 # Combining Square Root Functions & Domain 4 Examples

I am mr. Tarrou in this lesson we are
going to be doing four examples addition subtraction multiplication and then
division of combining square roots and then determining what the domain is of
our new function now I’m doing this video in addition to a pre calculus
playlist that I’ve already created I’ll have a link to that full playlist in the
description of this video because my student says you’re having a pretty
difficult time working with the domain finding domain of functions ones that
include these square roots and other ones that include rational functions
I’ll have a little bit of that in this video and then I’ll make another
additional video for an extra supplement we’re going to remember that with domain
you cannot square root a negative number and you cannot divide by zero now
there’s all kinds of functions we’ll be dealing with this year trig functions
natural exponential functions logarithmic functions and these all they
all bring in their little different twists into what you can use as far as
their domain goes but for right now we’re just dealing with division or
fractions and square roots and you cannot square root or indeed even roots
a negative number and one more time you cannot divide by zero so we’re going to
be dealing with and thinking about these restrictions on the domain as we go
through these problems which is going to start with f of X is equal to the square
root of x plus 3 and G of X is equal to the square root of x minus 1 just using
that function notation because we’re using two different functions at the
same time so we need a name you know to call them now we’re going to add these
functions and we’re going to subtract okay that just simply means take
function f which is the square root of x plus 3 add it with the function of G of
X which is the square root of x minus 1 and as far as simplifying goes with this
problem well there’s no simple fine to do remember like if you have the square
root of 2 plus the square root of 3 that is not equal to the square root of 5 you
well yeah the right are the same like if I have the square
root of two plus the square root of two maybe we’ll make this a little different
well then all you’re going to do is recognize that they both have the same
index they’re both you know square roots it’s not like a square root in a cube
root and they both have the same number underneath the radical well that that’s
the case to add these square roots we’re simply going to add the coefficients out
front and this becomes three times the square root of two but with of course
the algebra working I’m with right now and me writing the square root of x plus
three plus the square root of x minus one well we don’t have the same value or
same expression underneath the square root symbols so we’re not gonna be able
to add those together and so we kind of can easily set up there for the
subtraction of these two functions we have the square root of x plus three for
the F which is first and then minus the square root of x minus one so not too
much interesting going on with this first problem and as far as the domain
goes the addition of the addition or subtraction that is not going to make
any difference in determining the domain all we have to remember is you cannot
square root a negative number so for determining domain all we’re going to do
is make sure that that doesn’t happen so we’re going to see that each of these
equations have two conditions we have to make sure that the expressions that are
in each one of these radicals remains equal to 0 or is positive so we’re not
focusing on what X can be equal to we’re we’re focusing on making sure that these
expressions remain greater than or equal to zero so we have both X plus 3 has to
be greater than or equal to zero as our rate three and that’s a interval
notation for and if it was the other way around if it kind of like look like a
u-shape that would be or so that’s and X minus one also has to be greater than or
equal to zero so it’s kind of like where we’re calling how to solve compound
inequalities because both of these inequalities are in the same equation we
have X has to be greater than or equal to negative 3 as we subtract 3
from both sides and as we add 1 to both sides we get X has to be greater than or
equal to 1 now as far as you know which one of these are the answers well both
of these together have to pass because both of these conditions are coming from
the same equation whether it’s the plus or the minus so on a number line
we’re going to come through here and we’re going to recognize that when we
remember graphing inequalities on a number line we used a solid dot on that
value to represent the equal sign so we have X is equal to so a solid dot at
negative 3 equal to or greater than negative 3 we have this condition here
which is that X has to be greater than or equal to negative 1 excuse me
positive 1 and I need to look at this number line and make sure that I’m
choosing the part of the number line that satisfies both of these conditions
like if I pick a number out here like negative 4 well negative 4 plus 3 and
negative 4 minus 1 both of those expressions are going to come out to be
or both of those yeah these expressions are going to give us negative values and
of course again unless we’re talking about imaginary solutions you cannot
squirt a negative number if I pick a number like negative 1 you see that it’s
the only part the only in equality that’s over here at negative 1 it’s
satisfying the fact that this is going to be positive negative 1 plus 3 is 2
which is greater than equal to 0 you can see the shading on the number line here
but I don’t have the shading for both of these inequalities say here at negative
1 if I put negative 1 minus 1 I get negative 2 which of course is not
greater than or equal to 0 so I’m going to come through here neither one of
these conditions are satisfied here only this condition is satisfied nope this is
the parts that we want this is where we have the shading that is satisfying both
of these inequalities both of these expressions
need to be greater than or equal to zero so that we can take the square root of
those values so our final answer for the domain is that X needs to be greater
than or equal to one and that’s what the answer looks like with inequality
notation if you want to write it in interval notation you’ll use a square
bracket indicating equals two starting at one always go to the low value first
and then up to infinity and beyond and that’s what your domain looks like in
terms of interval notation so that is the combination of two functions two
square root functions using addition and subtraction of course let’s get on to
the division so we’re going to do the combining of these two functions with
division we have F over G of X and that just simply means you know take function
f square root of x plus 3 and put it over G which is the square root X minus
1 now we like to always tell our students that we want to never have a
square root in the denominator we need we want to rationalize that denominator
multiply the numerator and denominator in this case by the square root of x
minus 1 but you never want to determine domain when you have manipulated or
simplified your equation you want to find your domain as soon as you set up
that combination of functions or we’re going to be talking about later in this
section for compositions so I’m going to leave this in this format and understand
a couple of things we have two square roots and our equation again and with an
even root you cannot even root one more time a negative number. So we might think
that like before well the expression of X plus 3 which is in the square root in
our numerator has to be greater than or equal to 0 and with interval notation
because of my precalc class we use interval notation the expression of X
minus 1 has to also be greater than or equal to 0 so it looks like we’re going
to be setting up conditions for our domain
just like in the previous screen and we’re going to be talking about trig
functions logarithmic functions and all different types of functions this year
and we’ll have different situations of how we need to restrict the domain of
those functions you know what numbers can you basically restrict what numbers
can you not put into this so you get an undefined situation you cannot square
root a negative number but you also cannot divide by zero so I need to think
unlike before involves division so if I have a variable in the denominator I
can’t just think about oh you got you can’t square root a negative number I
also may have to make sure that that X minus 1 yeah this is a good condition if
it’s just you know something equals the square root of x minus 1 but now that
the square root of x minus 1 is in the denominator and I need to make sure that
it also does not equal zero because you can’t divide by zero and now this is
going to look very very similar to what we just got done doing the main
difference here again as we look through the number line and look for areas
within this number line that satisfy both of these conditions because they’re
both coming from the same equation and working together at the same time is we
have an open dot here at one because there’s no equal sign and it looks like
I’m indicating the same shading as before just you know going to the right
of one as my final answer for the domain that interval that satisfies both of
these conditions but we’re not including one so what that looks like with
interval notation now is that we’re going to start at our minimum value on
the number line of 1 we’re going to go to infinity again and beyond infinity is
a theoretical value we never stop going up so we can never use a square bracket
around infinity and this time because there’s an open dot at 1 we’re not
including the value of 1 we’re not going to use a square bracket again we’re
going to use a set of round parentheses and if your class is using just
any quality notation for the domain you can of course say that X has to be
greater than or equal to 1 and this problem was a challenge for some of my
students remembering to not include the equal part of this inequality from the
domain because you can’t divide by 0 but we really are working out this lesson
for the last example which involves multiplication of the square root
symbols all righty then we have the product of F and G well that means that
we’re going to take the square root of x plus 3 multiply it by the square root of
x minus 1 and voila now while we talked earlier about not being able to well
that was the final answer wasn’t it yeah well we talked about earlier you know
there’s a lot of restrictions involving the addition and subtraction of radicals
you need to make sure that you have the same index these are both square roots
little indexes of 2 we don’t need to write 1 we need right there don’t need
to right we have the same number inside the square root symbol and when we add
square root symbol we’re actually just adding the coefficients the 2 plus 1 is
equal to 3 square root of 2 but maybe you remember that you were able to
multiply and divide square roots like do you members simplifying things like this
the square root of 32 could you rewrite that so that there was a smaller number
inside the radical symbol well we are taking a square root so we can simplify
this aka make the number inside smaller by
factoring 32 and looking for a factor that you know we have that perfect
square perfect squares like 2 squared is 4 3 squared is 9 4 squared is 16 so this
32 the square root of 32 can be rewritten as the square root of 16 times
the square root of 2 and then indeed the square root of 16 is 4 square root of 2
trying to recall some of these older algebra and
medic skills that we’ve used in the past to help you figure out this particular
problem yes you can multiply numbers numerical values or expressions that are
inside you know like radicals in this case square roots so and I didn’t
rationalize the denominator for the previous questions simply because I was
okay with just the setup of the division and then talking about the domain so
here we have well the square roots we have x times X which is x squared x
times negative 1 is negative 1x we have 3 times X which is positive 3x and then
negative 3 times negative 1 is negative 3
excellent now how do we make sure that this expression is going to remain
positive probably the hardest domain question out of this entire section of
my precalculus book or probably your book whatever you’re using at this time
we need to make sure that x squared plus 2x minus 3 remains greater than or equal
to 0 so we need to remember how to solve basically quadratic equations you know
second-degree equations and we can do that by factoring which should be pretty
simple because we actually got this x squared plus 2x minus 3 from the two
factors of X plus 3 times X minus 1 and again that has to be remain become or
remain greater than or equal to 0 now if we were just solving a quadratic solving
a quadratic equation by factoring we would set each of these factors equal to
0 solve them so subtract by 3 and add by 1 and you’d be done you know because the
solution to equations equations normally only have like one or two you know maybe
they can have more than just one or two solutions or no solutions but there’s
usually like a limited number whereas with inequalities
there’s basically an infinite number of solutions for almost all of these
problems so this is not the solution to this inequality so there’s two ways to
do this you can solve what’s basically called solving polynomial inequalities I
have a video for that in my playlist again you’ll find that link in the
description and in my precalculus book we haven’t even talked about solving
rational or polynomial inequalities so this is what you’re going to do you can
look at it two ways you can put these values on the number line mark them as
sort of like test points if you will or important points along that number line
and set up three different intervals you’ve got interval here one an interval
here between negative three and one what’s your second interval and then you
have a third interval from everything to the right of one and you can think about
putting numerical values within this expression one from this interval this
interval in this interval and see if this expression would come out to be
positive or negative like if you go to negative four right if you go to
negative four and you think about putting a negative four into this
expression negative four plus three is negative one so that’s negative negative
four minus one is negative five so that’s negative and so we have a
negative number coming out of this factor a negative number coming out of
this factor and of course it’s X plus three times X minus one and I don’t
really care what that value is when you plug it in but I’m just trying to see if
it’s greater than zero and of course the negative times negative is positive any
positive number is greater than zero so this is an interval which will work we
can put a negative one negative one plus three is positive negative one minus one
is negative two or just a negative number
and a positive remember thinking about plugging a negative 1 or negative 2 or 0
a positive times negative is going to be a negative value which is not greater
than zero and then finally we can plug say 2 in 2 plus 3 is 5 2 minus 1 is 1
both positive and when you multiply two positive numbers you get a positive
answer so I can sort of test values along the number line set up those
intervals set up by the solutions to the equation if you all setting that equal
to 0 and I can say that my final answer my domain is that X cannot well at C I
was in st. not equal to but values from negative infinity up to negative 3
worked and we can include negative 3 because this isn’t like in a in a
denominator of a fraction and we can square root 0 so we go from negative
infinity to negative 3 use that square bracket because it’s ok if this value is
equal to 0 and then we’re gonna pick up okay so if we’re in this interval
negative times negative is positive pass if we’re in the middle no good if we’re
over here at 1 or farther to the right anything 2 3 4 decimals of course are
okay positive times positive is positive and that’s how I taught finding or
solving polynomial inequalities in my old video but we’re also going to be
talking about graphing polynomial functions and I don’t know if this is
review a preview or you know whatever kind of experience you have in your
class as you’re watching this video lesson but you can also come up with
this solution with just really kind of thinking about well what this graph
looks like and let’s highlight this and what color by not used yet green okay
here we go Green y equals x squared minus two X minus three this
is a second degree equation there’s this is a powerful one there’s one variable
just X second degree this is a quadratic equation this is a parabola and we have
functions and graphing we have a parabola opening up and this parabola
opening up if you think about member finding x-intercepts when you found an
x-intercept you let y oh yeah equal zero and you would finish solving
that by factoring so all this is all these skills were learning or tying
together so this parabola opening up has X intercepts at negative three and
positive one we’ve already kind of done that work of finding x-intercepts when
we were graphing parabolas so each of these tick marks are one so here’s an
x-intercept of one here is an x-intercept of negative 3 if you did
just cover graphing polynomial functions you will also hopefully remember this
and then if not I’m going to remind you this solution of negative 3 came from
this factor that only showed up once this answer of one came from this factor
that only showed up once remember talking about multiplicity of zeros if
the multiplicity is zero excuse me odd if the multiplicity is odd then the
graph passes through those x-intercepts I should have put x-intercepts those run
out of room so this parabola looks something like this
oh we can talk about letting letting X be zero and this is zero and your
y-intercept therefore is negative three and all kinds of stuff and you might
have thought well man that’s really complicated I would have never thought
about that well if you haven’t talked about graphing polynomials then yeah
drawing a picture or even visualize it in your mind to solve this inequality
maybe not be your you know might not be your first choice but if you do know
your polynomial functions very well on how they look graphically recognize that
parabola opening up then you might realize that oh yeah this function is
only going to give me positive values or values equal to zero over here starting at 1 and going to the right Co
the the graphs above the x-axis all those answers all those Y values are
positive so here’s your domain at least in terms of your solution at least in
terms of this inequality from one to the right your answers are greater than or
also again greater than equal to 0 so you can see the domain in this picture
those Y values that graph given you positive values just trying to tie in
all these things you’re learning through your algebra and our precalculus classes
I miss a true hey go to your homework you

## 23 thoughts on “Combining Square Root Functions & Domain 4 Examples”

1. Cute 私はとてもかわいい笑 says:

Great video. Thanks for the help!

2. Ethan F says:

Love the Intro. Short but cool.

3. Kidron Ramirez says:

@ProfRobBob. This was a really helpful video. I honestly never would've thoughts that there were these kind of functions where u have to find domains for them! thanks 4 the help!!!!!! have a great day!!!…

4. A1Sauce says:

5. sam yemen says:

robbiiiaaaaeee what you know dude

6. Weese Bowski says:

prof Rob helped me go from a failing college student in 5 weeks from an F to a C+. I was just about 1.9 points from getting a low B. thanks prof Rob!

7. Jorge B says:

THE MAN, THE MYTH, THE LEGEND… 🙌🏼

8. Sean McClanahan says:

YOURE BACK!!

9. Fadic4 says:

Guess who's back…back again…

10. Eric Chavez says:

Hey u a legend

11. Danrei Untalan says:

I was just wondering how to solve this equation, but my capability in solving this problem is limited, can you help me solve this problem sir??

Problem:
Solve: ( √4 )^√2 + (3)^√2 or simply (√4 +3)^√2

12. Jesus Canez says:

Welcome back, we missed you!

13. laurenphoenix says:

We missed you! Since my first year in college, you helped me professor! Thank you!

14. Ming Ming says:

15. Forrest Sweitzer says:

my professors expected us to work our own by using mymathlab.. I hateeeeee it. I wish he is more like you actually take time to explain it. but because of you.. i'm starting to understand it… thank you so mmmucccchh.

16. LBoltzyy says:

ProfRobBob you are my brother Matthew Bohl's teacher right

17. blustamove says:

Hey Mr. Tarrou, hope you're safe ahead of this storm!

18. leatherneck818 says:

BAM! Earned my B.S. Civil engineer last May! Thank you for your help early in my foundational math!

19. Srikar Valluri says:

YOUURR ALLLIIIIVVEEE!!!!!

20. Joshua Britt says:

Really enjoyed the Space theme intro💫☄️

21. Sarah DeAy says:

What would I do with a problem like 'simplify the square root of 50xy^2'?

22. دانية اناناس says:

WHAT?

23. Sara Ayd says:

Has anyone ever noticed the little fella in the bottom right corner of the board?