Math Lesson 16.4.2 - Properties of Composite Functions

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Welcome to our Math lesson on Properties of Composite Functions, this is the second lesson of our suite of math lessons covering the topic of Composite Functions, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Properties of Composite Functions

From the above example, it is clear that composite functions have the following properties:

1. For any function f(x)
   
   f(c · x) ≠ c ∙ f(x)
   where c is a constant.
   For example, if f(x) = 2x + 1 and c = 3, then
   f(c · x) = f(3x)
   = 2 ∙ 3x + 1
   = 6x + 1
   On the other hand,
   c ∙ f(x) = 3 ∙ (2x + 1)
   = 6x - 3
2. If f(x) and g(x) are functions, then in general
   f ∘ g(x) ≠ g ∘ f(x)
   We have explained this property through examples earlier in this tutorial. Thus, in general, the composition of functions is not commutative.
3. Composite functions are associative in the sense that if we have three functions f(x), g(x) and h(x), then
   f ∘ (g ∘ h) = (f ∘ g) ∘ h
   For example, if f(x) = 2x, g(x) = x^x and h(x) = 3/4x, then for f ◦ (g ◦ h) we begin with g ◦ h, where the entire function h(x) is inserted in the place of x in the function g(x), i.e.
   g ∘ h = g[h(x)]
   = (3/4x)²
   = 9/16x²
   Then, we insert the function f(x) in the place of x in the above function to find f ◦ (g ◦ h).
   f ∘ (g ∘ h) = f(g ∘ h)
   = 2 ∙ 9/16x²
   = 18/16x²
   = 9/8x²
   As for (f ◦ g) ◦ h we begin with f ◦ g by inserting the entire function g(x) in the place of x in the function f(x), i.e.
   f ∘ g = f[g(x)]
   = 2 ⋅ (x² )
   = 2x²
   Then, we find (f ◦ g) ◦ h, by inserting the entire function h(x) in the place of x in the above function, i.e.
   (f ∘ g) ∘ h = 2 ⋅ (3/4x)²
   = 2 ⋅ 9/16x²
   = 18/16x²
   = 9/8x²
   As you see, the composite functions obtained are the same in both cases. Hence, all the associative property of composite functions is true.
4. If f(x) and g(x) are inverse functions in the sense that f(x) = g-1(x), then
   f ∘ [g(x)] = g ∘ [f(x)] = xⁿ
   where n is an integer.
   Like in numbers, two functions are inverse if their product a coefficient equal to 1 followed by a power of x. For example, f(x) = x/2 is the inverse of g(x) = 2x because
   f(x) ∙ g(x) = x/2 ∙ 2x
   = 2x²/2
   = x²
   Now, let's check whether the fourth property of composite functions is true. Thus,
   f ∘ [g(x)] = 2x/2
   = x
   On the other hand,
   g ∘ [f(x)] = 2 ∙ x/2
   = 2x/2
   = x
   As you see, the results are the same, so the fourth property of composite functions is true.
5. The composition of one-to-one (injective) functions is also one-to-one (injective).
   For example, both functions f(x) = x - 1 and g(x) = 3x are one to one, as every value of the domain (input) has in correspondence a single value in the range (output). This is true for the two resulting combined functions as well. For f ◦ g (x) we have
   f ∘ g(x) = f[g(x)]
   = f(3x)
   = 3x - 1
   It is clear that this is an injective (one-to-one) function, as for every x-value there is a single f(x) value in correspondence.
   Likewise, for g ◦ f (x) we have
   g ∘ f(x) = g[f(x)]
   = g(x - 1)
   = 3 ⋅ (x - 1)
   = 3x - 3
   This is also an injective function, as for every x-value there is a single g(x) value in correspondence.
6. The composition of surjective (onto) functions is also surjective (onto).
   We have given the definition of surjective functions in tutorial 16.2. Thus, a function is surjective when all the output values have in correspondence at least one input value. For example, f(x) = x^x and g(x) = x3 - 3x + 1 are both surjective, as the horizontal line test gives more than one intercept for both functions, as shown below. Now, let's check whether the composite functions f ◦ g (x) and g ◦ f (x) are also surjective or not. We have
   f ∘ g(x) = f(x³-3x + 1)
   = (x³-3x + 1)²
   The horizontal line test for this function gives more than one intercept with the graph, so the composite function f ᵒ g (x) is also surjective, as shown in the figure below. As for g ◦ f (x), we have
   g ∘ f(x) = g(x²)
   = (x²)³ - 3 ∙ (x²) + 1
   = x⁶ - 3x² + 1
   This function is also sujective, as the horizontal line test gives more than one intercept, as shown in the figure below.
7. If f ◦ g(x) = g ◦ f(x), then f(x) = g(x).
   This is obvious, since in general f ◦ g(x) ≠ g ◦ f(x) as explained earlier. The only case when these two composite functions are equal is when the two individual functions are equal. For example, if f(x) = x^x - 2x + 1 and g(x) = (x - 1)2, then
   f ∘ g(x) = f[g(x)]
   = [(x - 1)²]² - 2 ∙ (x-1)² + 1
   = (x² - 2x + 1)² - 2(x² - 2x + 1) + 1
   = x⁴ - 2 ∙ (2x + 1) ⋅ 1 + 1² - 2x² + 4x - 2 + 1
   = x⁴ - 4x - 2 + 1 - 2x² + 4x - 2 + 1
   = x⁴ - 2
   On the other hand,
   g ∘ f(x) = g[f(x)]
   = [(x² - 2x + 1) - 1]²
   = (x² - 2x + 1)²-2(x² - 2x + 1) ∙ 1 + 1²
   = x⁴ - 2 ∙ (2x + 1) ∙ 1 + 1² - 2x² + 4x - 2 + 1
   = x⁴ - 4x - 2 + 1 - 2x² + 4x - 2 + 1
   = x⁴ - 2
   As you see, we obtained the same function in both compositions.

You have reached the end of Math lesson 16.4.2 Properties of Composite Functions. There are 9 lessons in this physics tutorial covering Composite Functions, you can access all the lessons from this tutorial below.

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