Question

In Exercises \(11-18,\) use the function \(f\) defined and graphed below to answer the questions. $$f(x)=\left\\{\begin{array}{ll}{x^{2}-1,} & {-1 \leq x<0} \\\ {2 x,} & {0 < x < 1} \\ {1,} & {x=1} \\ {-2 x+4,} & {1 < x < 2} \\ {0,} & {2 < x < 3}\end{array}\right.$$ (a) Does \(f(-1)\) exist? (b) Does \(\lim _{x \rightarrow-1^{+}} f(x)\) exist? (c) Does \(\lim _{x \rightarrow-1^{+}} f(x)=f(-1) ?\) (d) Is \(f\) continuous at \(x=-1 ?\)

Short answer

(a) Yes, \(f(-1) = 0\), (b) No, \(\lim _{x \rightarrow-1^{+}} f(x)\) does not exist, (c) No, \(\lim _{x \rightarrow-1^{+}} f(x)\) does not equal \(f(-1)\), and (d) No, \(f(x)\) is not continuous at \(x=-1\).

Step-by-step solution

Evaluation of \(f(-1)\)

To determine if \(f(-1)\) exists, simply look at the interval that contains -1 in the definition of the function. Here, it is \(-1 \leq x<0\), which corresponds to the segment \(x^{2}-1\). Therefore, to find \(f(-1)\), substitute -1 into this segment: \(f(-1) = (-1)^{2} - 1 = 0\). Therefore, \(f(-1)\) exists and equals 0.

Evaluation of \(\lim _{x \rightarrow-1^{+}} f(x)\)

The limit as \(x\) approaches -1 from the positive direction would be found by evaluating the function segment that meets at \(x=-1\) from the right side. In this case, that segment does not exist because there is no function defined for \(x > -1\) and \(x < 0\). So, \(\lim _{x \rightarrow-1^{+}} f(x)\) does not exist.

Checking if \(\lim _{x \rightarrow-1^{+}} f(x)=f(-1)\)

As from the previous steps, \(\lim _{x \rightarrow-1^{+}} f(x)\) does not exist while \(f(-1)\) equals 0. Therefore, \(\lim _{x \rightarrow-1^{+}} f(x)\) cannot equal \(f(-1)\) because one exists and the other does not.

Determining continuity at \(x=-1\)

A function \(f(x)\) is continuous at \(x=a\) if \(\lim _{x \rightarrow a^{-}} f(x), \lim _{x \rightarrow a^{+}} f(x), and f(a)\) all exist and are equal. From the previous steps, it was found that \(\lim _{x \rightarrow-1^{+}} f(x)\) does not exist, while \(f(-1)\) equals 0. Therefore, \(f(x)\) is not continuous at \(x=-1\).

Key concepts

Limits of Functions

Understanding the concept of limits is crucial when studying calculus, as it paves the way for defining both continuity and the derivative. A limit describes the behavior of a function as the input approaches a particular value. Formally, we say that the limit of a function f(x) as x approaches a is L, written as \( \lim_{x \rightarrow a} f(x) = L \), if f(x) can be made arbitrarily close to L by taking x sufficiently close to a.

In the context of piecewise functions, which are functions defined by different expressions over different intervals, the concept of the limit must be applied to each interval individually. If considering a limit as x approaches a point from the left (denoted by a-) or from the right (denoted by a+), these are termed one-sided limits. For a function to have a limit at a point, both the right-hand limit (as x approaches from the right) and the left-hand limit (as x approaches from the left) must exist and be equal.

For example, if we're trying to determine \( \lim_{x \rightarrow c} f(x) \) for some c where the function is defined piecewise, we need to look at the expressions that apply just before and just after c. In the exercise provided, when evaluating \( \lim_{x \rightarrow -1^{+}} f(x) \), we must consider the expression used for the interval immediately to the right of -1, which does not exist; therefore, the limit itself does not exist.

Continuity of Functions

Continuity is a fundamental property of functions in calculus, referring to whether a function is uninterrupted—without gaps, jumps, or abrupt changes—in its domain. A function f(x) is continuous at a point x=a if three conditions are met: firstly, \( \lim_{x \rightarrow a^{-}} f(x) \) exists, secondly, \( \lim_{x \rightarrow a^{+}} f(x) \) exists, and thirdly, both of these limits are equal to the actual value of the function at a, denoted f(a). This can be summarized as \( \lim_{x \rightarrow a} f(x) = f(a) \).

If any of these conditions are not met, the function f(x) is not considered continuous at x=a. In the exercise at hand, since the limit as x approaches -1 from the right does not exist, one of the three key conditions for continuity is violated. Thus, the function is not continuous at x=-1 even though f(-1) itself exists. This example nicely illustrates the difference between the existence of a function value at a point and the continuity of the function at that point.

Piecewise Function Evaluation

Piecewise functions are defined by different expressions over various intervals of their domain. Evaluating a piecewise function requires determining which piece of the function applies for the given input value. Once the correct piece is identified, you proceed to compute the function value using the corresponding formula.

Evaluating a piecewise function at a specific point also includes checking for the function's behavior around that point. When working with limits and continuity, as in the given exercise, one must pay attention not only to the value at the point but also to the function's behavior as it approaches that point from either side.

Evaluation Tips

Here are a few tips for the evaluation of piecewise functions:

• Identify the interval in which the input value falls.
• Substitute the input value into the correct formula.
• For limits, examine the behavior of the function as it approaches the value from both directions.
• Remember, the function's value and its limit as you approach that value can differ, affecting continuity.

Following these steps, in our exercise, the value of f(-1) was found by using the first formula specified for the interval -1 ≤ x < 0. However, evaluating the limit as x approaches -1 from the right requires us to examine the next interval, which does not exist in this case, leading to the conclusion that the limit as x approaches -1 from the right does not exist.