Problem 94 - A Graphical Approach to College Algebra

Question

Find the open interval in which \(x\) must lie in order for the given condition to hold. \(y=5 x+12,\) and the difference between \(y\) and 4 is less than 0.0001.

Step-by-Step Solution

Answer

\(-1.60002 < x < -1.59998\)

Step 1: Set Up the Inequality

First, we are given that the difference between \( y \) and 4 must be less than 0.0001. So we write the inequality as:\[ |y - 4| < 0.0001 \]

Step 2: Substitute for y

Using the given expression \( y = 5x + 12 \), substitute \( y \) into the inequality:\[ |5x + 12 - 4| < 0.0001 \]

Step 3: Simplify the Expression

Simplify the expression inside the absolute value by combining like terms:\[ |5x + 8| < 0.0001 \]

Step 4: Solve the Absolute Inequality

The inequality \( |5x + 8| < 0.0001 \) can be rewritten as two separate inequalities:1. \( 5x + 8 < 0.0001 \)2. \( 5x + 8 > -0.0001 \)

Step 5: Solve Inequality 1

Solve the inequality \( 5x + 8 < 0.0001 \):\[ 5x < 0.0001 - 8 \]\[ 5x < -7.9999 \]\[ x < -1.59998 \]

Step 6: Solve Inequality 2

Solve the inequality \( 5x + 8 > -0.0001 \):\[ 5x > -0.0001 - 8 \]\[ 5x > -8.0001 \]\[ x > -1.60002 \]

Step 7: Write the Solution as an Interval

Combine the two inequalities to find the open interval for \( x \):\[ -1.60002 < x < -1.59998 \]

Key Concepts

Absolute Value

In mathematics, absolute value is a way of describing how far a number is from zero on the number line. It's always expressed as a non-negative value. The absolute value of a number is denoted by vertical bars, such as \(|x|\). Essentially, \(|x|\) is equal to \(-x\) when \(x < 0\) and equal to \(x\) when \(x \geq 0\).

• If \(x = -3\), then \(|x| = 3\).
• If \(x = 5\), then \(|x| = 5\).

When applied to equations, absolute value plays a crucial role in expressing constraints and conditions like distances, as it indicates that we care only about the size of difference, not the direction. For example, the expression \(|y - 4| < 0.0001\) means that \(y\) can be as close as you like to \(4\) but never more than \(0.0001\) units away.

Inequality Solution

Solving inequalities is an essential part of understanding relationships between numbers. An inequality compares two values, showing if one is less than, greater than, or simply not equal to another. When it comes to absolute inequalities like this one: \(|5x + 8| < 0.0001\), the goal is to find the range of values for \(x\) that makes the inequality true.
To break it down:

• The inequality \(|5x + 8| < 0.0001\) can be split into two separate inequalities: \(5x + 8 < 0.0001 \) and \(5x + 8 > -0.0001 \).
• These inequalities can be solved individually, getting \(x < -1.59998\) and \(x > -1.60002\).

In this problem, the solution indicates the set of all possible \(x\) values creating this condition; a continuous range or interval. Solutions include not just single values but sometimes a whole spread of possibilities, as in open intervals.

Open Interval

An open interval defines a set of real numbers between two endpoints, where the endpoints themselves are not included in the set. This interval is denoted using parentheses. So, for the range from \(-1.60002\) to \(-1.59998\), the open interval would be written as \((-1.60002, -1.59998)\).
The significance of an open interval here is that \(x\) can be any number within this range but cannot be equal to \(-1.60002\) or \(-1.59998\). The attributes can be visualized on a number line as a segment that does not include the end points.

• An open interval is a common result when solving inequalities, especially absolute inequalities, signaling that solutions approach but do not touch endpoints.
• It indicates where the original condition, like an inequality or a differential constraint, holds true.

Graphical Approach

Solving inequalities often benefits from a graphical perspective. Visualizing can clarify understanding and confirm solutions. When dealing with absolute inequalities and intervals, plotting them on a number line or using graphing software can be helpful.
The graphical approach includes:

• Plotting the function \(y = 5x + 12\) on a Cartesian plane to see how it behaves relative to \(y = 4\).
• Showing the threshold lines \(y = 4 + 0.0001\) and \(y = 4 - 0.0001\) as horizontal guidelines.
• Indicating the region between these lines, defining where \(|y - 4| < 0.0001\) is satisfied.

This method offers an intuitive grasp of how the inequality narrows down the possible \(x\) values. A graph provides a visual confirmation of algebraic solutions, validating that \(x\) must reside within the calculated open interval.