Algebra- Finding the domain of certain functions
Domains of Functions
This hub will be about how to find the domain of certain functions. This concept may appear on the SAT and the ACT (ACT- more likely for this topic to appear), but it will eventually be integrated into further topics as well (ie. Precalculus, Calc I, etc.)
www.omegaprep.com and my other hubs. The website is still under construction, but we have started our online tutoring program and bookstore. We are still trying to add better services, but it may take a few months.
Now, what exactly is the domain? The domain is where the function exists in accordance to the x values. For example, the function y=x has a domain of all real numbers because it covers every x value. Actually, all real polynomials (functions with a non-negative exponent) will always have the domain of all reals.
Now, what about a function like y=x-1?
In this situation we must find when the total function will not exist (ie 1/0). We can rewrite x-1 as 1/x. Doing so, if the denominator is 0, then the function will not exist. 0 is actually what is called the vertical asymptote. The vertical asymptote is where the line approaches but will not reach.
NOW, in some instances, the function may have all reals as the domain even if there is an x value in the denominator of the fraction.
For example, 1/(x2+1). In fractions, we must find when the denominator will equal zero with real numbers, but is it possible for x2=-1? No. Therefore, there is no conflict with domain. This is also attributed to the fact that the degree is even. Even degrees will always result in positive values. Yet, if the bottom did not have the +1, then there would be a domain restriction at x=0.
The examples are seen slightly clearer on the bottom.
Example 1: y=(x-5)/(x-3).
The domain is all reals except for when x-3=0. Solving for x, the domain does not exist at x=3 (vertical asymptote).
Example 2: y=1/(x2-9)
The domain is when x2-9=0..... x2 =9......x=3 or x=-3 There could be two vertical asymptotes for a function.
Example 3: (x+3)/(x2-9)
This can be simplified into 1/(x-3) because (x+3)/((x+3)(x-3))=1/(x-3). But there are two domain 'problems'. The first one is the vertical asymptote, which is x=3 (from x-3=0). But the (x+3) is also a problem point, and it is a missing point on the function. At x=-3, there is simply a hole. It can be generalized that x values or expressions that are crossed out are the restrictions in terms of a point, not a vertical asymptote.
One of the most trickiest domain problems are sqaure roots and cubed roots. Many people know that the square root of a negative number would be imaginary, but it may not come to them during the exam.
Example 4 (Square Roots): Just remember: The domain is nonexistient when the function inside the square root is negative. (ie. sqrt(5x-10)) , Domain x is greater than or equal to 2. Square root of 0 is still a real number, yet if the square root of 0 is in the denominator, then the function is undefined.
For cubed roots, the domain is all reals if it is in the numerator and the function inside does not contain a squre root. If the cubed root is in the denominator, the x value where the function is 0 is another domain restriction.
More by this Author
I have compiled a growing list of SAT books that have been successful to the general student population as well as with tutors who work with students on the SAT. I have also included links for every book if you would...
Cuppa Tea Credit: Susan Eman Tea is becoming an increasingly popular drink as it is a refreshing alternative to the morning coffee. With that, tea usually does not contain as much caffeine as a cup of coffee and has a...