Expert: Josh - 8/6/2008

Question
QUESTION: Hi,

I was wondering how I could get started on this problem or what to do.

Write an equation with the domain (x/x is greater than or equal to 7)

Thank you for your time

ANSWER: Hi Sherry

The word "domain" basically means the set of values that x can take to generate a function y=f(x). In case you haven't seen this notation before, think of y as some expression which depends on x.

Example 1: For y=x+2 (a straight line), x can take any real value, so the domain is the set of all real numbers.

Your question asks for an expression which limits the set of values that x can take to any real number GREATER THAN OR EQUAL TO 7.
So, the answer is simply x >= 7 [note: >= is the "greater than or equal to" sign]

I'll give you several examples.

Example 2: For the inequality x^2-2x-3 > 0, the domain consists of all possible values that x can take, such that x^2-2x-3 is positive (i.e., strictly greater than zero). To find the domain, we factorize the left hand side (LHS) of the equation, to get (x+1)(x-3)>0. At this point, we recognize that y=(x+1)(x-3) describes an upright parabola with x-intercepts at x=-1 and x=3. The portion of x where y is positive is given by two intervals: x<-1 and x>3. This is the domain.

Example 3: Consider y^2=x. The domain is given by x>=0. This is because y^2 (anything squared) must be non-negative. Graphically, we have a C shape parabola with the turning point at the origin.

Example 4: Find the domain of y^2=(x-5).
As in example 3, y^2 must be greater than or equal to 0. The same condition is imposed on the right hand side of the equation. Therefore, we need x-5 >= 0. The domain is given by x>=5. [The parabola in example 3 has been shifted to the right by 5 units]

---------- FOLLOW-UP ----------

QUESTION: Hi Josh,

So how about the domain of each of the following?

a. y = the square root of (3-x)

b. y = 2 + the square root of the absolute value of x - 1

Thank you so much for your help!

Answer
The basic idea is still the same. We look for the set of values that x can take to cover the space of y.

For part a), y is the square root of something, so what goes inside the square root must be greater than or equal to 0 in order for the function to be defined. So, we need 3-x>=0. I'll leave you to rearrange this yourself.

For part b), y=2+sqrt(|x-1|). The constant "2" is irrelevant when it comes to finding the domain. The main constraint is imposed on the argument inside the square root. For the same reason as before, we need |x-1|>=0. This is true for all values of x. So, the domain comprises all real values. |x-1| is basically a V shape plot centered about x=1. It is always on or above zero, so it poses no problem.

On the other hand, if you were given y=sqrt(x-1), then we would need x-1>=0. In this case, the domain is x>=1.