/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 An equation of the form \(|f(x)|... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 69

An equation of the form \(|f(x)|=|g(x)|\) is given. (a) Solve the equation analytically and support the solution graphically. (b) Solve \(|f(x)|>|g(x)|\) (c) Solve \(|f(x)|<|g(x)|\) $$|-2 x+5|=|x+3|$$

Short Answer

Expert verified
Solutions: (a) \(x = \frac{2}{3}, 8\); (b) \((-\infty, \frac{2}{3}) \cup (8, \infty)\); (c) \((\frac{2}{3}, 8)\)."

Step by step solution

01

Remove Absolute Values

Given the equation \(|-2x + 5| = |x + 3|\), we need to address this by considering the two cases enabled by absolute values: one where both expressions are non-negative and one where both are non-positive. The two cases mean 1. a)\(-2x + 5 = x + 3\) b) \(-2x + 5 = -(x + 3)\)
02

Solve the First Case

For the first case, solve \(-2x + 5 = x + 3\):Subtract \(x\) from both sides:\(-2x - x + 5 = 3\)Combine like terms:\(-3x + 5 = 3\)Subtract 5 from both sides:\(-3x = -2\)Divide by \(-3\):\(x = \frac{2}{3}\).
03

Solve the Second Case

For the second case, solve \(-2x + 5 = -(x + 3)\):Distribute the negative sign on the right side:\(-2x + 5 = -x - 3\)Add \(2x\) to both sides to combine terms:\(5 = x - 3\)Add 3 to both sides:\(x = 8\).
04

Verify Solutions with Original Expression

Verify whether the solutions \(x = \frac{2}{3}\) and \(x = 8\) satisfy the original absolute value equation. Substitute back into \(|-2x + 5|\) and \(|x + 3|\):For \(x = \frac{2}{3}\):\(|-2(\frac{2}{3}) + 5| = |\frac{2}{3} + 3|\)Both sides simplify to equal amounts.For \(x = 8\):\(|-2(8) + 5| = |8 + 3|\)Both sides also simplify to equal amounts.
05

Analyze Inequality \(|-2x + 5| > |x + 3|\)

Consider both cases for the inequality:1. \(-2x + 5 > x + 3\)Solve by subtracting \(x\) and simplifying:\(-3x > -2\) Divide by \(-3\), flip inequality: \(x < \frac{2}{3}\).2. \(-x - 3 > -2x + 5\)Rearrange and solve:\(x > 8\).
06

Analyze Inequality \(|-2x + 5| < |x + 3|\)

Consider both cases for the inequality:1. \(-2x + 5 < x + 3\)Solve by subtracting \(x\): \(-3x < -2\)Divide by \(-3\), flip inequality: \(x > \frac{2}{3}\).2. \(-x - 3 < -2x + 5\)Rearrange: \(x < 8\).
07

Combine Results for Inequalities

Combine the solutions found to problem into intervals based on the inequalities.For \(|-2x + 5| > |x + 3|\), combine overlapping intervals:The solution is \((-\infty, \frac{2}{3}) \cup (8, \infty)\).For \(|-2x + 5| < |x + 3|\), the intersection of solutions gives:The solution is \((\frac{2}{3}, 8)\).

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Key Concepts

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

Inequality Solutions
When dealing with absolute value equations like \(|-2x + 5| = |x + 3|\), it's essential to understand how inequalities can arise and be solved. Absolute values can create two separate inequalities based on whether the expressions inside are positive or negative. For example, \(|-2x + 5| > |x + 3|\) leads to two cases: \(-2x + 5 > x + 3\) and \(-x - 3 > -2x + 5\). Solving these separately involves rearranging terms and isolating \(x\). Remember, dividing or multiplying by a negative reverses the inequality sign.\
  • For \(-2x + 5 > x + 3\), solving leads to \(x < \frac{2}{3}\).
  • For \(-x - 3 > -2x + 5\), solving results in \(x > 8\).
Combine these to get the solution set \((-\infty, \frac{2}{3}) \cup (8, \infty)\) for this specific inequality.
Graphical Representation
Graphing absolute value equations and inequalities can provide a visual intuition of the solution. For \(|-2x + 5| = |x + 3|\), you're essentially looking at the intersections of two V-shaped graphs. The graphs of \(|-2x + 5|\) and \(|x + 3|\) will intersect at points where the solutions lie, providing a visual reference for where the two expressions are equal.
  • Draw the graphs of both expressions on a coordinate plane.
  • The intersection points represent the solutions \(x = \frac{2}{3}\) and \(x = 8\).
For inequalities like \(|-2x + 5| > |x + 3|\), shading areas on the graph that satisfy the inequality will also highlight solution intervals, visually showing \((-\infty, \frac{2}{3})\) and \((8, \infty)\).
Analytical Methods
Analytically solving absolute value equations like \(|-2x + 5|=|x + 3|\) involves considering both scenarios: one where both terms are equal in magnitude but opposite in sign, and one where they are equal.
  • Break down the equation into two cases: \-2x + 5 = x + 3\ and \-2x + 5 = -(x + 3)\.
  • Solve each equation separately. The solutions \(x = \frac{2}{3}\) and \(x = 8\) are found.
Verifying such solutions ensures correctness. Plugging values back into the original equation confirms whether both sides are equal, affirming the solutions are accurate. Using these cases covers all necessary possibilities the equation might present.

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Most popular questions from this chapter

Functions such as the pairs in Exercises \(69-72\) are called inverse functions, because the result of composition in both directions is the identity function. (Inverse functions will be discussed in detail in Section 5.1.) In a square viewing window, graph \(y_{1}=\sqrt[3]{x-6}\) and \(y_{2}=x^{3}+6,\) an example of a pair of inverse functions. Now graph \(y_{3}=x .\) Describe how the graph of \(y_{2}\) can be obtained from the graph of \(y_{1},\) using the graph \(y_{3}=x\) as a basis for your description.

During the early years of the AIDS epidemic, cases and cumulative deaths reported for selected years \(x\) could be modeled by quadratic functions. For \(1982-\) 1994 , the numbers of AIDS cases are modeled by $$f(x)=3200(x-1982)^{2}+1586$$ and the numbers of deaths are modeled by $$g(x)=1900(x-1982)^{2}+619$$ $$\begin{array}{|l|c|c|}\hline \text { Year } & \text { Cases } & \text { Deaths } \\\\\hline 1982 & 1,586 & 619 \\ 1984 & 10,927 & 5,605 \\\1986 & 41,910 & 24,593 \\\1988 & 106,304 & 61,911 \\\1990 & 196,576 & 120,811 \\ 1992 & 329,205 & 196,283 \\\1994 & 441,528 & 270,533\end{array}$$ (a) Graph \(h(x)=\frac{g(x)}{f(x)}\) in the window \([1982,1994]\) by \([0,1] .\) Interpret the graph. (b) Compute the ratio \(\frac{\text { deaths }}{\text { cases }}\) for each year. Compare the results with those from part (a).

Solve each equation or inequality graphically. $$-|3 x-12| \geq-x-1$$

Find the open interval in which \(x\) must lie in order for the given condition to hold. \(y=4 x-8,\) and the difference between \(y\) and 3 is less than 0.001.

Solve each problem. The industrial process that is used to convert methanol to gasoline is carried out at a temperature range of \(680^{\circ} \mathrm{F}\) to \(780^{\circ} \mathrm{F}\). Using \(F\) as the variable, write an absolute value inequality that corresponds to this range.
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