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Chapter 5: Problem 15

\(f(x)= \begin{cases}2 x+1 & \text { if } x \leq 4 \\ 13-x & \text { if } 4

Short Answer

Expert verified
Evaluate using \(2x + 1\) for \(x \leq 4\) and \(13 - x\) for \(x > 4\).

Step by step solution

01

Understand the Piecewise Function

The function, denoted as \(f(x)\), is defined differently based on the value of \(x\). For \(x \leq 4\), the function is defined as \(2x + 1\). For \(x > 4\), the function is defined as \(13 - x\).
02

Identify the Domains

Identify the intervals for which each piece of the function is valid. For \(x \leq 4\), use the function \(2x + 1\). For \(x > 4\), use the function \(13 - x\).
03

Evaluate the Function at a Specific Point

To evaluate \(f(x)\) at any specific value, determine which part of the piecewise function to use based on the value of \(x\).
04

Example Evaluations

For example, to find \(f(3)\), since \(3 \leq 4\), use \(2(3) + 1 = 7\). For \(f(5)\), since \(5 > 4\), use \(13 - 5 = 8\).

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Key Concepts

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

Function Domains
In mathematics, a domain of a function is the set of all possible input values (usually represented by 'x') for which the function is defined. For the given piecewise function, the domain is split up based on different intervals:
  • For the function part defined as \(2x + 1\), the domain is all values of \(x \leq 4\).
  • For the function part defined as \(13 - x\), the domain is all values of \(x > 4\).

These specified intervals ensure that the function has clear input ranges for each piece, making it well-defined.
Evaluating Functions
When we talk about evaluating a function, we're referring to the process of finding the output value of the function for a specific input value of \(x\). To do this properly, follow these steps:
  • Determine which part of the piecewise function applies to the given value of \(x\).
  • Substitute the value of \(x\) into the corresponding part of the function.
  • Simplify, if necessary, to find the output value.

For example:
- To evaluate \(f(3)\), check that 3 is less than or equal to 4, so you use the function \(2(3) + 1 = 7\).
- For \(f(5)\), since 5 is greater than 4, you use \(13 - 5 = 8\).
Thus, by identifying which part of the function to use, evaluating it becomes straightforward.
Piecewise-Defined Functions
A piecewise-defined function is a function that is defined by different expressions based on different intervals of the input value ('x'). These functions are particularly useful for modeling scenarios where the behavior changes based on the value of \(x\).
To better understand piecewise-defined functions, consider the function:
  • \(f(x) = 2x + 1\) for \(x \leq 4\)
  • \(f(x) = 13 - x\) for \(x > 4\).

This structure allows for multiple formulas to represent the function's output based on intervals within its domain. Each 'piece' of the function handles a specific range of input values, making it effectively tailored to handle different conditions. Understanding how to decompose and evaluate these pieces is key to mastering piecewise functions.

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Most popular questions from this chapter

\(f(x)= \begin{cases}(x+9)^{2}-8 & \text { if } x<-7 \\ -\sqrt{25-(x+4)^{2}} & \text { if }-7 \leq x \leq 0 \\ (x-2)^{2}-7 & \text { if } 0

If \(f(x)=a x^{3}+b x^{2}\), determine \(a\) and \(b\) so that the graph of \(f\) will have a point of inflection at \((1,2)\).

The demand equation for a certain commodity produced by a monopolist is \(p=a-b x\), and the total cost, \(C(x)\) dollars, of producing \(x\) units is determined by \(C(x)=c+d x\), where \(a, b, c\), and \(d\) are positive constants. If the government levies a tax on the monopolist of \(t\) dollars per unit produced, show that in order for the monopolist to maximize his profits he should pass on to the consumer only one-half of the tax; that is, he should increase his unit price by \(\frac{1}{2} t\) dollars.

Find \(a, b, c\), and \(d\) so that the function defined by \(f(x)=a x^{3}+b x^{2}+c x+d\) will have relative extrema at \((1,2)\) and \((2,3)\).

\(f(x)=(x+2)^{2}(x-1)^{2}\)
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