Problem 38 - Calculus Early Transcendentals

Question

\(37-39\) Find the numbers at which \(f\) is discontinuous. At which of these numbers is \(f\) continuous from the right, from the left, or neither? Sketch the graph of \(f\) . \(f(x)=\left\\{\begin{array}{ll}{x+1} & {\text { if } x \leqslant 1} \\ {1 / x} & {\text { if } 1 < x < 3} \\ {\sqrt{x-3}} & {\text { if } x \geqslant 3}\end{array}\right.\)

Step-by-Step Solution

Answer

The function is discontinuous at \(x = 1\) and \(x = 3\), continuous from neither direction at both points.

Step 1: Identify Intervals and Functions

We begin by identifying the function definitions over different intervals. The function is piecewise defined as follows: for \(x \leq 1\), \(f(x) = x + 1\); for \(1 < x < 3\), \(f(x) = \frac{1}{x}\); and for \(x \geq 3\), \(f(x) = \sqrt{x - 3}\).

Step 2: Test for Discontinuities at Interval Bounds

We need to check the interval points \(x = 1\) and \(x = 3\) where the function definition changes. Discontinuities can occur at these points.

Step 3: Evaluate Limit from the Left at \(x = 1\)

As \(x\) approaches 1 from the left (\(x \to 1^-\)), \(f(x) = x + 1\). Hence, \(\lim_{x \to 1^-} f(x) = 1 + 1 = 2\).

Step 4: Evaluate Limit from the Right at \(x = 1\)

As \(x\) approaches 1 from the right (\(x \to 1^+\)), \(f(x) = \frac{1}{x}\). Therefore, \(\lim_{x \to 1^+} f(x) = \frac{1}{1} = 1\).

Step 5: Continuity at \(x = 1\)

Since the left-hand limit (2) and the right-hand limit (1) are not equal at \(x = 1\), the function \(f\) is discontinuous at \(x = 1\). It is continuous from neither direction.

Step 6: Evaluate Limit from the Left at \(x = 3\)

As \(x\) approaches 3 from the left (\(x \to 3^-\)), \(f(x) = \frac{1}{x}\). Hence, \(\lim_{x \to 3^-} f(x) = \frac{1}{3}\).

Step 7: Evaluate Limit from the Right at \(x = 3\)

As \(x\) approaches 3 from the right (\(x \to 3^+\)), \(f(x) = \sqrt{x - 3}\). Therefore, \(\lim_{x \to 3^+} f(x) = \sqrt{3-3} = 0\).

Step 8: Continuity at \(x = 3\)

The left-hand limit (\(\frac{1}{3}\)) and the right-hand limit (0) at \(x = 3\) differ, so \(f\) is discontinuous at \(x = 3\). It is continuous from neither direction.

Step 9: Sketch the Function

To sketch \(f\), plot the graph for each piece: a line \(y = x + 1\) for \(x \leq 1\), a hyperbola \(y = \frac{1}{x}\) for \(1 < x < 3\), and a square root curve \(y = \sqrt{x - 3}\) for \(x \geq 3\).

Key Concepts

Understanding Discontinuity

Discontinuity is a key aspect of piecewise functions like the one we're examining here. When a function jumps, breaks, or has a hole at certain points, it is said to be discontinuous. In the provided exercise, we check the points where the function's definition changes: specifically at \(x = 1\) and \(x = 3\). At these points, we investigate if the limit from the left and the right are the same. If they are not, there is a discontinuity. For our function, we determined that it is discontinuous at both \(x = 1\) and \(x = 3\) because the left and right limits at these points do not match. This means that the function "jumps" or "breaks" at these points, indicating discontinuity."

The Role of Limits in Discontinuity

Limits help us understand the behavior of a function as we approach a particular point from either side. In identifying discontinuity, calculating the limit from the left and right is crucial. In the case of \(x = 1\), we found that approaching from the left gives a limit of \(2\) since \(f(x) = x + 1\), while approaching from the right gives \(1\) because \(f(x) = \frac{1}{x}\). Since these are not equal, the function is discontinuous at \(x = 1\). Similarly, at \(x = 3\), the left limit yields \(\frac{1}{3}\) and the right limit gives \(0\), again showing a mismatch. Calculating these limits thus definitively reveals where the function does not smoothly continue, highlighting the discontinuities.

Graphical Representation of Piecewise Functions

The graphical representation of a piecewise function provides a visual insight into its behavior across different intervals. For this exercise, we sketch three distinct parts for the function:

• A linear section \(y = x + 1\) for \(x \leq 1\), showing a straight line.
• A hyperbolic curve \(y = \frac{1}{x}\) for \(1 < x < 3\), which dips sharply near the axes.
• A square root curve \(y = \sqrt{x-3}\) starting from \(x = 3\) and moving upwards.

By plotting these, we can visually identify breaks or jumps at the transition points of \(x = 1\) and \(x = 3\). These are the discontinuities where the sections do not connect, aiding in the understanding of where the function is not continuous.

Calculus and Continuity

Calculus is invaluable for analyzing piecewise functions, especially in identifying and understanding the implications of discontinuities. Concepts like limits and continuity are fundamental in this process. A function is continuous when it smoothly bridges across intervals without any jumps or holes. Using calculus, we calculate limits to check for continuity at transition points. When these limits differ, as they do at \(x = 1\) and \(x = 3\) in our problem, the function does not meet the criteria for continuity at those points. By applying calculus concepts, students can not only detect discontinuities but also understand the behavior of piecewise functions in depth, enhancing their appreciation for how calculus describes changes and transitions.