MAT 222 Algebra

Running head: COMPOSITION AND INVERSE 1

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Composition and Inverse

John Q. Student

MAT 222 Week 1 Assignment

Instructor’s Name

Date

COMPOSITION AND INVERSE 2

Composition and Inverse (title required on first line)

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) = 5x – 3 g(x) = x 2

+ 2 h(x) = 3 + x

7

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) = 5(4) – 3 The x was replaced with the 4 from the problem.

f(4) = 20 – 3 Order of operations was used to evaluate the function.

f(4) = 17

h(4) = (3 + 4)/7 The same process is used for h(4) and f(4).

h(4) = 7/7

h(4) = 1

(f – h)(4) = 17 – 1

(f – h)(4) = 16 This is the solution after substituting the values 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),will be 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)).

COMPOSITION AND INVERSE 3

(f° g)(x) = f(g(x))

(f° g)(x) = f(x 2

+ 2) f is now going to work on the rule of g. G replaces the x.

(f° g)(x) = 5(x 2

+ 2) – 3 The rule of f is applied to g.

(f° g)(x) = 5x 2

+ 10 - 3 Simplifying by using distributive property and order of

operations.

(f g)(x) = 5x 2

+ 7 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(x 2

+ 2) Substitute the g function in for the x.

(h g)(x) = 3 + (x 2

+ 2) Through distributive property, the rule of h is applied to g.

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(h g)(x) = 5 + x

2 The final results.

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The next task is to transform g(x) so the graph is placed 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) = x 2

+ 2

G(x) = (x – 6) 2

+ 2 – 7

G(x) = (x – 6) 2

– 5

COMPOSITION AND INVERSE 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) = 5x – 3 h(x) = 3 + x

Here we replace f(x) and h(x) with y:

y = 5x – 3 y = 3 + x

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Here we switch the y and the x:

x = 5y – 3 x = 3 + y

7

Now we solve for y:

Add 3 to both sides. Multiply both sides by 7.

x + 3 = 5y 7x = 3 + y

One more solving step:

Divide both sides of the problem on the left by 5; and subtract 3 from both sides of the problem on the

right.

x + 3 = y 7x – 3 = y

5

Presenting the inverse functions:

f -1

(x) = x + 3 h -1

(x) = 7x – 3

5

Conclusion paragraph would go here. Remember to include 4-5 sentences to make a

complete paragraph.

COMPOSITION AND INVERSE 5

Reference

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

McGraw-Hill Publishing.

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