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Chapter 1: Problem 47

Evaluate the function at each specified value of the independent variable and simplify. \(f(x)=|x| / x\) (a) \(f(2)\) (b) \(f(-2)\) (c) \(f(x-1)\)

Short Answer

Expert verified
The values of the function f at x=2, x=-2, and x=(x-1) are 1, -1, and the function is piecewise: 1 when x > 1, -1 when x < 1 and undefined when x = 1 respectively.

Step by step solution

01

Evaluate f(2)

Substitute x with 2 into the function \(f(x) = |x|/x\). Therefore, \(f(2) = |2|/2\), resulting in 1. Since both the numerator (|2|) and denominator (2) are positive, the division results in a positive value.
02

Evaluate f(-2)

Substitute x with -2 into the function \(f(x) = |x|/x\). Therefore, \(f(-2) = |-2|/-2\), resulting in -1. The numerator is positive (|-2| = 2) and the denominator is negative, therefore, the division results in a negative value.
03

Evaluate f(x-1)

Substitute x with x-1 into the function \(f(x) = |x|/x\). Therefore, \(f(x-1) = |x-1|/(x-1)\). Now, the function will be piecewise: it equals to 1 when \(x-1 > 0\) (or \(x > 1\)), and equals to -1 when \(x-1 < 0\) (or \(x < 1\)). When \(x = 1\), the function is undefined.

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Key Concepts

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

Absolute Value Function
The concept of the absolute value function is fundamental in mathematics, as it transforms a number into its non-negative counterpart. In essence, the absolute value of a number, denoted by vertical bars like this: \(|x|\), represents its distance from zero on the number line. This means:
  • The absolute value of a positive number is the number itself. For example, \(|2| = 2\).
  • The absolute value of a negative number is its positive equivalent, achieved by removing the negative sign. Hence, \(|-2| = 2\).
  • The absolute value of zero is simply zero, \(|0| = 0\).
This concept is especially useful in piecewise functions, where the behavior of the function changes based on the input values, such as the function given in the exercise: \(f(x) = \frac{|x|}{x}\). Understanding the absolute value is crucial because it dictates the function's output by maintaining non-negative values regardless of the sign of the input.
Evaluating Functions
Evaluating functions involves substituting specific values of the independent variable into the function to determine the output. For the function \(f(x) = \frac{|x|}{x}\), the process of evaluating involves plugging in the given x-values into the expression:
  • Start by replacing the variable with the specified number, such as evaluating \(f(2)\) by substituting \(x = 2\). This results in \(f(2) = \frac{|2|}{2} = 1\).
  • The same substitution is used for negative values like \(f(-2)\). Here, \(x = -2\) gives us \(f(-2) = \frac{|-2|}{-2} = -1\).
  • For expressions like \(f(x-1)\), replace \(x\) with \(x-1\) to evaluate the function, leading to \(f(x-1) = \frac{|x-1|}{x-1}\).
Evaluating each scenario helps understand how the function behaves for different values, revealing its piecewise nature. Recognizing how absolute values transform inputs is key in analyzing the outcome of the function.
Division of Integers
Division of integers is an essential arithmetic operation that determines how many times one integer fits into another. When dividing integers, it's important to consider their signs, which influence the result:
  • If both integers are positive, the result is positive, such as \(\frac{2}{1} = 2\).
  • If one integer is negative and the other is positive, the result is negative, as in \(\frac{-2}{2} = -1\) or \(\frac{2}{-2} = -1\).
  • Combining two negative integers results in a positive quotient, like \(\frac{-2}{-1} = 2\).
Understanding this principle is essential in evaluating functions like \(f(x) = \frac{|x|}{x}\), where the division of integers determines whether the output is positive, negative, or zero, significantly impacting the function's overall characteristics and graph behavior.

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Most popular questions from this chapter

The lengths (in feet) of the winning men's discus throws in the Olympics from 1920 through 2008 are listed below. (Source: International Olympic Committee) $$\begin{array}{llllll} 1920 & 146.6 & 1956 & 184.9 & 1984 & 218.5 \\ 1924 & 151.3 & 1960 & 194.2 & 1988 & 225.8 \\ 1928 & 155.3 & 1964 & 200.1 & 1992 & 213.7 \\ 1932 & 162.3 & 1968 & 212.5 & 1996 & 227.7 \\ 1936 & 165.6 & 1972 & 211.3 & 2000 & 227.3 \\ 1948 & 173.2 & 1976 & 221.5 & 2004 & 229.3 \\ 1952 & 180.5 & 1980 & 218.7 & 2008 & 225.8 \end{array}$$ (a) Sketch a scatter plot of the data. Let \(y\) represent the length of the winning discus throw (in feet) and let \(t=20\) represent 1920 (b) Use a straightedge to sketch the best-fitting line through the points and find an equation of the line. (c) Use the regression feature of a graphing utility to find the least squares regression line that fits the data. (d) Compare the linear model you found in part (b) with the linear model given by the graphing utility in part (c). (e) Use the models from parts (b) and (c) to estimate the winning men's discus throw in the year 2012 .

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