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 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) 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 ...view middle of the document...
(h°g)(x) = [7 – (x2 + 3)]/3 Through distributive property, the rule of ‘h’ is applied to ‘g’.
(h°g)(x) = (4 - x2 )/3 The final results.
The next task is to transform g(x) so the graph is placed 6 units to the right and 7 units down 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 down means, we need 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 – 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) = 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’:
Add -5 to...