Composition and Inverse - Speech or Presentation Example

We define the following functions: f(x) = 2x + 5 g(x) = x2 - 3 h(x) = (7 - x 3 Compute (f - h)(4)Solution: Substituting 4 into f(x) and h(x) separately,f(4) = 2(4) + 5 = 13 and h(4) = (7 - 4) / 3 = 1(f - h)(4) = 13 - 1 --- (f - h)(4) = 12Evaluate the following two compositions: (f o g)(x) and (h o g)(x) Solution: the expression for g(x) must be contained in f(x), so that g(x) replaces the ‘x’ in f(x) and upon substitution,(f o g)(x) = 2(x2 - 3) + 5 = 2x2 - 6 + 5 --- (f o g)(x) = 2x2 - 1Similarly, the expression for g(x) must be contained as well in h(x), in which case such expression must be substituted for ‘x’ in h(x), thus(h o g)(x) = [ 7 - (x2 - 3) ] / 3 = (7 - x2 + 3) / 3 --- (h o g)(x) = (10 - x2) / 3Transform the g(x) function so that the graph is moved 6 units to the right and 7 units down.

The original graph of g(x) = x2 - 3 is:By shifting g(x) 6 units to the right and 7 units down, the new function would be g(x) = (x - 6)2 - 10 whose graph looks:Find the inverse functions f-1(x) and h-1(x)To find the inverse of f(x), switch the variables ‘x’ and ‘y’ so thatf(x) = 2x + 5 --- y = 2x + 5 becomes x = 2y + 5 then solve express ‘y’ in terms of ‘x’ so upon isolating the ‘y’, 2y = x - 5 --- y = f-1(x) = (x - 5) / 2Then to find the inverse of h(x), likewise, switch the variables ‘x’ and ‘y’ so that h(x) = y = (7 - x) / 3 becomes x = (7 - y) / 3 and upon arranging the equation, this makes 3x = 7 - y.

Getting ‘y’ by itself on one side, y = h-1(x) = -3x + 7.ApplicationI love eating pizzas and a friend of mine once told me about a store in their place which sells different kinds of pizzas according to the topped components where each pizza contains three different toppings. Besides the cheese and the pepper which is optionally put on crust, the three toppings of one pizza are completely distinct from those of the other, so that every time a customer orders a number of pizzas, the total number of toppings T added may be obtained as 3P where P stands for the number of pizzas bought.

The storeowner allows a customer to have two separate toppings of choice for the first pizza in addition to the three originally present. Therefore, if I were to purchase from this store, my working equation for the toppings count would then be:T = 3P + 2 Moreover, functions create a great deal of advantage in business problems which involve making investments where two variables are assigned to refer to separate amounts or two kinds of investment at different rates of interest. Function may be used as well in relating costs to number of units purchased and fixed cost as in C(x) = 5x + 7 in dollars, where 7 is the fixed value and 5 is the rate at which cost changes per unit depending on ‘x’.

This way, it would be conducive for one to keep track of sales and profits generated since there exists a cost function that is associated to revenue (Waner, 2006). Mixture and rate problems can be worked out once details are set up as function or system of functions in linear and non-linear forms. It also becomes significant to understand how temperature conversion works such as between Celsius and Fahrenheit through °F = 1.8°C + 32 where temperature in °C appears to be a function of temperature in °F.

ReferenceWaner, Stefan (March 2006). “Functions and Linear Models.” Retrieved from http://people.hofstra.edu/stefan_waner/RealWorld/Calcsumm1.html on November 4, 2012.