Transcript for:
Operations with Functions Overview

In these examples, we're asked to find f plus g of x, f minus g of x, f times g of x, and f divided by g of x, given f of x and g of x. And we'll also give the domain in each case. Well, f plus g of x, looking at our definitions here, is equal to f of x plus g of x, where f of x is the quantity x squared minus three x. and g of x is the quantity x plus six.

So to perform this addition, let's go ahead and put the parentheses. If it's helpful, we can think of distributing a positive one, which really isn't going to change anything. This would give us just x squared minus three x plus x plus six. I think it's helpful to always put parentheses around each function, even though with addition it's not needed and is needed for subtraction.

So now we'll combine like terms. We would have x squared and then negative three x plus one x is negative two x or minus two x plus six. So f plus g of x is equal to x squared minus two x plus six.

The domain of the sum of these two functions is equal to the intersections of the domains of f of x and g of x. But since f of x is a quadratic function where the domain is all reals, and g of x is a linear function where the domain is all reals, the domain of the sum of the two functions is also all reals. So we can say all real numbers are all reals, or using interval notation we could say from negative infinity to positive infinity. Next we have f minus g of x, which is equal to f of x minus g of x.

So it's just like this sum here, except we have a difference. So we have x squared minus three x minus x plus six. And again, we can think of clearing the parentheses, so we can distribute a one here, but because of the minus, we can think of distributing a negative one.

So this would give us x squared minus three x, and then minus one x, or minus x. and then minus six. So combining like terms, we would have x squared, then minus four x, minus six. So, f minus g of x is equal to x squared, minus four x, minus six.

The domain of the difference of two functions is also the intersection of the domains of f of x and g of x. So once again, The domain of our difference here is all reals, or all real numbers. Or if we want, use the interval notation, the interval from negative infinity to positive infinity. Now let's take a look at the product and the quotient of f of x and g of x. F times g of x is equal to f of x times g of x.

So in this case, we would have the quantity x squared minus three x. times the quantity x plus six. So we'll have four products here, one, two, three, and four.

So x squared times x is x to the third. X squared times six, that's plus six x squared. And then we have negative three x times x, that's negative three x squared or minus three x squared.

And then finally we have negative three x times positive six. That's negative 18 x or minus 18 x. Last step, we'll combine like terms. These two terms here are both x squared terms. So we have x cubed plus three x squared.

minus 18x, so f times g of x is equal to x cubed plus 3x squared minus 18x. The domain of the product of f of x and g of x is the intersection of the domain of f of x and g of x once again. And because both domains are all reals, the domain of this product is also all reals. Or once again using interval notation, we can say the interval from negative infinity to positive infinity.

And finally, for f divided by g of x, this is equal to f of x divided by g of x, where f of x is x squared minus three x, and g of x is x plus six. Even though the numerator does factor with a common factor of x, nothing is going to simplify. And therefore we can go and leave it in this form, f divided by g of x is equal to the quantity x squared minus three x divided by x plus six. Of course if we wanted to, we could factor the numerator and write this as x times the quantity x minus three divided by the quantity x plus six. Just be careful, we cannot simplify these x's here because we cannot simplify across this addition, we also cannot simplify the three and the six once again because we cannot simplify across addition or subtraction.

We do need to be careful about the domain of this quotient though. The domain consists of the intersection of the domain of f of x and g of x for which g of x, the denominator, does not equal zero. And since the domain of f of x and g of x is all real numbers, we only have to exclude the value of x where x plus six would be equal to zero. which would be negative six.

So we can say the domain would be all reals, or all real numbers, except x equals negative six. We're using interval notation to exclude negative six. We'd have the interval from negative infinity to negative six, open on six, union, open on negative six to positive infinity. I hope you found this helpful.