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Chapter 4: Problem 5

Determine whether the function is continuous or discontinuous on each of the indicated intervals. $$ f(t)=\frac{|t-1|}{t-1} ;(-\infty, 1),(-\infty, 1],[-1,1],(-1,+\infty),(1,+\infty) $$

Short Answer

Expert verified
Continuous on \((-\infty, 1)\) and \((1, +\infty)\). Discontinuous on \((-\infty, 1]\), \([-1, 1]\), and \((-1, +\infty)\).

Step by step solution

01

- Analyze Function Behavior

The function given is \( f(t) = \frac{|t-1|}{t-1} \). To understand its behavior, let's simplify it in terms of intervals depending on the value of \( t \) relative to 1.
02

- Consider Interval \( t < 1 \)

When \( t < 1 \), \( |t-1| = -(t-1) \). Thus, the function in this interval becomes \( f(t) = \frac{-(t-1)}{t-1} = -1 \). This shows that \( f(t) = -1 \) for all \( t < 1 \). This function is continuous in \( (-\infty, 1) \) since it is a constant function.
03

- Consider Interval \( t > 1 \)

When \( t > 1 \), \( |t-1| = t-1 \), so \( f(t) = \frac{t-1}{t-1} = 1 \). This implies \( f(t) = 1 \) for all \( t > 1 \). This function is again continuous in \( (1, \infty) \) since it is a constant.
04

- Consider \( t = 1 \)

At \( t = 1 \), the function \( f(t) \) is not defined as it leads to division by zero ( \( \frac{|t-1|}{t-1} = \frac{0}{0} \)). Therefore, the function is discontinuous at \( t = 1 \).
05

- Analyze Interval \( (-\infty, 1] \)

This interval includes \( t = 1 \) where the function is not defined. Hence, \( f(t) \) is discontinuous in \( (-\infty, 1] \).
06

- Analyze Interval \( [-1, 1] \)

Similarly, this interval includes \( t = 1 \) where the function is not defined. Thus, \( f(t) \) is discontinuous in \( [-1, 1] \).
07

- Analyze Interval \( (-1, +\infty) \)

As \( t > 1 \) is included and there is no discontinuity for any value less than \( t = 1 \), the function is discontinuous in \( (-1, +\infty) \).
08

- Analyze Interval \( (1, +\infty) \)

Since the interval starts right after \( t = 1 \), and all points in \( (1, \infty) \) have a continuous value of 1, the function is continuous in \( (1, +\infty) \).

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Key Concepts

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

discontinuous functions
A function is considered discontinuous at a point if there is a sudden jump or there exists a value where the function is not defined. For our function, \( f(t) = \frac{|t-1|}{t-1} \), we observe that it behaves differently at \( t = 1 \). This point creates a disruption in the function because \( f(1) \) leads to \( \frac{0}{0} \), which is undefined. Therefore, \( f(t) \) is discontinuous at \( t = 1 \). This discontinuity might affect various intervals that include this point.
interval analysis
When examining the continuity of a function over an interval, we analyze the behavior and definition of the function across that range. For instance, evaluating \( f(t) = \frac{|t-1|}{t-1} \) over different intervals:
  • For \( t < 1 \), the function simplifies to \( f(t) = -1 \), which is constant and thus continuous in \( (-\infty, 1) \).
  • For \( t > 1 \), the function changes to \( f(t) = 1 \), again constant and continuous in \( (1, \infty) \).
  • When the interval includes \( t = 1 \) such as \( (-\infty, 1] \) or \( [-1, 1] \), the function becomes discontinuous due to the undefined nature at \( t = 1 \).
  • Intervals that do not include 1 but cross it, such as \( (-1, +\infty) \), also consider the discontinuity at 1 making the overall analysis critical to understanding continuity or discontinuity in specified ranges.
function behavior
Understanding a function's behavior helps in determining its continuity or discontinuity. Our function, \( f(t) = \frac{|t-1|}{t-1} \), shows different values based on the domain of \( t \). This change portrays how absolute value and division impact the function's output:
  • For \( t < 1 \), the function consistently outputs -1, showing constant behavior.
  • For \( t > 1 \), the function shifts to a constant output of 1.
  • At \( t = 1 \), both conditions merge causing an undefined value (\( \frac{0}{0} \)), resulting in a discontinuous behavior.
By breaking the function into respective intervals and understanding its behavior, we identify continuous segments (constant outputs) from discontinuous points (undefined values).

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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. $$ f(x)=\frac{x+1}{x-1} ; g(x)=\sqrt{x} $$

(a) Draw a sketch of the graph of the given function on the indicated interval; (b) test the three conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem and determine which conditions are satisfied and which, if any, are not satisfied; and (c) if the three conditions in part (b) are satisfied, determine a point at which there is a horizontal tangent line. $$ f(x)=\left|9-4 x^{2}\right| ;\left[-\frac{3}{2}, \frac{3}{2}\right] $$

Find the limits, and when applicable indicate the limit theorems being used. $$ \lim _{x \rightarrow+\infty} \frac{3 x^{4}-7 x^{2}+2}{2 x^{4}+1} $$

Find the absolute extrema of the given function on the given interval, if there are any, and find the values of \(x\) at which the absolute extrema occur. Draw a sketch of the graph of the function on the interval. $$ f(x)=\frac{3 x}{9-x^{2}} ;(-3,2) $$

(a) Draw a sketch of the graph of the given function on the indicated interval; (b) test the three conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem and determine which conditions are satisfied and which, if any, are not satisfied; and (c) if the three conditions in part (b) are satisfied, determine a point at which there is a horizontal tangent line. $$ f(x)=\frac{2 x^{2}-5 x-3}{x-1} ;\left[-\frac{1}{2}, 3\right] $$
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