+
3 Tutorials that teach Multiplying and Dividing Functions
Take your pick:
Multiplying and Dividing Functions

Multiplying and Dividing Functions

Description:

Multiplying and Dividing Functions

(more)
See More

Try Our College Algebra Course. For FREE.

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.*

Begin Free Trial
No credit card required

25 Sophia partners guarantee credit transfer.

221 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 20 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.

Tutorial

  • Multiplying Two Functions
  • Dividing Two Functions

Multiplying Two Functions

Oftentimes when working with functions, we will be asked to determine the result of multiplying two functions. When this occurs, there are two ways to approach the problem. One method is that we can evaluate each function for the value we are given and then multiply. In the second method, if the two functions are in terms of the same variable and are being evaluated for the same value, then we can multiply both functions together and substitute in the value we are given to find a solution. Let's look at both methods.

Method 1:

Find the value of f(3)g(2) when f left parenthesis x right parenthesis equals x squared minus 2 and g left parenthesis x right parenthesis equals x plus 7. To solve this problem first evaluate the functions f(x) when x equals 3 and g(x) when x equals 2, separately.

 

When given two functions evaluated at the different values simply multiply them together.

Method 2:

Now suppose we are trying to find the value of f(3)g(3) using the same functions given in Method 1, because each function is dependent on x and we are evaluating it for them for the same value we can multiply both functions first and then evaluate them for the given value. Let's look at how to do this.

When multiplying two functions given in terms of the same variable and evaluated at the same value we can multiply both equations first and then evaluate for the value we are given. Often times the notation used for this is f(a)g(a) = (f•g)(a).

Dividing Two Functions

When we divide two functions we go through a process very similar to that used when multiplying two functions; the only difference is that we have to perform division instead of multiplication. Let's look at an example.

 

When there is division between two functions you can first evaluate each functions for the given value and then perform the division.

Similar to Method 2 discussed for multiplying two functions, we can also perform division between two functions first and them evaluate for a given value if the two functions are dependent on the same variable and evaluated for the same value. For example, for the same functions given above, if we are asked to find f(2)/g(2) we could do the following.

As with multiplying two functions you can use either method shown above the first method works quite well for simple functions but the second method can be much more useful when working with more complex functions. Unlike with multiplication you have to be careful that you do not divide by 0 when dividing two functions. If we have the generic form f(x)/g(x) when g(x) equals 0 the function will be undefined.

When dividing two functions, you can evaluate each function, and then divide. If both functions are defined by the same variable and evaluated at the same value, we can divide both functions first and then evaluate for the given value. In the second case, the notation used is f(a)/g(a) = (f/g)(a).

We must be sure that the function in the denominator does not equal zero. If this does happen, then our function would be "undefined" or we would simply have "no solution."  Note that the function in the denominator may return an undefined solution for only a certain domain. In our previous example,  f(x)/g(x) would be undefined at x = 3, because g(3) = 0, however the function would have a value for other values of x.