Domains and Square Roots

Domains and Square Roots

When we want a domain, we want a list of x’s that can be used to calculate real numbers that can then be graphed or used…with square roots, then, since the square root of a negative number is an imaginary number, we want to restrict ourselves to x’s that are positive.

If our input is only x, then has the domain x 0.

What if our input is a line? Say, 3x  9. What is the domain for ?

Let’s look at the line y = 3x  9. The slope is 3, the y – intercept is 9 and the x intercept is 3…A graph of this line looks like

Let’s try it with another line and then move into using quadratics under a radical.

What’s the domain for ?

This is really the square root of a line. Let’s graph the line and find out which x’s give positive numbers when used in the calculation “3  x”.

Where do the y’s go from positive to negative (the x axis intercept)….Which x’s give positive y’s for the line?

Report these as the domain for the f(x) under question.

Now, what about the domain for ?

If we focus on the quadratic, we can figure out which x’s give us the square root of a positive number:

Looking at …what are the x axis intercepts?

= ( x + 2) ( x  5) so the intercepts (or zeros) are 2 and 5. Let’s graph those:

That’s right, below 2 and above 5.

Since these are the x’s that give positive values when used in the quadratic we’ve been given, then these are the x’s that are in the domain of the f(x) we started with. Those x’s between 2 and 5 give negative numbers (see the part of the graph that is below the x axis? Those are x as a first coordinate and a negative number as a second coordinate).

So the domain is .

You can do this without the graph by using test points. Take just the x axis and put the x intercepts of the quatratic on it. The two intercepts divide the x axis into 3 parts: x’s smaller than  2, x’s between  2 and 5, and x’s bigger than 5.

Now pick one point in each part of the line; we’ll call these test points

smaller than 2 …let’s use  5

between the two points…let’s use 0

bigger than 5…let’s use 10

(really, you can use any number so long as it’s from that part of the line)

Now calculate = ( x + 2) ( x  5) using each test point.

 5 gives us 3 times  8 which is positive 24…a positive number

0 gives us 2 time  5 which is  10…a negative number

10 gives us 12 times 5 which is positive 60…again, note positive

You should realize that we don’t really care what the number is from the calculation we only care about it’s sign.

Since we got a positive with numbers below  2, these x’s are in the domain.

The x’s between 2 and 5 give negative numbers so they’re NOT in the domain.

And x’s above 5 give positives so they are in the domain.

So the domain is .

Let’s look at another one: … what is the domain?

Focus on the quadratic: …what are the x axis intercepts? Put them on the axes below.

Sketch the quadratic:

Where are the second coordinates positive? Those sections of the x axis that produce positive numbers (that is, graph above the x axis) when used to calculate are the domain of our problem.

Now doing it without the graph…put the x axis intercepts on the x axis below and choose test points:

Using the factored form of the quadratic, figure out the sign of the number produced when each test point is used in the calculation.

Report the part or parts of the line that produce positive numbers as the domain.

Did you get

Here’s one that is slightly different in outcome from the preceding two. You can check with me before class or in MathLab (MW 9am – 11am) to see if you got it right…

Find the domain for

What are the x axis intercepts for the quadratic? Put them on the number line below.

Pick test points and calculate the sign of the result using each test point in .

What is the domain?