Hi, and welcome. This is Anthony Varela. And today, we're going to multiply and divide functions.
So we're working with two functions. We can call them f of x and g of x. And we're going to see how we can multiply two functions at a different value, so f of a times g of b. We'll go through an example with division as well.
And then we're going to take a look at when their arguments are the same and how we can write one function and then evaluate that. And we'll do it with division as well. So let's go through our multiplication examples first.
We'll say that f of x equals 5x plus 3, and g of x equals negative 2x plus 1. And we'd like to evaluate f of 3 times g of negative 1. So what we could do is evaluate each function separately. So we'll evaluate f of 3, and then we'll evaluate g of negative 1 and then multiply those two values together.
So to evaluate f of 3, we take f of x and we just replace x with 3. So we have 5 times 3 plus 3. Well, this is 15 plus 3 or 18. So we know that f of 3 equals 18.
Now we evaluate g of negative 1. So we take our function g of x-- so negative 2x plus 1 and rewrite in negative 1 for every instance of x. So negative 2 times negative 1 is a positive 2. And then when we add 1. We get g of negative 1 equals 3.
So now we know what f of 3 is. We know what g of negative 1 is. We can multiply those two together. 18 times 3 is 54.
Now we'd like to evaluate f of 2 times g of 2. And now the interesting thing here is they both have the same argument. So we could still multiply or evaluate f of 2 and evaluate g of 2 and then multiply them together, but what we could do is write one function.
And so we can define f times g of x and then evaluate it at x equals 2. So we write it with a dot in between f and g. And it's actually very important that it is a filled in dot. If it's an open circle, it means something entirely different. So make sure you're using a filled in dot-- so f times g of x.
And what we do is write our expression for f of x-- so 5x plus 3, and multiply that by our expression for g of x negative 2x plus 1. Well, this looks familiar. I can use FOIL to write this out.
So 5x times negative 2x, and then 5x times 1, then 3 times negative 2x, and finally 3 times 1. Combining all like terms, here is our single expression for this function, f times g of x. Now we can write f times g of 2.
So just replacing x with 2-- so negative 10 times 2 squared minus 2 plus 3. Well, this is negative 40 plus 1. So f times g of 2 is negative 39. Now let's go through our division examples.
So here f of x equals 3x squared minus 5x plus 1, and g of x equals 2x plus 3. And we would like to evaluate f of 1 divided by g of 3. So we are going to evaluate each function separately. And then we'll just divide them. And because we're dealing with division, we just have to make sure that g of 3 does not equal 0.
If it does, then we don't have a defined solution. So let's go ahead and evaluate f of x. So we're using our expression for f of x-- so 3x squared minus 5x plus 1 and substituting 1 in for x and evaluating. So we have 3 times 1 squared minus 5 times 1 plus 1. So 3 minus 5 plus 1.
So we see that f of 1 equals negative 1. Now, let's evaluate g of 3. So we're using g of x equals 2x plus 3 plugging in 3 for x. So 2 times 3 is 6. And then when we add 3, we get 9.
So g of 3 equals 9. So now that we have f of 1 and g of 3, we divide the 2 and we see that this equals negative 1 divided by 9. I could express that as a fraction negative 1/9.
Now let's evaluate f of 2 divided by g of 2. And once again, here, we see that same argument. So we could define f divided by g of x and then evaluate that at x equals 2. So what we're going to do is write out our expression then for f of x in a numerator. And g of x is then in the denominator.
And once again, we have to make sure then that our denominator does not equal 0. So now we can plug-in 2 for x and evaluate then our function here. So f/g of 2 equals-- let's see, when x equals 2, 3x squared is 12. Minus 5x is minus 10. And then we have our plus 1.
When x equals 2, 2x equals 4, and then we have plus 3. So our numerator simplifies to 3. The denominator simplifies to 7. So f of 2 divided by g of 2 equals 3/7.
So let's review multiplying and dividing functions. Well, in multiplying functions, we can evaluate each function separately and then multiply together. If we have the same argument, we can define one function, which we write as f times g of x and then evaluate it.
And same thing with dividing functions, we're just going to be dividing instead of multiplying. So you could evaluate each function separately and then divide. If you have the same argument, you can express it as one function, rewrite f/g of x, but you have to make sure with division that the whatever you're dividing by does not equal 0. So thanks for watching this tutorial on multiplying and dividing functions. Hope to see you next time.