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Chapter 2: Problem 36

Find the real zeros of the given function \(f\). $$ f(x)=-2 x+9 $$

Short Answer

Expert verified
The real zero of the function is \( x = \frac{9}{2} \).

Step by step solution

01

Understand the Problem

We need to find the real zeros of the function \( f(x) = -2x + 9 \). The real zeros are the values of \( x \) for which the function equals zero, i.e., \( f(x) = 0 \).
02

Set the Function Equal to Zero

To find the zeros, set the function \( f(x) \) equal to zero: \[ -2x + 9 = 0 \]
03

Isolate the Variable

Subtract 9 from both sides of the equation to isolate the term with \( x \): \[ -2x = -9 \]
04

Solve for x

Divide both sides by \(-2\) to solve for \( x \):\[ x = \frac{-9}{-2} \]
05

Simplify the Expression

Simplify the fraction:\[ x = \frac{9}{2} \]

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Key Concepts

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

Solving Linear Equations
Solving linear equations is a fundamental skill in algebra. A linear equation usually takes the form \( ax + b = 0 \), where \( a \) and \( b \) are constants. Our goal is to find the value of \( x \) that satisfies the equation. Let's consider the equation from our example, \( -2x + 9 = 0 \).

To solve this equation, we proceed as follows:
  • First, understand that the equation is essentially a balance. Whatever we do to one side of the equation, we must also do to the other side to maintain the equality.
  • In this example, the equation is set up to find the real zeros, which are values of \( x \) such that \( f(x) = 0 \).
Solving means breaking down the equation step-by-step until \( x \) is isolated on one side.
Isolating Variables
Isolating the variable is an important step to solve any equation effectively. When we solve \( -2x + 9 = 0 \), the initial goal is to get \( x \) by itself on one side of the equation. Here's how you accomplish that:
  • First, subtract 9 from both sides to eliminate the constant term on the left side. This gives us \( -2x = -9 \).
  • This step ensures that \( x \) is only multiplied by a number, making it easier to finally isolate \( x \) itself.
By doing this, we reverse all operations done to \( x \) and peer deeper into the structure of our equation. Properly isolating variables allows us to see the solution more clearly.
Function Roots
The roots of a function represent the values of \( x \) where the function equals zero. For a linear function like \( f(x) = -2x + 9 \), finding the root is equivalent to finding its zero.

This involves setting the function equal to zero and solving the equation. Here, after isolating \( x \), our equation \( -2x = -9 \) needs a final operation:
  • Divide both sides by \(-2\) to resolve \( -2x = -9 \) into \( x = \frac{-9}{-2} \).
  • Simplifying \( \frac{-9}{-2} \) results in our root, \( x = \frac{9}{2} \).
The root of the function \( f(x) \) indicates that \( x = \frac{9}{2} \) is where the function crosses the x-axis. Understanding how to determine function roots helps in graphing and predicting real-world models.

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Most popular questions from this chapter

The piecewise-defined function $$ f(x)=\left\\{\begin{array}{ll} 1, & x \text { a rational number } \\ 0, & x \text { an irrational number } \end{array}\right. $$ is called the Dirichlet function after the German mathematician Johann Peter Gustav Lejeune Dirichlet \((1805-1859) .\) Dirichlet is responsible for the definition of a function as we know it today. Find each of the following function values. (a) \(f\left(\frac{1}{3}\right)\) (b) \(f(-1)\) (c) \(f(\sqrt{2})\) (d) \(f(1.12)\) (e) \(f(5.72)\) (f) \(f(\pi)\)

The given function \(f\) is one-toone. The domain and range of \(f\) is given. Find \(f^{-1}\) and give its domain and range. $$ f(x)=\frac{2}{\sqrt{x}}, \quad x>0, y>0 $$

In Problems \(9-34\), sketch the graph of the given piece wise-defined function. Find any \(x\) - and \(y\) intercepts of the graph. Give any numbers at which the function is discontinuous. $$ y=\left\\{\begin{array}{ll} 1, & x<0 \\ |x-1|, & 0 \leq x \leq 2 \\ 1, & x>2 \end{array}\right. $$

The function \(f(x)=|2 x-4|\) is not one-to-one. How should the domain of \(f\) be restricted so that the new function has an inverse? Find \(f^{-1}\) and give its domain and range. Sketch the graph of \(f\) on the restricted domain and the graph of \(f^{-1}\) on the same coordinate axes.

Proceed as in Example 3 and verify that the inverse of the one-to-one function \(f\) is the function \(g\) by showing \(f(g(x))=x\) and \(g(f(x))=x\). $$ f(x)=\frac{1}{r^{3}} ; \quad g(x)=\frac{1}{3 \sqrt{x}} $$
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