Hi everyone.
Today we’re going to talk about how to calculate combinations of two functions and then define
the domain of those combinations.
To complete this problem, we will calculate our combinations, simplify and then define
the domain of each.
Let’s take a look.
In this particular problem, we’ve been given the two functions f of x equals x cubed plus
2 x squared and g of x equals 3 x squared minus 1 and asked to find the following four
functions and their domains.
So as you can see, we’re going to be finding f plus g, f minus g, f times g and f divided
by g and we’re just talking here about combinations of functions.
These are different combinations of f and g.
The first one is obviously just the sum of the functions, the difference of the functions,
the multiplication and the division of the functions or the product and the quotient
of the functions.
So let’s go ahead and start with f plus g of x and this is going to be really simple.
When we add functions together like this to find a combination of these functions, we
will just add f and g together.
So f here is x cubed plus 2 x squared and g of x is 3 x squared minus 1.
So we will add 3 x squared minus 1 and now we just need to simplify that.
As you can see, we will get x cubed.
When we add 2 x squared and 3 x squared together, we will get 5 x squared and then of course
our negative 1.
And that’s it.
That’s really as far as we can simplify.
We’ve also been asked to give the domain of each of these combinations of functions.
So we will go ahead and do that and we will say that the domain here, the domain is equal
to – the domain of any polynomial function like this is all real numbers so we can just
denote that with all real numbers like that.
As far as the difference of the two functions, really similar, we have f minus g of x here.
So we will just take f, just x cubed plus 2 x squared and we will subtract g of x.
Don’t forget to put g of x in parenthesis like that so that you make sure you distribute
the negative sign.
So when we subtract here, we will get x cubed. 2 x squared minus 3 x squared will give us
a negative x squared and then we have minus negative 1 which will give us a plus 1.
Again, we’ve really simplified as much as we can.
So now we just need to indicate the domain and in this case, again we have a polynomial
function so the domain is all real numbers.
If we look here at the product of these two functions f times g of x, similar concept,
we will just be multiplying the two together so x cubed plus 2 x squared and then multiply
that by g of x which is 3 x squared minus 1.
So now we will be expanding these with our FOIL method.
So we will have x cubed times 3 x squared which will give us 3 x to the fifth.
x cubed times a negative 1 will give us negative x cubed. 2 x squared times 3 x squared will
give us plus 6 x to the fourth and 2 x squared times the negative 1 will give us a negative
2 x squared.
So now when we reorder our terms, we will get 3 x to the fifth plus 6 x to the fourth
minus x cubed minus 2 x squared and again we have a polynomial which means that our
domain is all real numbers.
Finally our last function here, the quotient of f and g, we will take f of x and we will
divide it by g of x.
So we will see that our numerator here will be f of x which is x cubed plus 2 x squared
and we will divide that by g of x which we know to be 3 x squared minus 1 and there’s
really not a whole lot we can do here to simplify.
We could factor out an x squared from our numerator but that wouldn’t allow us to
cancel anything.
So this will be our quotient function.
We just need to figure out what the domain is.
Remember that the domain of a rational function like this, rational function being the quotient
of two polynomials, a polynomial and a numerator and a polynomial and a denominator, is undefined
wherever the denominator is equal to zero.
So the domain is going to be all real numbers except where the denominator equals zero.
So we need to figure out where that is the case.
So we will say over here, 3 x squared minus 1 equal to zero, and we need to solve this
to figure out which part of our domain is undefined.
So we will add 1 to both sides to get 3 x squared equals 1.
Divide both sides by 3 to get x squared equals one-third and then we want to take the square
root of both sides to solve for x.
So we will get x equals the square root of one-third.
So we will say the square root of one-third and remember that when we’re taking the
square root of a fraction like this, of a quotient, we can take the square root of the
numerator and the denominator separately.
So we will get x equals the square root of 1 over the square root of 3.
We know that the square root of 1 is just 1 and then we have the square root of 3 here
in the denominator.
Remember we have to include a positive or negative because we could get positive 1 over
the square root of 3 or negative 1 over the square root of 3.
Both of those would make our denominator equal to zero and therefore our quotient undefined.
So we can say that the domain of this function is equal to all real numbers except x equals
positive or negative 1 over the square root of 3.
So we can just say the domain is x not equal to positive or negative 1 over root 3 and
that indicates that x can be any value except these two and defines our domain.
That’s it.
That’s how you find the combinations of functions and their domains.
So I hope you found that video helpful.
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