Problem 3 - Calculus for Biology and Medicine

Question

Find the values of \(x\) such that $$ \left|x^{2}-9\right|<0.1 $$

Step-by-Step Solution

Answer

The values of \(x\) are between \(-3.0166\) and \(-2.9833\) or between \(2.9833\) and \(3.0166\).

Step 1: Understanding the Absolute Inequality

The inequality \(|x^2 - 9| < 0.1\) means that the expression \(x^2 - 9\) must lie between -0.1 and 0.1. This can be expanded into two inequalities to solve separately.

Step 2: Breaking Down the Inequality

Split the absolute inequality into two separate inequalities: 1. \(x^2 - 9 < 0.1\) 2. \(x^2 - 9 > -0.1\). Solving these will help us find the interval for \(x\).

Step 3: Solving the First Inequality

Starting with \(x^2 - 9 < 0.1\), we add 9 to both sides resulting in \(x^2 < 9.1\). Taking the square root on both sides, we get the range \(-\sqrt{9.1} < x < \sqrt{9.1}\).

Step 4: Solving the Second Inequality

For the inequality \(x^2 - 9 > -0.1\), add 9 to both sides to get \(x^2 > 8.9\). Again, taking square roots gives us the values \(x > \sqrt{8.9}\) or \(x < -\sqrt{8.9}\).

Step 5: Finding the Overlapping Range

To satisfy the original inequality, we need \(x\) to fall in the overlap of the solutions from both inequalities. Thus, the ranges are combined to find where both conditions are met: \(-\sqrt{9.1} < x < -\sqrt{8.9}\) and \(\sqrt{8.9} < x < \sqrt{9.1}\).

Step 6: Final Solution

The solution for the inequality includes two intervals where \(x\) could fall: 1. \(-3 < x < -3\) (where 3 indicates a small number just under -3). 2. 3 > x > 3, a small number just over 3.

Key Concepts

Absolute Inequality

An absolute inequality like \(|x^2 - 9| < 0.1\) provides a condition where the expression within the absolute value must remain less than 0.1 units away from zero. In simple terms, it means that the distance of the expression from zero is strictly small, specifically within 0.1 units.
This inequality breaks down into two separate inequalities:

• \(x^2 - 9 < 0.1\)
• \(x^2 - 9 > -0.1\)

By solving these two, we find the set of values for \(x\) where both conditions hold.
It allows us to locate the region in the number line where \(x\) keeps \(x^2 - 9\) close to zero, within the desired "distance" of 0.1. This is a crucial technique in calculus, often employed to assess limits and continuity.

Interval Notation

Interval notation is a convenient way to express a range of numbers that satisfy an inequality. It's efficient and allows for clear communication of solution sets.
The solution from the original exercise involves two conditions, forming two separate intervals:

• \(-\sqrt{9.1} < x < -\sqrt{8.9}\)
• \(\sqrt{8.9} < x < \sqrt{9.1}\)

In interval notation, the above conditions are written as the sets:

• \((-\sqrt{9.1}, -\sqrt{8.9})\)
• \((\sqrt{8.9}, \sqrt{9.1})\)

The round parenthesis "()" indicates that the endpoints are not included (open intervals).
This notation is particularly useful in calculus for quickly representing solution sets and is essential when working with functions and their domains.

Solving Inequalities

Solving inequalities means finding all values of a variable that make the inequality true. It involves similar steps to solving equations but with extra consideration due to the inequality sign.
Here’s a step-by-step process to solve the inequality from the exercise:

• First, break down the absolute inequality into two separate linear inequalities: \(x^2 - 9 < 0.1\) and \(x^2 - 9 > -0.1\).
• Solve each inequality individually by isolating \(x^2\) and then taking the square root. Remember to consider both positive and negative roots due to the square.
• The solutions from both parts are considered together to form the final solution. This involves looking for the intersection or union of the solutions, depending on the original inequality's requirements.

Solving inequalities is crucial in calculus, as it lays the groundwork for understanding more complex functions and their behaviors.
It helps to identify ranges of inputs that lead to desired outcomes, a fundamental concept in calculus analysis.