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Chapter 1: Problem 26

Discuss the continuity of each function. $$ f(x)=\frac{x^{2}-1}{x+1} $$

Short Answer

Expert verified
The function \( f(x)=\frac{x^{2}-1}{x+1} \) is continuous at all points except \(x=-1\).

Step by step solution

01

Understand the Concept of Continuity

In mathematics, a function \( f(x) \) is said to be continuous at a point \( x=a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) exists and is equal to \( f(a) \). In simple terms, a function is continuous if it has no gaps or jumps in its graph. So the given function is continuous wherever it is defined unless stated otherwise.
02

Simplify the function

Factor the function \( f(x)=\frac{x^{2}-1}{x+1} \) as \( \frac{(x-1)(x+1)}{x+1} \). If \( x \neq -1 \), the terms \((x+1)\) will cancel out and we have \( f(x) = x-1 \). If \(x = -1\), the function \(f(x)\) is undefined.
03

Identify Points of Discontinuity

A function is discontinuous at a point where it is not defined. Therefore, the function \( f(x)=\frac{x^{2}-1}{x+1} \) is discontinuous at the point \(x=-1\), as at \(x=-1\), the function is undefined.

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Key Concepts

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

Discontinuity
Discontinuity in calculus is an essential concept when analyzing the nature of functions. Simply put, a function is discontinuous at points where it isn't defined. This usually results in gaps, jumps, or holes in the graph of the function. For the given function, \( f(x)=\frac{x^{2}-1}{x+1} \), there's a key discontinuity at \( x = -1 \).
  • This happens because, at \( x = -1 \), the denominator \( x+1 \) becomes zero, causing the function to be undefined there.
  • The graph will show a hole at \( x = -1 \).
  • Understanding discontinuity helps in determining intervals where the function behaves normally and where it doesn't.
For any function, points of discontinuity are critical for complete analysis since these points affect both the graph and behavior of the function.
Limits
Limits play a crucial role in understanding the behavior of functions near points of discontinuity. They help in exploring what the value of a function might be, as the input approaches a specific point. For our function \( f(x)=\frac{x^{2}-1}{x+1} \), examining the limit as \( x \) approaches -1 is essential.
  • By simplifying \( f(x) = \frac{(x-1)(x+1)}{x+1} \), and canceling out \((x+1)\) when \( x eq -1 \), we get \( f(x) = x-1 \).
  • As \( x \to -1 \), although \( f(-1) \) is undefined, the function approaches \( -2 \). This suggests the closer \( x \) gets to -1, the closer \( f(x) \) gets to -2.
While the function may not be defined at \( x = -1 \), the limit tells us valuable information about how the function behaves around this point. Understanding limits helps in conceptualizing the continuity of functions and reveals insights into undefined points.
Rational functions
Rational functions are fractions where both the numerator and the denominator are polynomials. They are relevant in calculus for their non-linear characteristics and their behavior at points of discontinuity. Let's illustrate this with our function, \( f(x)=\frac{x^{2}-1}{x+1} \).
  • The numerator, \( x^2 - 1 \), is a polynomial that can be factored as \((x-1)(x+1)\).
  • The denominator, \( x+1 \), is a simple linear polynomial.
  • When \( x eq -1 \), the function simplifies to \( f(x) = x - 1 \), indicating basic linear behavior.
Rational functions are known for having vertical asymptotes or holes at points where the denominator equals zero. For \( f(x) \), at \( x = -1 \), the rational function does not exist, which leaves a hole in the graph. Thus, understanding rational functions is key to mastering various properties of calculus related to continuity and differentiability.

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Most popular questions from this chapter

Describe the interval(s) on which the function is continuous. $$ f(x)=\frac{x}{x^{2}+1} $$

Describe the interval(s) on which the function is continuous. $$ f(x)=\sec \frac{\pi x}{4} $$

Use a graphing utility to graph the function. Use the graph to determine any \(x\) -values at which the function is not continuous. $$ g(x)=\left\\{\begin{array}{ll} 2 x-4, & x \leq 3 \\ x^{2}-2 x, & x>3 \end{array}\right. $$

The signum function is defined by \(\operatorname{sgn}(x)=\left\\{\begin{array}{ll}-1, & x<0 \\ 0, & x=0 \\ 1, & x>0\end{array}\right.\) Sketch a graph of \(\operatorname{sgn}(x)\) and find the following (if possible). (a) \(\lim _{x \rightarrow 0^{-}} \operatorname{sgn}(x)\) (b) \(\lim _{x \rightarrow 0^{+}} \operatorname{sgn}(x)\) (c) \(\lim _{x \rightarrow 0} \operatorname{sgn}(x)\).

Use a graphing utility to graph the function. From the graph, estimate \(\lim _{x \rightarrow 0^{+}} f(x) \quad\) and \(\lim _{x \rightarrow 0^{-}} f(x)\) Is the function continuous on the entire real line? Explain. $$ f(x)=\frac{\left|x^{2}-4\right| x}{x+2} $$
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