Functions Part Two
Composite Functions
Consider the function f(x) = √(x + 4). It is quite possible that we can simplify this function into other smaller functions by segregating the operations that constitute the function. The above function can be separated into a square root of x and the addition of 4 to x. Thus, g(x) = √x & h(x) = x + 4. Now that we know the function has two evaluation steps, the function f(x) can be seen as a combination of g(x) and h(x). Thus now we can represent f(x) into:
The above is showing the simplification for the composite function and how we work from the inside of the function towards the outside. First we would substitute the innermost function given and then apply each evaluation on it from the outer layers. f is obtained by substituting h into g, this can be read as g of h of x. The mapping diagram for the same is:
We start with a number x in the domain of h and find its image in h(x). If this number h(x) is in the domain of g, we then compute the value of g(h(x)). The resulting composite function will be noted as:
Thus we can define a composite function as:
Example, given f(x) = 3x-6 and g(x) = (1/3)x+2, find the following:
The above example shows that f(g(x)) = g(f(x)) however this is only where there is a special relationship between the two function. But in general, f(g(x)) ≠ g(f(x)).
Decomposing a Composite Function
In the first example given for explaining composite functions, we created a single function using two functions. Now we will use that same idea to decompose a single function into its constituents.
Example 1:
k(x) = 2^(4x+1)
In order to evaluate k(x) you will first require to evaluate 4x+1 and then evaluate the expression 2^x. This divides our function into two parts, which can now be written as two different functions.
z(x) = 4x+1
d(x) = 2^x
k(x) = d(z(x))
Example 2:
p(x) = ∛(x²+4)
g(x) = ∛x (Cube rooting x)
h(x) = x² (Squaring x)
f(x) = x+4 (Adding 4 to x)
The above segregates the function into its parts and now we would be able to write it as a composite function again by carrying out the simplest operation first and then going up the chain.
h(x) (Innermost function, only squaring x)
f(h(x)) (Second level, adding 4 to x)
g(f(h(x))) (Third level, cube rooting the whole expression)
Therefore, g(f(h(x))) in words is, g of f of h of x. p(x) = g(f(h(x))).
Finding the Domain of a Composite Function
Referring to the mapping for the composite function above, it is important to note that in order for x to be in the domain of the composite function g o h two of the following conditions must be met:
- x must be in the domain of h.
- h(x) must be in the domain of g.
It is also worth noting that g(h(x)) is in the range of g o h only if x is in the domain of g o h. The following are the example to clarify the concept of the range and domain of a composite function.
g(x) = x²-8
h(x) = √x
Find the domain and range of the following:
g o h, firstly we would require to establish a domain and range for g and h. For g(x) = x²-8 the domain is x ∈ ℝ and the range is y ≥ -8. For h(x) = √x the domain is x ≥ 0 and the range is y ≥ 0.
To be in the domain of g o h, √x mus be defined for x as x ≥ 0. Therefore the domain of g of h is x≥0. The range of y = x-8 is y≥-8.
Thus we can conclude that (g o h)(x) = x-4 and its domain is x≥0 and is range is y≥-8.
This sums up the story, composite functions, decomposing them, finding their domain and range. In the next section of the functions series we will explore inverse functions and the transformation of functions.
