Problem 4 - Calculus for Biology and Medicine

Question

Find the values of \(x\) such that $$ |2 \sqrt{x}-5|<0.1 $$

Step-by-Step Solution

Answer

The values of \(x\) are within the interval \( (6.0025, 6.5025) \).

Step 1: Understand the Absolute Inequality

The given inequality is \(|2 \sqrt{x} - 5| < 0.1\). This means the expression \(2 \sqrt{x} - 5\) is within 0.1 units of 0. Therefore, to solve \(|A| < B\), we need to find the range of values for \(A\) such that \(-B < A < B\).

Step 2: Set up Two Inequalities

Given the inequality \(|2 \sqrt{x} - 5| < 0.1\), we can set it up as two separate inequalities: 1. \(2 \sqrt{x} - 5 < 0.1\)2. \(2 \sqrt{x} - 5 > -0.1\)

Step 3: Solve the First Inequality

For the inequality \(2 \sqrt{x} - 5 < 0.1\), add 5 to both sides to get:\[ 2 \sqrt{x} < 5.1 \]Now, divide by 2:\[ \sqrt{x} < 2.55 \]Square both sides to solve for \(x\):\[ x < 6.5025 \]

Step 4: Solve the Second Inequality

For the inequality \(2 \sqrt{x} - 5 > -0.1\), add 5 to both sides to get:\[ 2 \sqrt{x} > 4.9 \]Now, divide by 2:\[ \sqrt{x} > 2.45 \]Square both sides to solve for \(x\):\[ x > 6.0025 \]

Step 5: Combine the Results

From the two inequalities \(x < 6.5025\) and \(x > 6.0025\), the range for \(x\) is the intersection of these two solutions. Thus, the solution is:\[ 6.0025 < x < 6.5025 \]

Key Concepts

Inequality Solving

Inequality solving is about finding the range of numbers that satisfy a given inequality. Unlike equations that find specific values, inequalities describe a set of possible solutions that make the inequality true.

To tackle inequality solving, consider the symbol in the inequality:

• < Less than (<) or less than or equal to (≤)
• > Greater than (>) or greater than or equal to (≥)

The main task is to isolate the variable on one side, which helps in understanding what values make the inequality true. Solving inequalities can sometimes lead to multiple steps, such as:

• Adding or subtracting terms.
• Multiplying or dividing by a constant (be cautious with negatives).
• Considering special rules, such as when dealing with absolute values.

In the exercise provided, we explored a compound procedure where we had to first recognize the form of the inequality, then separate and solve it into two possible inequalities. This approach yields the range of values that satisfy the original expression, demonstrating that inequality solving is about understanding and manipulating conditions to reveal the solutions.

Absolute Value

The concept of absolute value is rooted in measuring how far a number is from zero on a number line, regardless of direction. It is denoted by vertical bars, like \(|x|\), meaning the absolute value of x. This reflects the non-negative value of x.

Absolute value inequalities, such as \(|A| < B\), suggest that A is within B units from zero. It means that:

• -B < A < B

This formulation sets up two separate inequalities to define the range of possible values for A. For absolute inequalities:

• The problem is split into two parts.
• Each part must be solved individually.
• The results are then combined to capture the full range of solutions.

In the example in question, the absolute value indicates a small range where \(2 \sqrt{x} - 5\) must sit very close to zero, indicating a small interval over which x is valid. Understanding absolute value in inequalities is fundamental for accuracy in range finding.

Interval Notation

Interval notation is a mathematical shorthand that describes a range of numbers between two endpoints. It gives a way for mathematicians to easily convey the set of solutions for inequalities.

Consider interval notation's specifics:

• Use of brackets or parentheses (") to denote whether endpoints are included.
• Parentheses, \( ( or ) \), describe boundaries where the endpoint is not included.
• Brackets, \( [ or ] \), include the endpoint as part of the solution.

For instance, the interval \( (3, 7) \) means 3 < x < 7, excluding 3 and 7. Conversely, \( [3, 7] \) would include 3 and 7.

For our problem, using interval notation to write the solution \( 6.0025 < x < 6.5025 \) results in \((6.0025, 6.5025)\). This concisely captures the valid range of x, maintaining clarity and precision.