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Composition and Inverse

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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. fx=2x+5 g(x)=x2-3 (x) = 7+x3 f-h(4) Use rule of composition and solve which is f(4)-h(4). f(x)-h(x)= 2x+5-(7-x)
2(4) +5- 7-43 Substitute 4 for x.
8+ 5- 33 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) = (7-x2+3)3 Substitute g(x) for x in h(x).
(hog)(x) = -x2+103 Answer.

x | g(x)= x2 - 3 | -2 | 1 | -1 | -2 | 0 | -3 | 1 | -2 | 2 | 1 |

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)= x2-12x+26

x | G(x) = x2-12x+6 | 4 | -6 | 5 | -9 | 6 | -10 | 7 | -9 | 8 | -6 |

The last step is to find the inverse of the functions f(x) and h(x). C. f-1(x) D. h-1(x)
To find f -1(x), replace x by y and f -1(x) by x so,f(x) = 2x +5 will equal x= 2y +5. Next solve for y by subtracting 5 from both sided then divide each side by 2 to isolate the y. x= 2y+5 x+ (-5)= (2y+5) + (-5) x-5= 2y+5-5 f -1(x) = (x-5)2
For finding h -1(x), switch the x and y values and substitute y for h(x) where: h(x)= (7-x)3= x= (7-y)3
Solve for y by multiplying 3 by each side and then subtract 7 from each side to isolate the y. Last multiply function by -1 to finalize y. x= (7-y)3 x= -y+73
3x= -y+7 y= -3x+7 This function graph would look like a parabola pointing downward. If this graph was used to determine the gross profit for a company, it would show that business profits are dropping due to production. Being able to use these unique functions ensures data is accurate. Transforming these functions to graphs is a better way for companies to track their gains and losses.

Reference:
Dugopolski, M., (2012) elementary and Intermediate Algebra (4th ed) New York: Mc-Graw-Hill Publishing.

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