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Chapter 4: Problem 8

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

Short Answer

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

Step by step solution

01

- 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.
02

- 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. \(2-x \eq 0 \rightarrow x \eq 2\)
03

- Determine the Valid Intervals

Combine the results of the inequalities: \(x \geq -2\) and \(x \eq 2\). The valid interval is \([-2, 2)\).
04

- 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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Functions \(f\) and \(g\) are defined. In each exercise define \(f \circ g\), and determine all values of \(x\) for which \(f \circ g\) is continuous. $$ f(x)=\frac{1}{x-2} ; g(x)=\sqrt{x} $$

Find the critical numbers of the given function. $$ f(x)=x^{4}+11 x^{3}+34 x^{2}+15 x-2 $$

Find the limits, and when applicable indicate the limit theorems being used. $$ \lim _{t \rightarrow \infty} \frac{\sqrt{t+\sqrt{t+\sqrt{t}}}}{\sqrt{t+1}} $$

Find the critical numbers of the given function. $$ f(x)=x^{3}+7 x^{2}-5 x $$

Determine whether the function is continuous or discontinuous on each of the indicated intervals. $$ f(x)=\sqrt{3+2 x-x^{2}} ;(-1,3),[-1,3],[-1,3),(-1,3] $$
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