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Chapter 2: Problem 21

Evaluate the function at the indicated values. $$\begin{aligned}&f(x)=\frac{1-2 x}{3}\\\&f(2), f(-2), f\left(\frac{1}{2}\right), f(a), f(-a), f(a-1)\end{aligned}$$

Short Answer

Expert verified
The evaluated values are: \(f(2) = -1\), \(f(-2) = \frac{5}{3}\), \(f\left(\frac{1}{2}\right) = 0\), \(f(a) = \frac{1-2a}{3}\), \(f(-a) = \frac{1+2a}{3}\), and \(f(a-1) = \frac{3-2a}{3}\).

Step by step solution

01

Evaluate f(2)

Substitute \(x = 2\) into the function \(f(x) = \frac{1-2x}{3}\). Then calculate \(f(2) = \frac{1-2(2)}{3} = \frac{1-4}{3} = \frac{-3}{3} = -1\).
02

Evaluate f(-2)

Substitute \(x = -2\) into the function \(f(x) = \frac{1-2x}{3}\). Then calculate \(f(-2) = \frac{1-2(-2)}{3} = \frac{1+4}{3} = \frac{5}{3}\).
03

Evaluate f(1/2)

Substitute \(x = \frac{1}{2}\) into the function \(f(x) = \frac{1-2x}{3}\). Then calculate \(f\left(\frac{1}{2}\right) = \frac{1-2\left(\frac{1}{2}\right)}{3} = \frac{1-1}{3} = \frac{0}{3} = 0\).
04

Evaluate f(a)

Substitute \(x = a\) into the function \(f(x) = \frac{1-2x}{3}\). Then calculate \(f(a) = \frac{1-2a}{3}\).
05

Evaluate f(-a)

Substitute \(x = -a\) into the function \(f(x) = \frac{1-2x}{3}\). Then calculate \(f(-a) = \frac{1-2(-a)}{3} = \frac{1+2a}{3}\).
06

Evaluate f(a-1)

Substitute \(x = a-1\) into the function \(f(x) = \frac{1-2x}{3}\). Then calculate \(f(a-1) = \frac{1-2(a-1)}{3} = \frac{1-2a+2}{3} = \frac{3-2a}{3}\).

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Key Concepts

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

Substitution
Substitution is a key technique when dealing with functions in algebra. It involves replacing the variable in a function with a specified value. This aids in determining the function's output for particular inputs. In the context of our problem, we have the function \( f(x) = \frac{1-2x}{3} \). By substituting different values for \( x \), we can evaluate \( f \) at those points.
  • For \( f(2) \), we replace \( x \) with \( 2 \), which results in \( \frac{1-2(2)}{3} \).
  • For \( f(-2) \), substituting \( -2 \) for \( x \) gives us \( \frac{1-2(-2)}{3} \).
  • For \( f(\frac{1}{2}) \), \( x \) becomes \( \frac{1}{2} \), resulting in \( \frac{1-2(\frac{1}{2})}{3} \).
Substitution allows us to systematically explore how changes in \( x \) affect the function's output. This practice builds a strong foundation for solving more complex equations and understanding mathematical relationships.
Rational Function
A rational function is any function of the form \( \frac{p(x)}{q(x)} \) where both \( p(x) \) and \( q(x) \) are polynomials and \( q(x) eq 0 \). In our example, the function \( f(x) = \frac{1-2x}{3} \) is a simple rational function. The numerator, \( 1-2x \), is a first-degree polynomial and the denominator, \( 3 \), is a constant, meaning it's actually a very basic polynomial (a zero-degree polynomial).
  • The whole expression defines how the function behaves across different values of \( x \).
  • Since the denominator is non-zero for all real numbers, this function has no vertical asymptotes or undefined points.
Understanding these properties of rational functions is essential because it provides insights into their continuity and behavior. These functions also exhibit interesting characteristics, such as horizontal and oblique asymptotes, which come into play in more advanced settings.
Algebraic Manipulation
Algebraic manipulation involves changing the form of an algebraic expression to simplify a problem. This skill is crucial in both solving and understanding mathematical functions. For the function \( f(x) = \frac{1-2x}{3} \), while evaluating it for different values, algebraic manipulation helps to simplify expressions:
  • When evaluating \( f(a) \), algebraic manipulation directly gives \( \frac{1-2a}{3} \), which represents the function's value without extensive calculation.
  • Similarly, for \( f(a-1) \), careful manipulation provides \( \frac{3-2a}{3} \) as the simplified expression.
This manipulation helps in reducing potential errors and streamlining calculations when handling complex problems. Through manipulation, large or complicated expressions can become manageable, making it easier to draw conclusions and find solutions in various mathematical contexts.

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Most popular questions from this chapter

When a bowl of hot soup is left in a room, the soup eventually cools down to room temperature. The temperature \(T\) of the soup is a function of time \(t .\) The table below gives the temperature (in "F) of a bowl of soup \(t\) minutes after it was set on the table. Find the average rate of change of the temperature of the soup over the first 20 minutes and over the next 20 minutes. During which interval did the soup cool off more quickly? $$\begin{array}{|c|c||c|c|} \hline t \text { (min) } & T\left(^{\circ} \mathrm{F}\right) & t \text { (min) } & T\left(^{\circ} \mathrm{F}\right) \\ \hline 0 & 200 & 35 & 94 \\ 5 & 172 & 40 & 89 \\ 10 & 150 & 50 & 81 \\ 15 & 133 & 60 & 77 \\ 20 & 119 & 90 & 72 \\ 25 & 108 & 120 & 70 \\ 30 & 100 & 150 & 70 \\ \hline \end{array}$$

DISCUSS: Determining When a Linear Function Has an Inverse For the linear function \(f(x)=m x+b\) to be one-to-one, what must be true about its slope? If it is one to-one, find its inverse. Is the inverse linear? If so, what is its slope?

Graphing Functions Sketch a graph of the function by first making a table of values. $$k(x)=-\sqrt[3]{x}$$

Sketch a graph of the piecewise defined function. $$f(x)=\left\\{\begin{array}{ll} 1 & \text { if } x \leq 1 \\ x+1 & \text { if } x>1 \end{array}\right.$$

DISCUSS: Solving an Equation for an Unknown Function In Exercises \(69-72\) of Section 2.7 you were asked to solve equations in which the unknowns are functions. Now that we know about inverses and the identity function (see Exercise 104 ), we can use algebra to solve such equations. For instance, to solve \(f \circ g=h\) for the unknown function \(f,\) we perform the following steps: \(f^{\circ} g=h \quad\) Problem: Solve for \(f\) \(f^{\circ} g^{\circ} g^{-1}=h^{\circ} g^{-1} \quad\) Compose with \(g^{-1}\) on the right \(f \circ I=h \cdot g^{-1} \quad \text { Because } g^{\circ} g^{-1}=I\) \(f=h e g^{-1} \quad\) Because \(f \circ I=f\) So the solution is \(f=h \circ g^{-1} .\) Use this technique to solve the equation \(f \circ g=h\) for the indicated unknown function. (a) Solve for \(f,\) where \(g(x)=2 x+1\) and \(h(x)=4 x^{2}+4 x+7.\) (b) Solve for \(q,\) where \(f(x)=3 x+5\) and \(h(x)=3 x^{2}+3 x+2.\)
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