Composition and Inverse
MAT 222 Intermediate Algebra
June 23, 2014
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 this week’s assignment.
1. f (x) = 2x + 5 2. g (x) = x2 – 3 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 operation 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 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 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)) Rewrite so the rule of f will apply.
(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 Rule of f 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.
The third problem also involves composing two pairs into each other (h ◦ g) (x) = h (g(x)).
(h ◦ g)(x) = h (g(x)) The rule of h will work on g.
(h ◦...