How to Find the Domains of (f+g), (f-g), (fg), (f/g), and (g/f), The Sum, Difference, Product and Quotient of Functions

This Article will use two Functions as an example to show how easy it is to find the Domain of the Sum, Difference, Product and Quotient of two Functions. It is the hope, that this Article will give a much clearer Understanding to a much misunderstood topic in the Algebra of Functions.

Things You'll Need

• Paper
• and Pencil

Instructions

1. 

   In this article, we will be finding the Domains of (f+g), (f-g), (fg), (f/g), and (g/f), when f(x)=√(3x-1) and g(x)=√(5-3x). Please click on the image for a better understanding.

2. 

   The Domains of the Sum, (f+g), Difference, (f-g), and Product, (fg), are the same. In order to find that domain, we will first find the domain of f(x) and g(x), seperately. To do this, we will graph the Domains of
   f(x) and g(x). To graph the Domain of f(x), we should set the Radicand, the function inside the radical sign, to be greater than or equal to zero. That is 3x-1>0 or 3x=1 = 0. Then x > 1/3 or x=1/3. This is the initial point of the graph. To graph the Domain of g(x), we set: 5-3x>0 or 5-3x = 0. then x<5/3 or x=5/3. This is the endpoint of the graph of the Domain of g(x). Please click on the image for a better understanding.

3. 

   The Domains of (f+g), (f-g), and (fg), are the intersection of the Domains of f(x) and g(x). If we take the graphs of the domain of f(x) and the Domain of g(x) and put one graph on top of the other graph, where ever all the points of f(x) and g(x) coincide, that is where the intersection is. For these graphs, it is [1/3,5/3]. This is the domain of (f+g), (f-g), and (fg). Please click on the image for a better understanding.

4. 

   Now, we will find the Domain of (f/g). (f/g)=[√(3x-1)/√(5-3x)]. We will first substitute x=(1/3),in both the Numerator and the Denominator functions and see if the Rational function is Real, if it is real, then (1/3) is a part of the Domain of (f/g). If x=(1/3) causes (f/g) to be undefined or imaginary, then, x=(1/3) is not a part of the Domain of (f/g). When we substitute 1/3, we get (f/g)=[√(3x-1)/√(5-3x)] = (f/g)=[√(3(1/3)-1)/√(5-3(1/3))] = (f/g)=[√(1-1)/√(5-1)] = (f/g)=[[√(0)/√(4)] = (0/2) = (0). Since this is a real number, it can be equal to 1/3. When we substitute 5/3, we get:(f/g)=[√(3x-1)/√(5-3x)] = (f/g)=[√(3(5/3)-1)/√(5-3(5/3))] = (f/g)=[√(5-1)/√(5-5)] = [√(5-1)/√(5-5)] = [√(4)/√(0)] = (2/0). This is undefined, and so (5/3) can not be a part of the Domain of (f/g). The domain for (f/g) is [1/3,5/3). Please click on the image for a better understanding.

5. 

   Finally, we will find the domain of (g/f). (g/f)=[√(1-3x)/√(3x-1)]. When we substitute 1/3, we get (g/f)=[√(5-3x)/√(3x-1)] = (g/f)=[√(5-3(1/3))/√(3(1/3)-1)] = (g/f)=[√(5-1)/√(1-1)] = [√(4)/√(0)] = (2/0). This is undefined, and so (1/3) cannot be a part of the Domain of (g/f). When we substitute 5/3, we get (g/f)=[√(5-3x)/√(3x-1)] = (g/f)=[√(5-3(5/3))/√(3(5/3)-1)] = (g/f)=[√(5-5)/√(5-1)] = [√(0)/√(4)] = (0/2) =0. This is a real number, and so (5/3) can be a part of the Domain of (g/f). The domain for (g/f) is (1/3,5/3]. Please click on the image for a better understanding.