Problem 19 - Calculus Early Transcendentals

Question

\(15-20\) Explain why the function is discontinuous at the given number a. Sketch the graph of the function. \(f(x)=\left\\{\begin{array}{ll}{\cos x} & {\text { if } x<0} \\ {0} & {\text { if } x=0} \\ {1-x^{2}} & {\text { if } x>0}\end{array}\right. \quad a=0\)

Step-by-Step Solution

Answer

The function is discontinuous at \( x = 0 \) because \( f(0) \neq \lim_{{x \to 0^+}} f(x) = \lim_{{x \to 0^-}} f(x) = 1 \).

Step 1: Clarify the Problem

We need to determine if the function \( f(x) \) is discontinuous at \( a = 0 \), using \( f(x) = \begin{cases}\cos x, & \text{if } x < 0 \0, & \text{if } x = 0 \1 - x^2, & \text{if } x > 0 \\end{cases} \). To prove discontinuity, we should check whether the left-hand limit, right-hand limit, and the function value at \( x = 0 \) are equal.

Step 2: Evaluate the Left-Hand Limit

The left-hand limit as \( x \to 0^- \) is the limit of \( \cos(x) \) as \( x \) approaches 0 from the left. The limit is the value of \( \cos(0) = 1 \). Thus, \( \lim_{{x \to 0^-}} f(x) = 1 \).

Step 3: Evaluate the Right-Hand Limit

For the right-hand limit as \( x \to 0^+ \), we evaluate \( 1 - x^2 \) as \( x \to 0^+ \). This limit is \( 1 - 0^2 = 1 \). Thus, \( \lim_{{x \to 0^+}} f(x) = 1 \).

Step 4: Check the Function Value at 0

The function value \( f(0) \) is given as 0. Thus, \( f(0) = 0 \).

Step 5: Verify Discontinuity

For \( f(x) \) to be continuous at \( x = 0 \), both the left-hand limit, right-hand limit, and \( f(0) \) must be equal.However, while both limits (left and right) at 0 are 1, \( f(0) \) itself is 0. Hence, the limits do not equal \( f(0) \), indicating a discontinuity at \( x = 0 \).

Step 6: Sketch the Graph

To sketch the function:1. For \( x < 0 \), plot \( f(x) = \cos x \).2. At \( x = 0 \), plot the point as (0,0) as \( f(0) = 0 \).3. For \( x > 0 \), sketch the parabola \( f(x) = 1 - x^2 \) opening downwards starting from \( (0, 1) \). The graph shows discontinuity at \( x = 0 \) where the two sections do not connect fluidly.

Key Concepts

Limits

Limits are a fundamental concept in calculus, particularly when discussing discontinuities at specific points in a function. A limit describes what a function approaches as its input (or "x" value) gets arbitrarily close to a certain point, say "a". It can be perceived as the behavior of a function at the brink of entering a boundary.

The left-hand limit and right-hand limit refer to the values that a function approaches from the left side and the right side of point "a", respectively. In our exercise, we evaluated the left-hand limit of the function as it approaches zero from the negative side, noted as \( \, \lim_{{x \to 0^-}} f(x) = 1 \, \). Similarly, the right-hand limit when approaching from the positive side was calculated to be \( \, \lim_{{x \to 0^+}} f(x) = 1 \, \).

These two limits help identify the expected behavior of the function at the point of interest. If the left and right limits are not equal, or if either of these does not match the function's value at that point, this is a clue to a potential discontinuity.

Continuity

Continuity in functions occurs when there are no breaks, jumps, or holes at any point on the graph of the function. More formally, a function is continuous at a point "a" if the limit of the function as it approaches "a" from both sides is equal to its value at "a". This implies three conditions:

• The function must be defined at the point \( a \).
• The left-hand limit as \( x \) approaches \( a \) and the right-hand limit as \( x \) approaches \( a \) must exist.
• The function's value at the point must be equal to both these limits.

In our exercise, although both the left-hand and right-hand limits of the function at \( x = 0 \) are 1, the value of the function \( f(x) \) at \( x = 0 \) is given as 0.

Because the actual function value does not match the limits, the function is not continuous at this point, showing a type of discontinuity known as a jump discontinuity.

Piecewise Functions

Piecewise functions are defined using multiple sub-functions, each of which applies to a certain interval within the function's domain. These functions switch rules at specific points, often leading to characteristics like abrupt jumps or edges in their graphs.

In the given exercise, the function \( f(x) \) is defined in three parts:

• For \( x < 0 \), \( f(x) = \cos{x} \).
• For \( x = 0 \), \( f(x) = 0 \).
• For \( x > 0 \), \( f(x) = 1 - x^2 \).

Because of these different rules within distinct sections of \( x \), the function may change dramatically at each boundary, potentially causing discontinuities. This is evident at \( x = 0 \), where the rule switches and leads to a different function behavior.

Understanding how each segment of a piecewise function behaves in its section is crucial to determining how the overall function might behave as a whole. Graphing each piece with careful attention to its boundaries allows for a visual confirmation of where discontinuity occurs and how the graphs of different parts fit together.