugged in the -6 and -7. g(x) = (x-6)^2 – 10This is my solution after using the order
of operation and simplifying.
My last problem is to find the inverse of functions of f and h. The function is going to be y instead of its own name. The x and y will then be switched to solve for y. The functions are: f^-1(x) = 2x + 5andh^-1(x) = 7-x/3

-y = 2x + 5and-y = 7 – x/3
I replaced the f^-1(x) and h^-1(x) with –y.
Now I will switch the –y and the x:
x = -2y + 5and-x = 7 – y/3
I will solve for y now:
x – 5 = -2yand-3x = 7 – y
x – 5/2 = yand-3x – 7 = -y
This is the solution with the inverse function:
f(x) = x – 5/-2andh(x) = -3x – 7
Using the composition and inverse method, I was able to use expressions with different values. People who have their own business or work for a company, can use this method to compare with other informations. This method can also be graphed and used for companies and businesses to see where they stand and if they need to improve something or not.


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