Composition and Inverse
Composition and Inverse
Alicia Frambro
MAT 222: Intermediate Algebra
Prof: Michael Smith
September 22, 2013
Composition and Inverse
When using functions, there are different ways to solve various values. Many companies use these functions to monitor business pertaining gross profit and many other operations such as adjusting productivity. Visual examples of these functions are graphs. They provide a visual relationship between composition and inverse solutions as well and the difference between profit gain and profit loss for a business. The following functions will be used to solve certain problems. ?(x) = Use rule of composition and solve which is f(4)-h(4). f(x)-h(x)= 2x+5-(7-x)
2(4) +5- Substitute 4 for x.
8+ 5- Use order of operations to solve.
8+5-1= 12 Therefore (f-h)(4)= 12 Evaluate two compositions of the above functions.
A. (fog)(x)
B. (fog)(x)
The method that should be used to solve is to find the solution of one function and then substitute the solution for x in the other solution. For A we would first solve for g(x) and substitute the solution for x into f(x).
A. (fog)(x)= f(g(x))
(fog)(x)= f(x2-3)
(fog)(x)= 2(x2-3) +5 Substitute g(x) for x in f(x).
(fog)(x) = 2x2-6+5 Simplify by using the distributive property and order of operations.
(fog)(x)= 2x2 -1 Answer.
B. (h0g)(x) = h(g(x)) Use the same concept of A but (hg)(x)= h(g(x))
(hog)(x) = h(x2 – 3)
(hog)(x) = Substitute g(x) for x in h(x).
(hog)(x) = Answer.
The third section is to graph the g(x) function and transform it so that the graph is moved 6 units to the right and 7 units down. When graphing g(x), substitute the values x for g(x). Moving the graph 6 units to the right replaces x with -6 and should be squared. Moving the graph 7 units down means g(x) is replaced by -7 and is outside the squaring. g(x)= x2-3 g(x) = (x-6)2-3-7 g(x) = (x-6)2-10 g(x)= x2-12x=36
g(x)=
Alicia Frambro
MAT 222: Intermediate Algebra
Prof: Michael Smith
September 22, 2013
Composition and Inverse
When using functions, there are different ways to solve various values. Many companies use these functions to monitor business pertaining gross profit and many other operations such as adjusting productivity. Visual examples of these functions are graphs. They provide a visual relationship between composition and inverse solutions as well and the difference between profit gain and profit loss for a business. The following functions will be used to solve certain problems. ?(x) = Use rule of composition and solve which is f(4)-h(4). f(x)-h(x)= 2x+5-(7-x)
2(4) +5- Substitute 4 for x.
8+ 5- Use order of operations to solve.
8+5-1= 12 Therefore (f-h)(4)= 12 Evaluate two compositions of the above functions.
A. (fog)(x)
B. (fog)(x)
The method that should be used to solve is to find the solution of one function and then substitute the solution for x in the other solution. For A we would first solve for g(x) and substitute the solution for x into f(x).
A. (fog)(x)= f(g(x))
(fog)(x)= f(x2-3)
(fog)(x)= 2(x2-3) +5 Substitute g(x) for x in f(x).
(fog)(x) = 2x2-6+5 Simplify by using the distributive property and order of operations.
(fog)(x)= 2x2 -1 Answer.
B. (h0g)(x) = h(g(x)) Use the same concept of A but (hg)(x)= h(g(x))
(hog)(x) = h(x2 – 3)
(hog)(x) = Substitute g(x) for x in h(x).
(hog)(x) = Answer.
The third section is to graph the g(x) function and transform it so that the graph is moved 6 units to the right and 7 units down. When graphing g(x), substitute the values x for g(x). Moving the graph 6 units to the right replaces x with -6 and should be squared. Moving the graph 7 units down means g(x) is replaced by -7 and is outside the squaring. g(x)= x2-3 g(x) = (x-6)2-3-7 g(x) = (x-6)2-10 g(x)= x2-12x=36
g(x)=