Problem 8 - The Calculus with Analytic Geometry

Question

Determine whether the function is continuous or discontinuous on each of the indicated intervals. $$ f(x)=\sqrt{\frac{2+x}{2-x}},(-2,2),[-2,2],[-2,2),(-2,2],(-\infty,-2),[2,+\infty) $$

Step-by-Step Solution

Answer

Continuous in \(((-2,2))\) and \(([-2,2))\). Discontinuous otherwise.

Step 1 - Understand the Domain of the Function

The function given is \(f(x)=\sqrt{\frac{2+x}{2-x}}\). For the square root to be defined, the expression inside must be non-negative. Hence, \(\frac{2+x}{2-x} \geq 0\). Also, the denominator cannot be zero, so \(2-x \eq 0\). Solve to find the intervals where the function is defined.

Step 2 - Solve the Inequality

To solve \(\frac{2+x}{2-x} \geq 0\), set up the inequality and solve for \x\. Solve for \2+x \geq 0\ and \2-x \eq 0\.\1. \((2+x) \geq 0 \rightarrow x \geq -2\)\2. \(2-x \eq 0 \rightarrow x \eq 2\)

Step 3 - Determine the Valid Intervals

Combine the results of the inequalities: \(x \geq -2\) and \(x \eq 2\). The valid interval is \([-2, 2)\).

Step 4 - Check Given Intervals for Continuity

Evaluate each given interval to see where the function is continuous:\ \(((-2,2))\): \(x=-2\) is not included so it's part of the interval. \(x = 2\) is not part of the interval. \Thus, the function is continuous on \(((-2,2))\).\ \(([-2,2])\): \(x=2\) is not included in the valid interval, so the function is discontinuous here.\ \(([-2,2))\): This interval fits within the valid interval, so the function is continuous.\ \(((-2,2])\): Since \(x=2\) is included, the function is discontinuous.\ \(((-\infty,-2))\): The function is not defined in this interval, so it is discontinuous.\ \(([2,+\infty))\): The function is not defined in this interval, so it is discontinuous.

Key Concepts

Domain of a Function

The domain of a function includes all the input values (x-values) for which the function is defined. For the function given, \( f(x)=\frac{2+x}{2-x} \), we must carefully determine where it is defined and valid. The primary conditions to consider are:

• The expression inside the square root must be non-negative, which makes sure the square root is a real number.
• The denominator cannot be zero, as division by zero is undefined.

To find the domain, solve the inequality \( \frac{2+x}{2-x} \geq 0 \) and ensure the condition \( 2 - xeq 0 \) is met. By solving through these constraints, we find the valid interval is \([-2, 2)\).

Solving Inequalities

Solving inequalities involves finding the range of values that satisfy a given condition. We solve \( \frac{2+x}{2-x} \geq 0 \) by breaking it into two parts:

• Make sure the numerator is non-negative: \( 2 + x \geq 0 \), which simplifies to \( x \geq -2 \).
• Ensure the denominator does not become zero: \( 2 - x eq 0 \), leading to \( x eq 2 \).

Combining these results, we derive that \( x \) must be between \(-2\) and \(2\), excluding \( 2 \). Hence the interval is \([-2, 2)\).

Interval Notation

Interval notation provides a compact way to represent the range of values that satisfy certain conditions. It uses brackets and parentheses to show whether endpoints are included or excluded. For example:

• \([-2, 2)\): The interval includes \( -2 \) but not \( 2 \).
• \((a, b] \): The interval includes \( b \) but not \( a \).
• \((c, d)\): Neither \( c \) nor \( d \) are included.

In our example, the valid domain of the function is represented as \([-2, 2)\) since \( -2 \) is part of the domain, but \( 2 \) is not allowed due to the denominator becoming zero.

Discontinuous Points

A function is continuous on an interval if there are no breaks, jumps, or holes in the graph within that interval. Checking for continuous and discontinuous points involves:

• Determining where the function is undefined or singular.
• Examining given intervals to see where the function maintains continuity.

For the example function, we evaluated different intervals:

• \(((-2, 2))\) and \([-2, 2)\): The function is continuous as \( x = 2 \) is excluded.
• \(([-2, 2])\) and \(((-2, 2])\): The function is discontinuous as \( x = 2 \) is included, which makes the denominator zero.
• \(((-\infty, -2))\) and \(([2, + \infty))\): The function is discontinuous outside the valid domain.