Composing Functions at a Point
Purplemath
Composition of functions is the process of plugging one function into another, and simplifying or evaluating the result at a given x-value.
Suppose you are given the two functions f(x) = 2x + 3 and g(x) = −x2 + 5. Composition means that you can plug g(x) into f(x), (or vice versa).
The composition of f(x) with g(x) is written as "(f ∘ g)(x)", which is pronounced as "f-compose-g of x". And "(f ∘ g)(x)" means "f(g(x))", which is "f-of-g of x".
When composing functions f and g, you plug something in for x, then you plug that value into g, simplify, and then plug the result into f. The process here is just like what we saw on the previous page, except that now we will be using formulas to find the output values, rather than just reading the values from lists of points or from graphs.
What is an example of composing two functions and evaluating at a given x-value?
- Given f (x) = 2x + 3 and g(x) = −x2 + 5, find (g ∘ f )(1).
When I work with function composition, I usually convert "(f ∘ g)(x)" to the more intuitive "f (g(x))" form. This is not required, but I certainly find it helpful. In this case, I get:
(g ∘ f )(1) = g(f (1))
This means that, working from right to left (or from the inside out), I am plugging x = 1 into f(x), evaluating f(x), and then plugging the result into g(x). I can do the calculations bit by bit, like this:
f (1) = 2(1) + 3 = 2 + 3 = 5
g(5) = −(5)2 + 5 = −25 + 5 = −20
(g ∘ f )(1) = g(f (1)) = g(5) = −20
Doing the calculations all together (which you may find useful later on when we're doing things symbolically) looks like this:
(g ∘ f )(1) = g(f (1))
= g(2( ) + 3)
= g(2(1) + 3)
= g(2 + 3)
= g(5)
= −( )2 + 5
= −(5)2 + 5
= −25 + 5
= −20
(Note how I wrote each function's rule clearly, leaving open parentheses for where the input [x or whatever] would go. This is a useful technique.)
Whichever method I use (bit-by-bit or all-in-one), my answer is:
(g ∘ f )(1) = g(f (1)) = −20
Above, I worked with the composition (g ∘ f )(1); the composition can also work with the functions in the opposite order:
- Given f (x) = 2x + 3 and g(x) = −x2 + 5, find (f ∘ g)(1).
First, I'll convert this from function-compositional form to the more intuitive form, and then I'll simplify:
(f ∘ g)(1) = f (g(1))
Working the composition bit-by-bit, my steps are the following:
g(1) = −(1)2 + 5 = −1 + 5 = 4
f (4) = 2(4) + 3 = 8 + 3 = 11
(f ∘ g)(1) = f (g(1)) = f (4) = 11
On the other hand, working all-in-one (right to left, or from the inside out), I get this:
(f ∘ g)(1) = f (g(1))
= f (−( )2 + 5)
= f (−(1)2 + 5)
= f (−1 + 5)
= f (4)
= 2( ) + 3
= 2(4) + 3
= 8 + 3
= 11
Either way, my answer is:
(f ∘ g)(1) = f (g(1)) = 11
What are "fog" and "gof"?
These are mispronunciations based on treating the circle character, ∘, as a lowercase letter O: despite what some might say, "f ∘ g" is not pronounced as "fogg" and "g ∘ f " is not pronounced as "goff". They are pronounced as "f-compose-g" and "g-compose-f", respectively.
Don't make yourself sound ignorant by pronouncing these wrongly.
As you have seen above, you can plug one function into another. You can also plug a function into itself.
What is an example of a function composed with itself?
- Given f (x) = 2x + 3 and g(x) = −x2 + 5, find (f ∘ f )(1).
I will work this composition bit-by-bit:
f (1) = 2(1) + 3 = 5
f (5) = 2(5) + 3 = 13
f (f (1)) = f (5) = 13
Writing out all the steps in one go:
(f ∘ f )(1) = f (f (1))
= f (2( ) + 3)
= f (2(1) + 3)
= f (2 + 3)
= f (5)
= 2( ) + 3
= 2(5) + 3
= 10 + 3
= 13
Either way, my answer is:
f (f (1)) = 13
- Given f (x) = 2x + 3 and g(x) = −x2 + 5, find (g ∘ g)(1).
Working bit-by-bit (which I find is the easier way to go), I get:
g(1) = −(1)2 + 5 = 4
g(4) = −(4)2 + 5 = −16 + 5 = −11
g(g(1)) = g(4) = −11
Doing it all in one go (which is easier to read), the steps look like this:
(g ∘ g)(1) = g(g(1))
= g(−( )2 + 5)
= g(−(1)2 + 5)
= g(−1 + 5)
= g(4)
= −( )2 + 5
= −(4)2 + 5
= −16 + 5
= −11
Either way, my answer is:
g(g(1)) = −11
In each of these exercises, using either method of doing the computations (whether bit-by-bit or all in one go), I wrote out my steps carefully, and used parentheses to indicate where my input went with respect to a function's formula. You should use the method — bit-by-bit or all at once — that works better for you. Whichever method you pick, though, take the time to be clear in each of your steps.
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