Operations on Functions
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What does it mean to operate on functions?
"Operations on functions" is the process of taking two functions and applying binary arithmetical operations to those functions. This is a fancy way of saying that, in addition to evaluating and graphing functions, you can also do arithmetic with them.
What are the four function operations?
The four function operations are the same as the four operations in basic arithmetic; namely, addition, subtraction, multiplication, and division. These are called "binary" operations because you're taking two things (functions, in this case) and putting the operation symbol between them. You can add one function to another, subtract one function from another, multiply one function against another, or divide one function into another.
Thinking back to grammar school, you learned that you can add, subtract, multiply, and divide numbers. Then you learned that you can add, subtract, multiply, and divide variables and polynomials. Now you will learn that you can also add, subtract, multiply, and divide functions.
Performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.
What is the notation for function operations?
The notation for function operations is the same as the notation for number (that is, arithmetical) operations:
- addition: (g + h)(x) = g(x) + h(x)
- subtraction: (g − h)(x) = g(x) − h(x)
- multiplication: (g × h)(x) = g(x) × h(x)
- division: (g ÷ h)(x) = g(x) ÷ h(x)
Since you're in algebra now, you are familiar with multiplication "by juxtaposition"; that is, you know that you can multiply things by simply putting them next to each other. For instance, 2 can be multiplied by x by simply writing 2x; there is no need for a "times" symbol between them. In the same way, you can indicate the multiplication of functions by simply putting the two functions next to each other:
multiplication: (gh)(x) = g(x)h(x)
Also, you've learned that fractions are division, and that division can be indicated by creating a fraction. So the division of functions can also be written as:
division:
Working exercises on function operations is really nothing more than applying the arithmetic you've always used.
- Given f (x) = 3x + 2 and g(x) = 4 − 5x, find (f + g)(x), (f − g)(x), (f × g)(x), and (f / g)(x)
To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.
(f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 − 5x]
= 3x + 2 + 4 − 5x
= 3x − 5x + 2 + 4
= −2x + 6
(f − g)(x) = f (x) − g(x)
= [3x + 2] − [4 − 5x]
= 3x + 2 − 4 + 5x
= 3x + 5x + 2 − 4
= 8x − 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 − 5x)
= 12x + 8 − 15x2 − 10x
= −15x2 + 2x + 8
![(f/g)(x) = [f(x)]/[g(x)] = [3x + 2]/[4 − 5x]](fcns/fcnops01.gif){kind=small}
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = −2x + 6
( f − g ) (x) = 8x − 2
( f × g ) (x) = −15x2 + 2x + 8
- Given f (x) = 2x, g(x) = x + 4, and h(x) = 5 − x3, find (f + g)(2), (h − g)(2), (f × h)(2), and (h / g)(2)
This exercise differs from the previous one in that I not only have to do the operations with the functions, but I also have to evaluate at a particular x-value. To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or else I can find the values of the functions at x = 2 and then work from there. It's probably simpler in this case to evaluate first, so:
f (2) = 2(2) = 4
g(2) = (2) + 4 = 6
h(2) = 5 − (2)3 = 5 − 8 = −3
Now I can evaluate the listed expressions:
(f + g)(2) = f (2) + g(2)
= 4 + 6 = 10
(h − g)(2) = h(2) − g(2)
= −3 − 6 = −9
(f × h)(2) = f (2) × h(2)
= (4)(−3)= −12
(h / g)(2) = h(2) ÷ g(2)
= −3 ÷ 6 = −0.5
Then my answer is:
(f + g)(2) = 10
(h − g)(2) = −9
(f × h)(2) = −12
(h / g)(2) = −0.5
If you work symbolically first, and plug in the x-value only at the end, you'll still get the same results. Either way will work. Evaluating first is usually easier, but the choice is up to you.
- Given f (x) = 3x2 − x + 4, find the simplified form of the following expression, and evaluate at h = 0:
![[f(x + h) − f(x)] / h](fcns/fcnops02.gif){kind=small}
This isn't really a functions-operations question, but something like this often arises in the functions-operations context. This looks much worse than it is; it's completely do-able, as long as I'm willing to take the time to be careful and complete.
The simplest way for me to proceed with this exercise is to work in pieces, simplifying as I go; then I'll put everything together and simplify at the end.
For the first part of the numerator, I need to plug the expression x + h in for every x in the formula for the function, using what I've learned about function notation, and then simplify:
f (x + h)
= 3(x + h)2 − (x + h) + 4
= 3(x2 + 2xh + h2) − x − h + 4
= 3x2 + 6xh + 3h2 − x − h + 4
The expression for the second part of the numerator is just the function itself:
f (x) = 3x2 − x + 4
Now I'll subtract and simplify:
f (x + h) − f (x)
= [3x2 + 6xh + 3h2 − x − h + 4] − [3x2 − x + 4]
= 3x2 + 6xh + 3h2 − x − h + 4 − 3x2 + x − 4
= 3x2 − 3x2 + 6xh + 3h2 − x + x − h + 4 − 4
= 6xh + 3h2 − h
All that remains is to divide by the denominator. Factoring lets me simplify:
Now I'm supposed to evaluate at h = 0, so:
6x + 3(0) − 1 = 6x − 1
simplified form: 6x + 3h − 1
value at h = 0: 6x − 1
That's pretty much all there is to operations on functions, at least until you get to function composition. Don't let the notation for this topic worry you; it means nothing more than exactly what it says: add, subtract, multiply, or divide; then simplify and evaluate as necessary. Don't overthink this. It really is this simple.
Oh, and that last example? They put that in there so you can "practice" stuff you'll be doing in calculus. Of course, you likely won't remember this by the time you actually get to calculus (I didn't), but, once you get to calculus, you'll follow a very similar process for finding something called "derivatives".
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