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Chapter 2: Problem 20

For each piecewise-defined function, find (a) \(f(-5),\) (b) \(f(-1),\) (c)\(f(0),\) and ( \(d\) ) \(f(3)\) See Example 2. $$f(x)=\left\\{\begin{array}{ll} -2 x & \text { if } x < -3 \\ 3 x-1 & \text { if }-3 \leq x \leq 2 \\ -4 x & \text { if } x > 2 \end{array}\right.$$

### Short Answer

Expert verified

a) 10, b) -4, c) -1, d) -12

## Step by step solution

01

## - Evaluate at x = -5

Since \(-5 < -3\), use the first piece of the function: \(f(x) = -2x\). Substituting \(-5\) in, \(f(-5) = -2(-5) = 10\).

02

## - Evaluate at x = -1

Since \(-3 \leq -1 \leq 2\), use the second piece of the function: \(f(x) = 3x - 1\). Substituting \(-1\) in, \(f(-1) = 3(-1) - 1 = -3 - 1 = -4\).

03

## - Evaluate at x = 0

Since \(-3 \leq 0 \leq 2\), use the second piece of the function: \(f(x) = 3x - 1\). Substituting \0\ in, \(f(0) = 3(0) - 1 = 0 - 1 = -1\).

04

## - Evaluate at x = 3

Since \x > 2\, use the third piece of the function: \(f(x) = -4x\). Substituting \3\ in, \(f(3) = -4(3) = -12\).

## Key Concepts

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

###### Evaluating Functions

Evaluating functions involves finding the value of a function at specific points. For piecewise-defined functions, we first determine which piece of the function to use based on the given input value. Then we substitute the value into the corresponding expression.

In the original problem, we were given a piecewise-defined function:

\[ f(x) = \begin{cases} -2x & \text{if } x < -3 \ 3x - 1 & \text{if } -3 \leq x \leq 2 \ -4x & \text{if } x > 2 \end{cases} \]

To evaluate this function at different points, follow these steps:

- Find the interval in which the given input value falls.
- Use the corresponding expression to compute the function value.

For example, to evaluate \(f(-5)\), we note that \(-5 < -3\). Therefore, we use the first piece, \(f(x) = -2x\), and substitute \(-5\) to get \(f(-5) = -2(-5) = 10\). Follow similar steps for other values of \(x\).

###### Precalculus

Precalculus lays the groundwork for understanding calculus concepts. It involves studying functions, graphs, and limits. Piecewise-defined functions, such as the function given in the exercise, are a key topic in precalculus. These functions have different expressions based on the domain intervals.

Here are some core concepts in precalculus that relate to piecewise functions:

**Domain:**The set of all possible input values for the function.**Range:**The set of all possible output values.**Continuity:**Whether the function is continuous (unbroken) over its domain.

Understanding these concepts helps students handle more complex topics in calculus, such as integration and differentiation of piecewise functions. Mastering precalculus is crucial for building a strong mathematical foundation.

###### Function Notation

Function notation is a way to express relationships between inputs and outputs in a concise form. The general form \(f(x)\) is read as 'f of x' and represents a function named \(f\) with \(x\) as the input. The value of the function at \(x\) is the output.

In our given exercise, the function is defined as:

\[f(x) = \begin{cases} -2x & \text{if } x < -3\ 3x - 1 & \text{if } -3 \leq x \leq 2\ -4x & \text{if } x > 2 \end{cases} \]

Here's how to interpret and use function notation:

**Identify the function name:**In \(f(x)\), the name of the function is \(f\).**Recognize the input variable:**The input variable here is \(x\).**Understand the output:**The output is what you get after evaluating the function for a particular value of \(x\).

Using function notation makes it easier to communicate and solve problems involving functions. It provides a clear and systematic way to show relationships and perform calculations.

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