We are given f[x] and g[x],
and asked to perform function arithmetic.
We're asked to determine f plus g[x],
f minus g[x], f times g[x], and f divided by g[x].
For the first step, notice how f[x]
is given in factored form.
We will need to expand this.
To expand this, we have two factors of x plus two,
and then we multiply, which gives us four products.
X times x plus x times two plus two times x
plus two times two, which after simplifying
gives us x squared plus 4x plus four.
When performing the function arithmetic,
we will use the expanded form of f[x].
To begin, f plus g[x] is equal to f[x] plus g[x]
where f[x] is equal to the quantity x squared
plus 4x plus four.
Then we have plus g[x], which is equal to
the quantity two minus 3x.
Now, because we have addition, we simply combine like terms,
or if it's helpful, we can think of distributing
a positive one here and a positive one here
to clear the parenthesis.
We know multiplying by positive one doesn't change anything,
and therefore, this gives us x squared plus 4x plus four,
and then plus two minus 3x.
And now we combine like terms.
We have two x terms, and we have two constants.
Simplifying, we have x squared,
and then 4x minus 3x is 1x, or just x, giving us plus x.
And four plux two is six, giving us plus six.
F plus g[x] equals x squared plus x plus six.
Sometimes we are asked to give the domain
after performing function arithmetic.
In this case, because we have a quadratic function,
the domain would be all reals.
Next, we have f minus g[x],
which is equal to f[x] minus g[x],
which gives us the quantity x squared plus 4x plus four
minus the quantity two minus 3x.
Because we are subtracting g[x] here,
we do have to have g[x] in parenthesis
so that we subtract the entire function.
And again here, because of the subtraction
we have to be more careful.
We can think of distributing a positive one here,
but because of the subtraction,
we would think of distributing a negative one here.
So, distributing a positive one here
doesn't change anything.
We have x squared plus 4x plus four.
And then distributing negative one,
we have negative one times two, which is negative two,
giving us minus two, and then negative one
times a negative 3x is positive 3x, giving us plus 3x.
If we don't want to think of distributing,
we just have to be careful and subtract each term.
We have minus positive two, and then minus negative 3x,
which gives us plus 3x.
And now, we combine like terms.
We have two x terms and we have two constants.
So, we have x squared, and then 4x plus 3x is 7x,
giving us plus 7x.
And then four minus two is two, giving us plus two.
So again, f minus g[x] is a quadratic function,
and therefore, the domain again is all reals.
F minus g[x] is equal to x squared plus 7x plus two.
And now determine f times g[x].
F times g[x] is equal to f[x] times g[x]
which is equal to the quantity x squared
plus 4x plus four times the quantity two minus 3x.
Here because we have a trinomial times a binomial,
we will have six products.
We distribute the x squared first,
which gives us two products.
And then we distribute 4x, which gives us two more products.
And then finally we distribute the four,
giving us the last two products.
So, multiplying, we have x squared times two,
which is 2x squared.
And then we have x squared times negative 3x,
which is equal to negative 3x cubed,
giving us minus 3x cubed.
And now we distribute 4x.
4x times two is 8x, giving us plus 8x.
And then we have 4x times negative 3x,
which is negative 12x squared, giving us minus 12x squared.
And now we distribute four.
Four times two is eight, giving us plus eight,
and then four times negative 3x is negative 12x,
giving us minus 12x.
And notice how we have two x squared terms
and we have two x terms.
Simplifying, starting with the highest degree term,
we have negative 3x cubed, and then 2x squared
minus 12x squared is negative 10x squared,
giving us minus 10x squared.
And then 8x minus 12x is negative 4x, giving us minus 4x,
and then finally we have plus eight.
F times g[x] is equal to negative 3x cubed
minus 10x squared minus 4x plus eight,
and because we have a polynomial,
the domain of f times g[x] again, is all real numbers.
And then for part d, we have f divided by g[x],
which is equal to f[x] divided by g[x],
where we know f[x] is equal to
the quantity x squared plus 4x plus four,
and g[x] is equal to the quantity two minus 3x.
At this point, we would normally try to
factor the numerator and then simplify,
but we know the factored form of x squared plus 4x plus four
is the quantity x plus two squared,
so we can also give this quotient as
the quantity x plus two squared
divided by the quantity two minus 3x.
But again, it doesn't simplify,
and therefore, either form is acceptable for f minus g[x].
But because we have quotient,
the domain is not going to be all reals
because division by zero is undefined.
To determine which value to exclude from the domain,
we have to determine which value of x
makes the denominator equal to zero.
To do this, let's set two minus 3x equal to zero
and solve for x.
Solve for x, we subtract two on both sides,
giving us negative 3x equals negative two,
and then dividing both sides by negative three.
And simplifying, we have x equals positive 2/3.
So, when x equals positive 2/3
we have division by zero, which is undefined,
and therefore, the domain is all reals except x equals 2/3.
So, we'll say the domain
is all reals
except x equals 2/3.
But sometimes we just say x can't equal 2/3.
I hope you found this helpful.