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Let \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) be defined by \(f(m)=3-m\). (a) Evaluate \(f(-7), f(-3), f(3),\) and \(f(7)\) (b) Determine the set of all of the preimages of 5 and the set of all of the preimages of 4 (c) Determine the range of the function \(f\). (d) This function can be considered a real function since \(\mathbb{Z} \subseteq \mathbb{R}\). Sketch a graph of this function. Note: The graph will be an infinite set of points that lie on a line. However, it will not be a line since its domain is not \(\mathbb{R}\) but is \(\mathbb{Z}\)

Short Answer

Expert verified
(a) \(f(-7) = 10\), \(f(-3) = 6\), \(f(3) = 0\), and \(f(7) = -4\). (b) The set of all preimages of 5 is \({-2}\), and the set of all preimages of 4 is \({-1}\). (c) The range of the function is \(\mathbb{Z}\). (d) The graph is an infinite set of points that lie on a line with slope -1 and y-intercept 3, but it is not a continuous line.

Step by step solution

01

Part (a): Evaluating the function for given values

To determine the value of the function for the given inputs, we will plug the values into the expression and calculate the result. 1. \(f(-7)\): \(f(-7) = 3 - (-7) = 3 + 7 = 10\) 2. \(f(-3)\): \(f(-3) = 3 - (-3) = 3 + 3 = 6\) 3. \(f(3)\): \(f(3) = 3 - 3 = 0\) 4. \(f(7)\): \(f(7) = 3 - 7 = -4\)
02

Part (b): Preimages of 5 and 4

To find the preimages of 5 and 4, we need to find the inputs (m values) that give us the outputs 5 and 4. For preimage of 5: \[f(m) = 5\] \[3 - m = 5\] \[m = -2\] For preimage of 4: \[f(m) = 4\] \[3 - m = 4\] \[m = -1\] The set of all preimages of 5 is \({-2}\) and the set of all preimages of 4 is \({-1}\).
03

Part (c): Determine the range of the function

The range of a function is the set of all possible outputs. For our function, the input is any integer (\(\mathbb{Z}\)), and the output is also any integer (\(\mathbb{Z}\)). Since subtracting an integer from 3 can cover all integers, the range is \(\mathbb{Z}\).
04

Part (d): Sketching the graph of the function

Since the function has domain \(\mathbb{Z}\) and not \(\mathbb{R}\), the graph will consist of an infinite set of points that lie on a line but will not be a line. The slope of the graph will be -1, and it will pass through 3 on the y-axis. To sketch the graph: 1. Plot the y-intercept (0, 3) on the graph. 2. Plot some other points using the function like \((-7, 10), (-3, 6), (3, 0) \textrm{ and }(7, -4)\). 3. Label these points, which lie on a line with a slope of -1. 4. Keep in mind that the graph consists of just these labeled points and other similar points on the line that result from integer inputs; it is not a continuous line. The graph for the function will look like a dashed line with dots at each integer point.

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Key Concepts

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

Integer Functions
Integer functions are special mathematical functions where both the input and output values are whole numbers. In this context, we are looking at a function denoted as \(f: \mathbb{Z} \rightarrow \mathbb{Z}\), which tells us that any integer input \(m\) will yield another integer as the output \(f(m)\).
For instance, let's consider the function \(f(m) = 3 - m\). This equation reveals that for every integer \(m\) you plug into the function, you’ll subtract \(m\) from 3 to find the result. Integer functions, such as this one, are great for modeling real-world scenarios where only whole numbers make sense, like counting objects or steps.
Understanding integer functions helps build a solid foundation in maths, enabling one to handle problems involving sequences, series, and discrete structures efficiently.
Function Range
The range of a function describes all potential results it can produce. In simpler terms, it's the set of all possible outputs.
For the function \(f(m) = 3 - m\), since \(m\) can be any integer \(\mathbb{Z}\), the resulting outputs will also be integers. Therefore, the range of this function is all integers \(\mathbb{Z}\).
This occurs because subtracting an integer from 3 can result in any other integer. When dealing with function ranges in general, it’s essential to consider all possible ways the input domain affects the outputs.
Function Preimages
Function preimages are about finding the set of all inputs that can produce a specific output when fed into the function. It's like working backward from the output to determine which inputs are possible.
For the function \(f(m) = 3 - m\), let’s find the preimage of 5 and 4. To do this, solve the equations for \(m\):
  • For the preimage of 5: Set \(f(m) = 5\), leading to \(3 - m = 5\). Solving gives \(m = -2\), so \(-2\) is the preimage of 5.
  • For the preimage of 4: Set \(f(m) = 4\), leading to \(3 - m = 4\). Solving gives \(m = -1\), so \(-1\) is the preimage of 4.

By understanding preimages, one gains insights into how inputs and outputs are connected, providing a deeper understanding of a function's behavior.
Graphing Discrete Functions
Graphing discrete functions involves plotting points that correspond to integer inputs rather than drawing a continuous line.
For the function \(f(m) = 3 - m\) with integer inputs, we plot only those points where both input and output are integers. This results in a series of isolated points lined up along a line with a slope of \(-1\).
To sketch this graph:
  • Mark the y-intercept, which is the point \((0, 3)\).
  • Plot additional points at coordinates \((-7, 10), (-3, 6), (3, 0)\), and \((7, -4)\).
  • These points align with a line's path and illustrate the linear progression, yet, they are discreet, not forming a continuous line.

Graphing discrete functions is vital for visualizing them and understanding the relationship between inputs and outputs when only specific values are applicable.

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Most popular questions from this chapter

Constructing an Inverse Function. If \(f: A \rightarrow B\) is a bijection, then we know that its inverse is a function. If we are given a formula for the function \(f,\) it may be desirable to determine a formula for the function \(f^{-1}\). This can sometimes be done, while at other times it is very difficult or even impossible. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f(x)=2 x^{3}-7 .\) A graph of this function would suggest that this function is a bijection. (a) Prove that the function \(f\) is an injection and a surjection. Let \(y \in \mathbb{R}\). One way to prove that \(f\) is a surjection is to set \(y=f(x)\) and solve for \(x\). If this can be done, then we would know that there exists an \(x \in \mathbb{R}\) such that \(f(x)=y .\) For the function \(f,\) we are using \(x\) for the input and \(y\) for the output. By solving for \(x\) in terms of \(y,\) we are attempting to write a formula where \(y\) is the input and \(x\) is the output. This formula represents the inverse function. (b) Solve the equation \(y=2 x^{3}-7\) for \(x\). Use this to write a formula for \(f^{-1}(y),\) where \(f^{-1}: \mathbb{R} \rightarrow \mathbb{R}\) (c) Use the result of Part (13b) to verify that for each \(x \in \mathbb{R}, f^{-1}(f(x))=\) \(x\) and for each \(y \in \mathbb{R}, f\left(f^{-1}(y)\right)=y\) Now let \(\mathbb{R}^{+}=\\{y \in \mathbb{R} \mid y>0\\} .\) Define \(g: \mathbb{R} \rightarrow \mathbb{R}^{+}\) by \(g(x)=e^{2 x-1}\) (d) Set \(y=e^{2 x-1}\) and solve for \(x\) in terms of \(y\). (e) Use your work in Exercise (13d) to define a function \(h: \mathbb{R}^{+} \rightarrow \mathbb{R}\). (f) For each \(x \in \mathbb{R},\) determine \((h \circ g)(x)\) and for each \(y \in \mathbb{R}^{+},\) determine \((g \circ h)(y)\) (g) Use Exercise (6) to explain why \(h=g^{-1}\).

The Proof of Theorem \(6.21 .\) Use the ideas from Exercise (9) to prove Theorem \(6.21 .\) Let \(A, B,\) and \(C\) be nonempty sets and let \(f: A \rightarrow B\) and \(g: B \rightarrow C\) (a) If \(g \circ f: A \rightarrow C\) is an injection, then \(f: A \rightarrow B\) is an injection. (b) If \(g \circ f: A \rightarrow C\) is a surjection, then \(g: B \rightarrow C\) is a surjection.

Let \(f: S \rightarrow T,\) let \(A\) and \(B\) be subsets of \(S,\) and let \(C\) and \(D\) be subsets of \(T\). For \(x \in S\) and \(y \in T\), carefully explain what it means to say that (a) \(y \in f(A \cap B)\) (b) \(y \in f(A \cup B)\) (c) \(y \in f(A) \cap f(B)\) (d) \(y \in f(A) \cup f(B)\) (e) \(x \in f^{-1}(C \cap D)\) (f) \(x \in f^{-1}(C \cup D)\) (g) \(x \in f^{-1}(C) \cap f^{-1}(D)\) (h) \(x \in f^{-1}(C) \cup f^{-1}(D)\)

Let \(A\) and \(B\) be two nonempty sets. Define $$p_{1}: A \times B \rightarrow A \text { by } p_{1}(a, b)=a$$ for every \((a, b) \in A \times B .\) This is the first projection function introduced in Exercise (5) in Section 6.2 (a) Is the function \(p_{1}\) a surjection? Justify your conclusion. (b) If \(B=\\{b\\}\), is the function \(p_{1}\) an injection? Justify your conclusion. (c) Under what condition(s) is the function \(p_{1}\) not an injection? Make a conjecture and prove it.

Prove Part (2) of Corollary 6.28. Let \(A\) and \(B\) be nonempty sets and let \(f: A \rightarrow B\) be a bijection. Then for every \(y\) in \(B,\left(f \circ f^{-1}\right)(y)=y\).

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