Problem 18 - Algebra for College Students

Question

If \(f(x)=\left\\{\begin{array}{rll}2 & \text { for } x<0 & \\ x^{2}+1 & \text { for } 0 \leq x \leq 4, & \text { find } f(3), f(6) \\ -1 & \text { for } x>4 & f(0), \text { and } f(-3)\end{array}\right.\)

Step-by-Step Solution

Answer

\( f(3) = 10 \), \( f(6) = -1 \), \( f(0) = 1 \), \( f(-3) = 2 \).

Step 1: Understanding the Piecewise Function

The function \( f(x) \) is defined piecewise, which means its output depends on the range in which the input \( x \) falls. The function comprises three parts:1. \( f(x) = 2 \) when \( x < 0 \).2. \( f(x) = x^2 + 1 \) when \( 0 \leq x \leq 4 \).3. \( f(x) = -1 \) when \( x > 4 \).To determine \( f(3), f(6), f(0), \) and \( f(-3) \), we need to identify which part of the piecewise function applies to each input.

Step 2: Calculating f(3)

Since 3 lies within the interval \( 0 \leq x \leq 4 \), we use the second part of the function: \( f(x) = x^2 + 1 \).Substitute \( x = 3 \):\[ f(3) = 3^2 + 1 = 9 + 1 = 10. \]

Step 3: Calculating f(6)

Since 6 is greater than 4, we use the third part of the function: \( f(x) = -1 \).Thus, \( f(6) = -1. \)

Step 4: Calculating f(0)

Since 0 lies on the boundary of the interval \( 0 \leq x \leq 4 \), we use the second part of the function: \( f(x) = x^2 + 1 \).Substitute \( x = 0 \):\[ f(0) = 0^2 + 1 = 1. \]

Step 5: Calculating f(-3)

Since -3 is less than 0, we apply the first part of the function: \( f(x) = 2 \).Therefore, \( f(-3) = 2. \)

Key Concepts

Function Evaluation

When dealing with piecewise functions, the key is to evaluate the function based on the different rules defined for various intervals of the input variable. This process is known as function evaluation. For our function, we have three intervals:

• For inputs less than 0, the function outputs a constant value of 2.
• For inputs between 0 and 4 (inclusive), the function is defined by an algebraic expression, \( f(x) = x^2 + 1 \).
• For inputs greater than 4, the function outputs -1.

To understand function evaluation better, let’s look at an example. To find \( f(3) \), we notice that 3 falls within the interval \( 0 \leq x \leq 4 \). Thus, we use the rule \( f(x) = x^2 + 1 \):\[ f(3) = 3^2 + 1 = 9 + 1 = 10 \]This illustrates how different values of \( x \) require deciding which "piece" of the function we use.

Algebraic Expressions

Understanding algebraic expressions is crucial in evaluating piecewise functions. An algebraic expression is a mathematical phrase that can include numbers, variables, and operation symbols. In our piecewise function, one part is given by the expression \( x^2 + 1 \).
To handle these:

• Identify any operations involving the variable. Here, the operations include squaring \( x \), then adding 1.
• Substitute the value of \( x \) you are evaluating into the expression. For instance, to calculate \( f(0) \), substitute 0 into \( x^2 + 1 \), giving \( f(0) = 0^2 + 1 = 1 \).

Using algebraic expressions helps in capturing the changes in function behavior over specific intervals of the variable. This makes solving for specific function values simply a step-by-step substitution process.

Mathematics Education

Teaching piecewise functions effectively is an important part of mathematics education. The goal is to ensure students grasp how functions can change behavior over different intervals. Here are some tips:

• Use visual aids like graphs to show how each segment of the function maps to its corresponding interval.


• Encourage practice with various examples, to familiarize with the approach of evaluating inputs according to the correct piece of the function.


• Introduce real-world examples where piecewise functions naturally occur, such as tax brackets or volume discounts.

Through consistent practice and varied examples, students can deepen their understanding of how to evaluate functions using piecewise definitions, making them adept at interpreting and solving real-world and theoretical problems.