Compositions and Inverses

Topics: Inverse function, Mathematics, Injective function Pages: 5 (662 words) Published: July 24, 2014
Running Header: INVERSE 1

Composition and Inverse

Esther Bakeberg

MAT222: Week 5 Assignment

Donna Wall

July 21, 2014


Composition and Inverse

Functions give us an opportunity for using expressions with different values. The

values can help business owners, (small or large), data collectors and 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 functions I will be using in the required problems are

f(x) = 5x – 3 g(x) = x2 +2 h(x) = 3 +x / 7

The first task is to compute (f – h) (4).

(f – h) (4) = f(4) – (h - 4) The rules of composition allow each function to 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 The order of operations is used to evaluate the function.

f(4) = 17

h(4) = (3 +7)/7 Here, the same process was used for h(4) and f(4).

h(4) – 7/7

h(4) = 1

(f - h)(4) = 17 -1

(f – h)(4) = 16 Resulting solution after substituting the values and

subtracting. Two pairs of the functions will be composed into each other next. One way 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 would be to replace the x in the f function with the g function. This way, the

rule of f will work on the problem (f° g)(x) = f(g(x).

(f°g) (x) = f(x2 + 2) f is not working on the rule of g. G replaces the x. INVERSE 3

(f° g)(x) = 5(x2 + 2) – 3 Rule of f applied to g.

(f° g)(x) = 5x + 10 -3 Simplification done using distributive property and

order of operations.

(f g)(x) = 5x2 + 7 Final answer.

Now, I will compose the following: (hg)(x) = h(g(x) in the same manner.

(h g)(x) = h(g(x)) The rule of h will work on g.

(h g)(x) = h(x2 +2) Next, I will substitute the g function for the x.

(H g)(x) = 3 + (x2 + 2)/ 7 Distributive property used to remove parentheses.

Rule of h is applied to g.

(h g)(x) = 5 + x2/7 Final result.

My next task is to transform g(x) 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 that a -6 would be included with x to be squared.

Seven units downward means to put -7 outside of the squaring. The new functions will look like this using the order of operations to simplify:

g(x) = x2 + 2

g(x) = (x -6)2 + -7

g(x) = (x – 6)2 – 5

The last requirement is to find the inverse of two functions f and h. To find the inverse

the functions are written with y instead of the function name, then the places of x and y

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