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Chapter 3: Problem 62

Find, without graphing, where each of the given functions is continuous. $$ f(x)=\left\\{\begin{array}{ll} x & \text { if } x<1 \\ 2 & \text { if } x=1 \\ x-1 & \text { if } x>1 \end{array}\right. $$

Short Answer

Expert verified
The function is continuous on \\((-\infty, 1) \, \cup \, (1, \infty)\\).

Step by step solution

01

Understand the Meaning of Continuity at a Point

A function is continuous at a point if the function is defined at that point, the limit of the function exists as it approaches the point, and the limit is equal to the function value at that point. Mathematically, \(\lim_{x \to c} f(x) = f(c)\) where \(c\) is the specific value we are checking continuity for.
02

Identify the Different Intervals in the Function

The function \(f(x)\) is defined in three different pieces: For \(x < 1\), \(f(x) = x\); for \(x = 1\), \(f(x) = 2\); and for \(x > 1\), \(f(x) = x - 1\). We will check the continuity at \(x = 1\) where the function's formula changes.
03

Check Continuity for \\(x < 1\\) and \\(x > 1\\)

In the intervals where \(x < 1\) and \(x > 1\), the function is composed of simple linear expressions \(x\) and \(x - 1\) respectively. Linear functions are continuous everywhere in their domain, so \(f(x)\) is continuous on the intervals \((-\infty, 1)\) and \( (1, \infty)\).
04

Evaluate Continuity at \\(x = 1\\)

1. First, find \(f(1) = 2\).2. Next, calculate the limit from the left: \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1} x = 1\).3. Then, calculate the limit from the right: \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1} (x - 1) = 0\).Since \(\lim_{x \to 1^-} f(x) eq \lim_{x \to 1^+} f(x)\), the limit does not exist at \(x = 1\) and the function is not continuous at this point.
05

Conclude the Continuity of the Function

Since the function is continuous for all intervals except at \(x = 1\), \(f(x)\) is continuous on \((-\infty, 1) \, \cup \, (1, \infty)\).

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Key Concepts

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

Mathematical Analysis of Piecewise Functions
Mathematical analysis provides us a framework to study different aspects of functions, such as limits, continuity, and differentiability. When we deal with piecewise functions, we must carefully examine each segment to determine its properties.
For the function \[f(x) = \begin{cases} x & \text{if } x < 1 \ 2 & \text{if } x = 1 \ x-1 & \text{if } x > 1 \end{cases}\] we analyze each piece:
  • The segment where \(x < 1\) is governed by \(f(x) = x\). Since this is a simple linear function, it is continuous in the interval \((-\infty, 1)\).
  • The segment where \(x = 1\) is a constant value \(f(x) = 2\). Here, we must ensure this matches the behavior from either side to check continuity.
  • Finally, for \(x > 1\), the function is given by \(f(x) = x - 1\), another linear function, continuous in \((1, \infty)\).
Putting it all together, we need to especially verify the point of transition \(x = 1\), where the rule for the function changes.
Understanding Limits in Context
Limits are fundamental in determining the behavior of functions as they approach a particular point. For continuity, it's not just about whether the function exists at a given point, but also whether it behaves nicely as it gets close to that point from both sides.
To check limit behavior in continuity problems, you should:
  • Calculate the limit from the left: \(\lim_{x \to c^-} f(x)\).
  • Calculate the limit from the right: \(\lim_{x \to c^+} f(x)\).
  • Compare both limits and ensure they are equal for the function to have a limit \(\lim_{x \to c} f(x)\).
For our function at \(x=1\):
- The limit from the left \(\lim_{x \to 1^-} f(x) = 1\) because as \(x\) approaches 1 from the left, \(f(x) = x\).
- The limit from the right \(\lim_{x \to 1^+} f(x) = 0\) since \(f(x) = x - 1\) for \(x>1\).
These limits do not match, indicating a discontinuity at \(x = 1\).
Discontinuous Functions and Their Characteristics
Discontinuous functions are those where there are one or more points where the function does not behave in a predictable manner. This can happen for various reasons, most typically where the limit from the left and right of a particular point do not agree.
The function we studied has a discontinuity at \(x = 1\). Let's summarize why discontinuity occurs:
  • The value of the function \(f(x)\) at \(x = 1\) is defined as 2. Therefore, \(f(1) = 2\).
  • We found differing limits as \(x\) approaches 1. The left limit is 1 and the right limit is 0.
  • For a function to be continuous at a point, these must be equal and also equal to \(f(c)\). Here, neither condition is met.
So, despite being defined everywhere from \((-\infty, \infty)\), the point \(x = 1\) is a break in the function's continuity. The segments before and after this point are independently continuous, giving us \((-\infty, 1) \cup (1, \infty)\) as the intervals of continuity.

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Most popular questions from this chapter

Find, without graphing, where each of the given functions is continuous. $$ 5 x^{7}-3 x^{2}+4 $$

Barton and colleagues \(^{61}\) studied how the size of agricultural cooperatives were related to their variable cost ratio (variable cost over assets as a percent). The data is given in the following table. $$ \begin{array}{|l|lllllll|} \hline x & 2.48 & 3.51 & 2.50 & 3.37 & 3.77 & 6.33 & 4.74 \\ \hline y & 11.7 & 16.8 & 25.4 & 16.9 & 17.9 & 22.8 & 16.7 \\ \hline \end{array} $$ Here \(x\) is the assets in millions of dollars, and \(y\) is the variable cost ratio. a. Use quadratic regression to find the best-fitting quadratic function using least squares. b. Graph on your computer or graphing calculator using a screen with dimensions [0,9.4] by \([0,30] .\) Have your grapher find the derivative when \(x\) is \(1.5,3,4.5,6,\) and 6.6. Note the slope, and relate this to the rate of change. On the basis of this model, describe what happens to the rate of change of the variable cost ratio as assets increase.

Use the rules of limits to find the indicated limits if they exist. Support your answer using a computer or graphing calculator. $$ \lim _{x \rightarrow 2} \frac{x+3}{x-3}\left(2 x^{2}-1\right) $$

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Suppose \(y=f(x)\) is defined on \((-1,1) .\) Of what importance is knowledge of \(f(0)\) to finding \(\lim _{x \rightarrow 0} f(x)\) ? Explain.
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