Arithmetic Combinations of Functions (Add, Subtract, Multiply and Divide)

Video Series: Functions and Their Graphs (Step-by-Step Tutorials for Precalculus)

(D4M) — Here is the video transcript for this video.

00:00
in this episode you’ll learn how to add and subtract functions we’ll also
practice multiplying and dividing them let’s do some math [Music]
hi everyone welcome back so we’re going to begin by talking about the sum
difference product and quotient functions so here’s the sum
and here’s the difference and so for example how these work is
we’re going to be given two functions f and g
and so we know what the domains are a and b so f has domain a
and the function g has domain b and we’re going to be defining a new function
based upon these two given functions here so
f plus g is a new function now in order to define this function right here

00:01
so we’re going to need to specify the inputs and outputs so the way that we do
this is that we’re going to give f plus g and input
and so let’s give it an input so let’s say our input is x
and then we’re going to specify how to find that output
so the way to find the output from f plus g function is we input x into f
and we input x into g and then we add those two real numbers
together and then we get the output so that’s the sum function so we put x into
the sum function by substituting it into f into g and then adding those two real
numbers together so in order for this function to work f plus g
x needs to be in the domain of f and x also needs to be in the domain of
g if you try to input a real number into the sum
and that real number x isn’t in the domain of f well then you have no way to
get an output so this x has to be in the domain of f and the domain of g

00:02
and the same thing for minus the only difference is we’re getting the two
outputs and then we’re subtracting this in this order here f f of x minus g of x
for f minus g and so then we’ll look at f times g
which is f of x times g of x so there’s implied multiplication right here
and then we’re looking at the last function right here f the quotient of f and g
to input an x into this function we simply take the outputs here and divide
now that of course assuming that not only is x in the domain of f and x is in
the domain of g but also g of x here is non-zero so we can say
that right here so the input is in the intersection has to the input has to be
in a and has to also be in the domain b and we say that this denominator here is
non-zero so then we have the quotient defined so this right here means i’m

00:03
defining this new function right here using what’s already already known over
here on the right so let’s look at lots of examples so
you know just for example let’s say here we have f of x is x squared minus 5x
and let’s say g of x here is 2x minus 9 and so what will these four functions
right here be so let’s find the summation function
and let’s do that right here so f plus g evaluated at some input x
is going to be uh f and then plus the g of x so plus g of x and this is
all g of x right here 2x minus 9. all right so very good so this what does
this come out to be x squared and then we have a minus 3x here

00:04
and then we have a minus 9. and so we can substitute in now we can
substitute in values right here for example f plus g at say 1
will be what 1 squared minus 3 minus 9. so what’s that minus 11 right and so we
can find these other four functions very similarly so what is f minus g of x
this will be f of x all of that minus the g of x and so this becomes x squared
and then we have minus five x minus two x so that’s minus seven x and we have
minus minus so we have a plus nine so this is the difference between f and g
this is the difference function notice that that is different than g of
g minus f of x so we could go find this function as well
but that but these two functions are not the same so this would be g of x

00:05
minus f of x and so this would be 2x minus 9 minus and then f of x
and so f of x is x squared minus 5x and so this would be minus x squared
and then this would be a positive 5x along with the 2x so that’s 7x
and then we have minus 9 here so you can see the relationship between
these two functions right here they’re not the same and they are different
um so what about uh the product function we can find the product function right
here f times g is f of x times g of x which is just x squared minus 5x times the
2x minus 9. and if you wanted to we could expand that out and simplify that
all right and so what would the quotient function be here

00:06
now before i say that what is the domain and range of these so far that we’ve
done so far so the domain of f here is all real
numbers and the domain of g is all real numbers
and so the intersection is just going to be all real numbers in other words
i can input any x i want into the summation any real number here because i
can input any real number into f and i can input any real number to g
and well we can always add any two real numbers together to get another real
number so there’s not going to be any restriction upon the domain for f plus g
and similarly for f minus g and f times g
there’s no restriction the domain is all
real numbers for these two so the domain is going to be all real numbers for
these three also we can substitute in any x we want into
here and we’ll get an output so let’s look at f divided by g now the quotient
so let’s just make a little room right here and

00:07
so here’s f the quotient of f and g evaluated at some input and so by
definition this is just f f over g and so we get f of x
and then over 2x minus nine and so this is the
quotient function right here this is this this is a rational function
notice here that because we’re dividing uh by 2x minus 9
that the domain here is not all real numbers so the domain is all real numbers
such that x is not equal to whatever makes this denominator 0 if
anything so it turns out that 9 over 2 makes the denominator 0. so 9 over 2.
so this is the domain all real numbers except 9 over 2. okay so there’s uh

00:08
an example of using these four right here and let’s perhaps look at some more so
let’s say we have evaluate the sum difference product and quotient
so let me give you some more let’s get some more functions going up here
so now let’s say we have a function here f of x is um x squared plus one
and our g of x is x minus four all right and so now let’s look at some
different things that we can do so let’s find so for part a here we’ll
say f minus g of zero let’s try to find that and f minus g at three t
and f times g at six and f over g at five and for the last part let’s say f

00:09
over g evaluated at minus one minus g of three
so let’s find uh all of these right here giving these two functions right here
all right so let’s do that all right so let’s find f minus g at
zero now there’s two ways that we can do this we can find
this function right here for all input for all input x in other words we can
find a formula for it and then we can go plug in x into that formula right there
but you know we don’t really need to do that it’s just f of 0 minus g of 0
just by definition of what the difference function is
it’s f of 0 minus g of 0. so now i can just go plug in 0 into each one of these
so if i plug in 0 here i get 1 minus and then i plug in g i plug into 0 into g
and so i get minus four and so this is just one plus four or in other words five

00:10
all right what about f minus g of three t so this is f of three t
minus g of three 3t and let’s see here what is f of 3t so
i’m going to put in 3t here so this would be 9t squared plus 1
and then a minus sign right here in fact i’ll put parentheses just to
help with the form and then now g of 3t so g of 3t so i need to put 3t in here
so it’s 3t minus 4 and we can carry this out a little bit so nine t squared
and then minus three t and then let’s see here we have plus five
all right looks good yeah plus five all right so now let’s do f
times g at six so this will be f of six times g of six and f of six is what um

00:11
36 plus once that’s 37 and then g of two is sorry g of 6 is 2
so that is 37 times 2 which is 4 carry the 1
74 right and so let’s look at f of 5 divided by g of 5.
now if g of 5 is 0 then we’ll have to erase this and say does not exist
but let’s see if we can do it so f of five is 25 plus one so 26 over
and now let’s see if we get zero here or not
g of five so five five minus four is one so we’re good shape if this turned out
to be zero for example if they had asked us what is what is this at four in fact
i’ll just do that over here let’s put a 5 back in here that’s just

  1. what if they asked us to find g of 4 sorry f over g at 4

00:12
so this would have been 16 plus 1 which is 17 and then this would be zero
so you know you wouldn’t leave your answer like this this is not something
you would write down because this this this is not a number
so you’re not going to be able to use it with an equal sign that’s not the way
the equal sign makes any sense so you cannot write this down so what we’ll
just say is this right here is is undefined it’s not defined
all right so now let’s do this one right here so f over g at minus one
and so that’ll be f of minus one over g at minus one and then minus g of three
all right so i just want to write it out a little bit better here now let’s go
substitute everything in so g at my f at minus one so that would be a two
and then g at minus one will be minus five
minus and then g at three which will be another minus one so that’ll be you know

00:13
minus two over five plus one which will be five over five
so that’ll be three fifths all right looks good
all right so let’s see what else we can do let’s use graphs to evaluate now
yeah let’s use graphs to evaluate so i’m going to give you some graphs and
we’re going to have to evaluate now instead of giving rules so let’s
let’s do something like that so let’s write a graph for f
and so it’s going to be shaped like a v it’s going to come down and then go back
up and this is going to be two this is going to be about a one here
about a three and then about a four um and the height here is a four
three and then two um and then a one let me see if i can space them better

00:14
so let’s say one two three all right that’s good enough and
so now let’s say here we have a uh g over here and let’s just say this one goes
straight down let’s put this out of 4 also
and i’ll say there’s a three and a two and a one right there
and then this is a four and then halfway is a two and then a
three and then a one all right there we go that’s good enough
there’s an f and there’s a g and so let’s evaluate some
combinate function combination so let’s do f plus g at three
and let’s do f minus g at one and let’s do f over g at two and then
we have room for one more here let’s do f times g at four so let’s find these
let’s evaluate these functions right here in these places right here so f plus g

00:15
at three right so that’s f of three plus g of three what is f of three
f of three here is three g at three is one so that’s a 4.
and so f of 1 minus g at 1 so f of 1 is 3 and g at 1 is a three so this is zero
f over g the quotient here that’s f of two divided by g of two if possible
so f of two is 0 so that’s 0 over g of 2 is not 0. it’s a 2 right here

00:16
so 0 over 2 is 0. and then here we have f of 4 times g of 4 f 4 is 4 g of 4 is 0
so this one is also 0 here so sometimes you might have a graph you
can evaluate these functions right here without having to go actually make a
whole graph of these right here if you’re just looking to evaluate these
functions right here we can just evaluate them right here
but you know what let’s go ahead and use uh some functions now and let’s change
this up here let’s go using graphs to evaluate now let’s use using graphs to
graph let’s and let’s pick on the sum right here so let’s do something like
that real quick let me get this out of the way real quick
and here we go so i’m going to give you um this um
yeah so here we go so i’m going to give you an f and a g and then we’re going to
graph the sum let’s see if we can do this so here’s going to be a function right

00:17
here i’m going to need a 1 2 3 1 2 3 and i’m going to need a 2 about here
and a 4 about here so i’m going to go and let’s say one’s about here
so i’m going to go here up to right there
and then i’m going to come down to about right there
and this is going to be the graph of my f so this is going to go up to about 2
right there so this is a graph of f now let’s look at the graph of g
so g is going to go from right here that’s going to be a minus 1
to 0 right there and it’s going to go [Music]
up right here also and hit right there and then it’s going to come back down
and hit right there okay so there’s g right there it’s just
coming straight up right there it goes through one

00:18
and then it comes back down and goes right there so that’s that’s g now to
see g are different from f let’s put g in red right here so there’s g right
there and it’s coming down through there so there’s g i’ll put g right there
all right so maybe that’s hard to read maybe i can just
make that a little bit redder and this part here a little bit redder
and so yeah there we go so let’s look at what would the graph of f plus g look
like all right so let’s do that over here so um let’s find f plus g at
zero f plus g at one f plus g at two and f plus g
at three let’s find these right here and then connect them with lines so what
we’re going to do is say f plus g so here’s f and g right here so what is f

00:19
of 0 right that’s f of 0 plus g of 0. so f of 0 is 2 and g at 0 is minus 1
so that comes out to be a one and what is f plus g at one it’s three plus zero
right so f of one is three and then g of one is zero so that gives us three
and then f of two plus g of two so f of two is one and g of two is one
so it’s one plus one is two and this one right here f of f of 3 right here is
is so f is coming right through here f bounces at 2 and
comes back up so that’s f right there so f of 3 is 2 and g at three is zero
so this is also two so at zero we’re going to be at one so

00:20
at zero we’re going to be height of one and at one we’re gonna be a height of 3
and then at 2 we’re going to be a height of 2.
well i kind of made that look messy all right and then at 3 we’re still at 2
so it’s constant right there that kind of makes sense because
when we pick a number between two and three say like this one right here
we’re a little bit short of two but how much short of two are we that part right
there so if you combine those together and add them together so this is the
graph of f plus g all right so that looks good all right so
let me know if you guys had fun in this video let me know if you want to see

00:21
some more examples i look forward to seeing you
in the next video and i’ll see you then have a great day
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About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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