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Chapter 6: Problem 2

Let \(h: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(h(x)=3 x+2\) and \(g: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(g(x)=x^{3} .\) Determine formulas for the composite functions \(g \circ h\) and \(h \circ g .\) Is the function \(g \circ h\) equal to the function \(h \circ g\) ? Explain. What does this tell you about the operation of composition of functions?

Short Answer

Expert verified
The composite functions \(g(h(x))\) and \(h(g(x))\) are given by: \(g(h(x)) = (3x + 2)^3\) and \(h(g(x)) = 3x^3 + 2\). Since they are different functions, we conclude that \(g(h(x)) \neq h(g(x))\). This tells us that the operation of composition of functions is not commutative, i.e., the order in which functions are composed matters.

Step by step solution

01

Find g(h(x))

To find the composite function g(h(x)), substitute the function h(x) into g(x) wherever there is an 'x' in the function g: \(g(h(x)) = (3x + 2)^3\)
02

Simplify g(h(x)) if possible

In this case, we don't need to simplify g(h(x)) since it is already in its simplest form: \(g(h(x)) = (3x + 2)^3\)
03

Find h(g(x))

To find the composite function h(g(x)), substitute the function g(x) into h(x) wherever there is an 'x' in the function h: \(h(g(x)) = 3(x^3) + 2\)
04

Simplify h(g(x)) if possible

In this case, we can simplify h(g(x)): \(h(g(x)) = 3x^3 + 2\)
05

Compare g(h(x)) and h(g(x))

Now, compare the composite functions to see if they are equal: \(g(h(x)) = (3x + 2)^3\) \(h(g(x)) = 3x^3 + 2\) As the two composite functions are different, we can conclude that \(g(h(x)) \neq h(g(x))\).
06

Explain the results

The fact that the composite functions are not equal tells us that the operation of composition of functions is not commutative. That is, the order in which functions are composed matters and swapping the order of composition may result in a different function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Composition
Function composition is a fundamental concept in mathematics. It involves taking two functions and combining them to form a new one. When you compose two functions, you take the output of one function and use it as the input for another. In the exercise, we define two functions:
  • Function \( h(x) = 3x + 2 \)
  • Function \( g(x) = x^3 \)
We then create two new functions by composing \( g \) and \( h \):
  • \( g \circ h \), where you replace \( x \) in \( g \) with \( h(x) \), giving \( g(h(x)) = (3x + 2)^3 \)
  • \( h \circ g \), where you replace \( x \) in \( h \) with \( g(x) \), resulting in \( h(g(x)) = 3x^3 + 2 \)
This concept is crucial as it lays the foundation for many advanced mathematical topics and applications.
Commutative Property
Unlike addition or multiplication of real numbers, function composition is not always commutative. This means changing the order in which you compose two functions can lead to different results. In our exercise, we observed:
  • \( g(h(x)) = (3x + 2)^3 \)
  • \( h(g(x)) = 3x^3 + 2 \)
Since these two results are not equal, it demonstrates that \( g \circ h eq h \circ g \). Thus, composition of functions depends on the order.Understanding the non-commutative nature of function composition is of great importance. Many mathematical operations and algorithms rely on this property, guiding the correct sequence of function applications.
Algebraic Functions
Algebraic functions like the ones in our exercise are functions that involve polynomials. These are expressions that include variables raised to whole number powers. In the exercise:
  • \( h(x) = 3x + 2 \) is a linear polynomial function.
  • \( g(x) = x^3 \) is a cubic polynomial function.
Understanding how to work with algebraic functions is essential. It forms the basis for more complex calculations involving function composition. Each function's degree influences the composition's complexity and resulting polynomial form. This knowledge is vital for solving many mathematical problems.
Mathematical Proofs
In mathematics, it's not just enough to state things; we prove them to ensure our statements hold true. Proofs are step-by-step explanations detailing why a mathematical fact or theorem is true.In this exercise, we have demonstrated via computation:
  • Finding \( g(h(x)) = (3x + 2)^3 \)
  • Finding \( h(g(x)) = 3x^3 + 2 \)
  • Concluding \( g(h(x)) eq h(g(x)) \)
This provides a proof that composition is not commutative for these functions. It shows how each function can be systematically substituted and calculated to verify the final result. Mathematical proofs like this one are crucial. They add rigor and reliability to mathematical arguments and solutions.

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Most popular questions from this chapter

Let \(A\) and \(B\) be two nonempty sets. There are two projection functions with domain \(A \times B,\) the Cartesian product of \(A\) and \(B .\) One projection function will map an ordered pair to its first coordinate, and the other projection function will map the ordered pair to its second coordinate. So we define \(p_{1}: A \times B \rightarrow A\) by \(p_{1}(a, b)=a\) for every \((a, b) \in A \times B ;\) and \(p_{2}: A \times B \rightarrow B\) by \(p_{2}(a, b)=b\) for every \((a, b) \in A \times B\) Let \(A=\\{1,2\\}\) and let \(B=\\{x, y, z\\}\) (a) Determine the outputs for all possible inputs for the projection function \(p_{1}: A \times B \rightarrow A\) (b) Determine the outputs for all possible inputs for the projection function \(p_{2}: A \times B \rightarrow B\) (c) What is the range of these projection functions? (d) Is the following statement true or false? Explain. For all \((m, n),(u, v) \in A \times B,\) if \((m, n) \neq(u, v),\) then \(p_{1}(m, n) \neq p_{1}(u, v)\)

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Proposition. The function \(f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}\) defined by \(f(x, y)=\) \((2 x+y, x-y)\) is an injection. Proof. For each \((a, b)\) and \((c, d)\) in \(\mathbb{R} \times \mathbb{R},\) if \(f(a, b)=f(c, d),\) then $$(2 a+b, a-b)=(2 c+d, c-d)$$ We will use systems of equations to prove that \(a=c\) and \(b=d\). $$\begin{aligned} 2 a+b &=2 c+d \\ a-b &=c-d \\ 3 a &=3 c \\ a &=c \end{aligned}$$ Since \(a=c,\) we see that $$(2 c+b, c-b)=(2 c+d, c-d)$$ So \(b=d\). Therefore, we have proved that the function \(f\) is an injection. (b) Proposition. The function \(f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}\) defined by \(f(x, y)=\) \((2 x+y, x-y)\) is a surjection Proof. We need to find an ordered pair such that \(f(x, y)=(a, b)\) for each \((a, b)\) in \(\mathbb{R} \times \mathbb{R}\). That is, we need \((2 x+y, x-y)=(a, b)\), or $$\begin{array}{ll} 2 x+y=a & \text { and } \quad x-y=b \text { . } \end{array}$$ Treating these two equations as a system of equations and solving for \(x\) and \(y,\) we find that $$x=\frac{a+b}{3} \quad \text { and } \quad y=\frac{a-2 b}{3}$$. Hence, \(x\) and \(y\) are real numbers, \((x, y) \in \mathbb{R} \times \mathbb{R},\) and $$\begin{aligned} f(x, y) &=f\left(\frac{a+b}{3}, \frac{a-2 b}{3}\right) \\ &=\left(2\left(\frac{a+b}{3}\right)+\frac{a-2 b}{3}, \frac{a+b}{3}-\frac{a-2 b}{3}\right) \\ &=\left(\frac{2 a+2 b+a-2 b}{3}, \frac{a+b-a+2 b}{3}\right) \\ &=\left(\frac{3 a}{3}, \frac{3 b}{3}\right) \\ &=(a, b) \end{aligned}$$ Therefore, we have proved that for every \((a, b) \in \mathbb{R} \times \mathbb{R},\) there exists an \((x, y) \in \mathbb{R} \times \mathbb{R}\) such that \(f(x, y)=(a, b)\). This proves that the function \(f\) is a surjection.

Let \(A=\\{1,2,3\\}\) and \(B=\\{a, b, c\\}\). (a) Construct an example of a function \(f: A \rightarrow B\) that is not a bijection. Write the inverse of this function as a set of ordered pairs. Is the inverse of \(f\) a function? Explain. If so, draw an arrow diagram for \(f\) and \(f^{-1}\). (b) Construct an example of a function \(g: A \rightarrow B\) that is a bijection. Write the inverse of this function as a set of ordered pairs. Is the inverse of \(g\) a function? Explain. If so, draw an arrow diagram for \(g\) and \(g^{-1}\).

Let \(f: A \rightarrow B\) and \(g: B \rightarrow A\). Let \(I_{A}\) and \(I_{B}\) be the identity functions on the sets \(A\) and \(B\), respectively. Prove each of the following: (a) If \(g \circ f=I_{A},\) then \(f\) is an injection. (b) If \(f \circ g=I_{B},\) then \(f\) is a surjection. (c) If \(g \circ f=I_{A}\) and \(f \circ g=I_{B},\) then \(f\) and \(g\) are bijections and \(g=f^{-1}\)

Let \(s: \mathbb{N} \rightarrow \mathbb{N},\) where for each \(n \in \mathbb{N}, s(n)\) is the sum of the distinct natural number divisors of \(n\). This is the sum of the divisors function that was introduced in Preview Activity 2 from Section 6.1. Is \(s\) an injection? Is \(s\) a surjection? Justify your conclusions.
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