Download 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 f ( x )  x has the domain x  0.
What if our input is a line? Say, 3x  9. What is the domain for f ( x )  3x  9 ?
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
( 3, 0)
Notice that the second
coordinates are negatives for
x’s smaller than 3 and are
positive for x’s larger than
3….
This observation is what
gives us the domain:
[ 3, ) .
You can also solve the
inequality 3x  9  0 to find
this.
( 9, 0)
Let’s try it with another line and then move into using quadratics under a radical.
What’s the domain for f ( x )  3  x ?
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 f ( x )  x 2  3x  10 ?
If we focus on the quadratic, we can figure out which x’s give us the square root of a
positive number:
Looking at x 2  3x  10 …what are the x axis intercepts?
x 2  3x  10 = ( x + 2) ( x  5) so the intercepts (or zeros) are 2 and 5. Let’s graph
those:
There’s two choices – cupped up
or down, which is it?
Now, graph the parabola
and study the y
values…where are the
positive numbers?
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 (,2)  (5, ) .
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.
2
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  3x  10 = ( 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 (,2)  (5, ) .
Let’s look at another one:
f ( x)  x 2  2x  3 … what is the domain?
Focus on the quadratic: x 2  2 x  3 …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 x 2  2 x  3 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
(,3)  (1, ) ?
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 f ( x )   x 2  x  12
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  x 2  x  12 .
What is the domain?