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Chapter 1: Problem 6

Consider the functions $$ f(x)=\left\\{\begin{array}{ll}{1,} & {0 \leq x} \\ {0,} & {x<0}\end{array} \text { and } g(x)=\left\\{\begin{array}{ll}{0,} & {0 \leq x} \\ {1,} & {x<0}\end{array}\right.\right. $$ In each part, is the given function continuous at \(x=0 ?\) $$ \begin{array}{llll}{\text { (a) } f(x)} & {\text { (b) } g(x)} & {\text { (c) } f(-x)} & {\text { (d) }|g(x)|} \\ {\text { (e) } f(x) g(x)} & {\text { (f) } g(f(x))} & {\text { (g) } f(x)+g(x)}\end{array} $$

Short Answer

Expert verified
\( f(x), g(x), f(-x), |g(x)| \) are not continuous. \( f(x)g(x), g(f(x)), f(x)+g(x) \) are continuous at \( x = 0 \).

Step by step solution

01

Analyze \( f(x) \) for Continuity at \( x = 0 \)

A function is continuous at a point if the left-hand limit, right-hand limit, and the function's value at that point are all equal. For \( f(x) \), to check continuity at \( x=0 \), we find:- **Left-hand limit**: \( \lim_{x \to 0^-} f(x) = 0 \)- **Right-hand limit**: \( \lim_{x \to 0^+} f(x) = 1 \)- **Function value**: \( f(0) = 1 \)Since the left-hand limit is not equal to the right-hand limit, \( f(x) \) is **not continuous** at \( x = 0 \).
02

Analyze \( g(x) \) for Continuity at \( x = 0 \)

Similarly, for \( g(x) \):- **Left-hand limit**: \( \lim_{x \to 0^-} g(x) = 1 \)- **Right-hand limit**: \( \lim_{x \to 0^+} g(x) = 0 \)- **Function value**: \( g(0) = 0 \)The left-hand and right-hand limits are not equal, so \( g(x) \) is **not continuous** at \( x = 0 \).
03

Analyze \( f(-x) \) for Continuity at \( x = 0 \)

Consider the function \( f(-x) \):- **Left-hand limit**: \( \lim_{x \to 0^-} f(-x) = \lim_{x \to 0^+} f(x) = 1 \)- **Right-hand limit**: \( \lim_{x \to 0^+} f(-x) = \lim_{x \to 0^-} f(x) = 0 \)- **Function value**: \( f(-0) = f(0) = 1 \)Since the left-hand limit is not equal to the right-hand limit, \( f(-x) \) is **not continuous** at \( x = 0 \).
04

Analyze \(|g(x)|\) for Continuity at \( x = 0 \)

Since \( g(x) \) can only be 0 or 1, \(|g(x)|\) maintains the same values:- **Left-hand limit**: \( \lim_{x \to 0^-} |g(x)| = 1 \)- **Right-hand limit**: \( \lim_{x \to 0^+} |g(x)| = 0 \)- **Function value**: \(|g(0)| = 0 \)The limits are different, confirming \(|g(x)|\) is **not continuous** at \( x = 0 \).
05

Analyze \( f(x)g(x) \) for Continuity at \( x = 0 \)

Consider \( f(x)g(x) \):- **Left-hand limit**: \( \lim_{x \to 0^-} f(x)g(x) = 0 \cdot 1 = 0 \)- **Right-hand limit**: \( \lim_{x \to 0^+} f(x)g(x) = 1 \cdot 0 = 0 \)- **Function value**: \( f(0)g(0) = 1 \cdot 0 = 0 \)The left-hand limit, right-hand limit, and the function value are the same, making \( f(x)g(x) \) **continuous** at \( x = 0 \).
06

Analyze \( g(f(x)) \) for Continuity at \( x = 0 \)

Evaluate \( g(f(x)) \), where for \( x \geq 0 \), \( f(x) = 1 \) and for \( x < 0 \), \( f(x) = 0 \):- **Left-hand limit**: \( \lim_{x \to 0^-} g(f(x)) = g(0) = 0 \)- **Right-hand limit**: \( \lim_{x \to 0^+} g(f(x)) = g(1) = 0 \)- **Function value**: \( g(f(0)) = g(1) = 0 \)All values match, so \( g(f(x)) \) is **continuous** at \( x = 0 \).
07

Analyze \( f(x) + g(x) \) for Continuity at \( x = 0 \)

Evaluate \( f(x) + g(x) \):- **Left-hand limit**: \( \lim_{x \to 0^-} (f(x) + g(x)) = 0 + 1 = 1 \)- **Right-hand limit**: \( \lim_{x \to 0^+} (f(x) + g(x)) = 1 + 0 = 1 \)- **Function value**: \( f(0) + g(0) = 1 + 0 = 1 \)The left-hand limit, right-hand limit, and the function value are all equal, thus \( f(x) + g(x) \) is **continuous** at \( x = 0 \).

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Key Concepts

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

Piecewise Functions
Piecewise functions are special types of functions that have different expressions based on the input value. This means the function behaves differently in various intervals of the domain. For the functions in our exercise, we use them to break down complex behaviors into simple, manageable parts.

  • Function Definition: A piecewise function is defined by multiple sub-functions, where each applies to a certain interval of the main function's domain.
In our example:
  • f(x) is defined as: \(f(x)=\begin{cases} 1, & \text{if } x \geq 0\ 0, & \text{if } x < 0 \end{cases}\)
  • g(x) behaves oppositely: \(g(x)=\begin{cases} 0, & \text{if } x \geq 0\ 1, & \text{if } x < 0 \end{cases}\)
Piecewise functions allow us to describe real-world situations where one rule may apply under certain conditions, and another rule under different conditions. Understanding each piece and their intervals is crucial for analyzing these functions.
Continuity at a Point
Continuity at a point means that the graph of a function at that point is unbroken; the function can be drawn without lifting the pencil from the paper. Mathematically, a function is continuous at a point if three conditions are met:

  • The left-hand limit at that point is equal to the right-hand limit.
  • The function value at the point is defined and equals these limits.
This is essential when investigating continuity in real-world applications, such as ensuring safety in engineering designs where seamless transitions are required. Consider f(x) and g(x) at x = 0. Neither function is continuous at this point because their left-hand and right-hand limits are different, although they do have defined function values at x = 0. In contrast, the composition g(f(x)) and the sum f(x) + g(x) at x = 0 are continuous, exemplifying that sometimes functions derived from discontinuous functions can be continuous.
Left-hand and Right-hand Limits
Left-hand and right-hand limits help us understand what happens to functions as we approach a particular point from either direction on the x-axis. This concept is essential for determining continuity of functions which govern real-world systems and models.

  • Left-hand limit: Approaching the point from the left, represented mathematically as \( \lim_{x \to a^-} f(x) \).
  • Right-hand limit: Approaching from the right, shown as \( \lim_{x \to a^+} f(x) \).
To check if a function at x = 0 is continuous, you compare these limits with the function's value at that point. For example,
  • In f(x) at x = 0, the left-hand limit is 0 and right-hand limit is 1, showing a disconnect.
  • In g(x) at x = 0, the left-hand limit is 1 and right-hand limit is 0, illustrating another type of discontinuity.
When these two limits are equal and match the function value, the function is continuous. Understanding each side's behavior helps solve complex problems and determine function stability.

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Most popular questions from this chapter

A function \(f\) is said to have a removable discontinuity at \(x=c\) if lim \(_{x \rightarrow c} f(x)\) exists but \(f\) is not continuous at \(x=c\), either because \(f\) is not defined at \(c\) or because the definition for \(f(c)\) differs from the value of the limit. This terminology will be needed in these exercises. (a) The terminology removable discontinuity is appropriate because a removable discontinuity of a function \(f\) at \(x=c\) can be "removed" by redefining the value of \(f\) appropriately at \(x=c .\) What value for \(f(c)\) removes the discontinuity? (b) Show that the following functions have removable dis- continuities at \(x=1,\) and sketch their graphs. $$ f(x)=\frac{x^{2}-1}{x-1} \quad \text { and } \quad g(x)=\left\\{\begin{array}{ll}{1,} & {x>1} \\ {0,} & {x=1} \\ {1,} & {x<1}\end{array}\right. $$ (c) What values should be assigned to \(f(1)\) and \(g(1)\) to remove the discontinuities?

Writing In some population models it is assumed that a given ecological system possesses a carrying capacity \(L .\) Populations greater than the carrying capacity tend to decline toward \(L,\) while populations less than the carrying capacity tend to increase toward \(L .\) Explain why these assumptions are reasonable, and discuss how the concepts of this section apply to such a model.

Find the limits. $$ \lim _{x \rightarrow+\infty} \ln \left(\frac{x+1}{x}\right) $$

The notion of an asymptote can be extended to include curves as well as lines. Specifically, we say that curves \(y=f(x)\) and \(y=g(x)\) are asymptotic as \(x \rightarrow+\infty\) provided $$ \lim _{x \rightarrow+\infty}[f(x)-g(x)]=0 $$ In these exercises, determine a simpler function \(g(x)\) such that \(y=f(x)\) is asymptotic to \(y=g(x)\) as \(x \rightarrow+\infty\) or \(x \rightarrow-\infty\) Use a graphing utility to generate the graphs of \(y=f(x)\) and \(y=g(x)\) and identify all vertical asymptotes. $$ f(x)=\frac{-x^{3}+3 x^{2}+x-1}{x-3} $$

(a) Use the Intermediate-Value Theorem to show that the equation \(x=\cos x\) has at least one solution in the interval \([0, \pi / 2] .\) (b) Show graphically that there is exactly one solution in the interval. (c) Approximate the solution to three decimal places.
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