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Chapter 2: Problem 19

Evaluate the function at the indicated values. $$\begin{array}{l} f(x)=2 x+1 ; \\ f(1), f(-2), f\left(\frac{1}{2}\right), f(a), f(-a), f(a+b) \end{array}$$

Short Answer

Expert verified
\( f(1)=3, f(-2)=-3, f\left(\frac{1}{2}\right)=2, f(a)=2a+1, f(-a)=-2a+1, f(a+b)=2a+2b+1 \).

Step by step solution

01

Understanding the Function

The function given is a linear function, \( f(x) = 2x + 1 \). This means that for any input \( x \), we multiply it by 2 and then add 1 to find the output.
02

Evaluating \( f(1) \)

Substitute \( 1 \) into the function: \( f(1) = 2(1) + 1 = 2 + 1 = 3 \). So, \( f(1) = 3 \).
03

Evaluating \( f(-2) \)

Substitute \( -2 \) into the function: \( f(-2) = 2(-2) + 1 = -4 + 1 = -3 \). So, \( f(-2) = -3 \).
04

Evaluating \( f\left(\frac{1}{2}\right) \)

Substitute \( \frac{1}{2} \) into the function: \( f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2 \). So, \( f\left(\frac{1}{2}\right) = 2 \).
05

Evaluating \( f(a) \)

Substitute \( a \) into the function: \( f(a) = 2a + 1 \). The output is in terms of \( a \).
06

Evaluating \( f(-a) \)

Substitute \( -a \) into the function: \( f(-a) = 2(-a) + 1 = -2a + 1 \). The output is also in terms of \( a \).
07

Evaluating \( f(a+b) \)

Substitute \( a+b \) into the function: \( f(a+b) = 2(a+b) + 1 = 2a + 2b + 1 \). The output is in terms of both \( a \) and \( b \).

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Key Concepts

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

Understanding Linear Functions
A linear function is a type of mathematical function that can be represented by the equation \( f(x) = ax + b \). This equation describes a straight line when graphed on a coordinate plane.

Each linear function has a few unique characteristics:
  • **Slope (a):** This indicates how steep the line is. In our example, \( f(x) = 2x + 1 \), the slope is \(2\), meaning for every unit increase in \(x\), the output \( f(x)\) increases by 2.
  • **Y-intercept (b):** This is the point where the line crosses the y-axis. In our example, the y-intercept is \(1\), indicating the output value when \( x = 0 \).
Linear functions are straightforward due to their constant rate of change. This means they are predictable, as they won't curve or have sudden steep increases or decreases.
The Substitution Method in Function Evaluation
The substitution method is an effective way to find the output of a function by replacing the variable with actual values. This is particularly useful for evaluating functions at specific points.

Here's how it works with the function \( f(x) = 2x + 1 \):
  • You "substitute" a number for \(x\). For instance, if you want to find \( f(1) \), you replace \(x\) with \(1\), giving \( f(1) = 2(1) + 1 \).
  • After substituting, you simplify the expression: \( f(1) = 2 + 1 = 3 \).
Repeating this simple substitution and evaluation process allows you to find the function's output for various inputs, such as given numbers or parameters like \(a\) and \(b\). This method is crucial for answering more complex algebraic questions involving functions.
Algebraic Expressions and Function Evaluation
When dealing with a linear function, evaluating it often involves working with algebraic expressions. An algebraic expression is a mathematical phrase that includes constants, variables, and operations (like addition, multiplication).

Consider the function \( f(x) = 2x + 1 \). Evaluating this at \( x = a \) involves substituting \(a\) into the equation, resulting in the expression \( 2a + 1 \). This expression represents all possible outputs of the function when the input is \( a \).
  • **Complex Inputs:** When the input is a more complex expression, like \(a+b\), you substitute the entire expression: \( f(a+b) = 2(a+b) + 1 = 2a + 2b + 1 \).
  • These algebraic results can help in further studying the behavior of the function for different inputs, making it versatile in application.
Understanding how to manipulate and evaluate algebraic expressions is foundational in algebra and essential for solving equations and inequations.

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Most popular questions from this chapter

At the beginning of this section we discussed three examples of everyday, ordinary functions: Height is a function of age, temperature is a function of date, and postage cost is a function of weight. Give three other examples of functions from everyday life.

When we seek a minimum or maximum value of a function, it is sometimes easier to work with a simpler function instead. (a) Suppose $$g(x)=\sqrt{f(x)}$$ where \(f(x) \geq 0\) for all \(x .\) Explain why the local minima and maxima of \(f\) and \(g\) occur at the same values of \(x .\) (b) Let \(g(x)\) be the distance between the point \((3,0)\) and the point \(\left(x, x^{2}\right)\) on the graph of the parabola \(y=x^{2} .\) Express \(g\) as a function of \(x\) (c) Find the minimum value of the function \(g\) that you found in part (b). Use the principle described in part (a) to simplify your work.

Evaluate the function at the indicated values. $$\begin{aligned} &f(x)=x^{2}+2 x ; \\ &f(0), f(3), f(-3), f(a), f(-x), f\left(\frac{1}{a}\right) \end{aligned}$$

A tank holds 50 gallons of water, which drains from a leak at the bottom, causing the tank to empty in 20 minutes. The tank drains faster when it is nearly full because the pressure on the leak is greater. Torricelli's Law gives the volume of water remaining in the tank after \(t\) minutes as $$V(t)=50\left(1-\frac{t}{20}\right)^{2} \quad 0 \leq t \leq 20$$ (a) Find \(V(0)\) and \(V(20)\) (b) What do your answers to part (a) represent? (c) Make a table of values of \(V(t)\) for \(t=0,5,10,15,20\) (image cannot copy)

Express the function (or rule) in words. $$k(x)=\sqrt{x+2}$$
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