# 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

cannot add radicals unless the numbers inside the radicand or the radical looks

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

add another two as a coefficient and if I say we’re going to add these radicals

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

about this other condition here and remember that oh well now this question

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

those radicals we actually don’t add those numerical values underneath the

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

a positive leading coefficient so if you know a little bit about your polynomial

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

equal to 0 and from negative three to the left your Y values your answers are

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

Great video. Thanks for the help!

Love the Intro. Short but cool.

@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!!!…

Whaaaaaaaaat!!!! An upload!?!?!?!?!?!?!?!

robbiiiaaaaeee what you know dude

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!

THE MAN, THE MYTH, THE LEGEND… 🙌🏼

YOURE BACK!!

Guess who's back…back again…

Hey u a legend

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

Welcome back, we missed you!

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

So glad u r back!

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.

ProfRobBob you are my brother Matthew Bohl's teacher right

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

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

YOUURR ALLLIIIIVVEEE!!!!!

Really enjoyed the Space theme intro💫☄️

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

WHAT?

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