Problem 28 - Calculus for Biology and Medicine

Question

(a) Show that $$ f(x)=\sqrt{x^{2}-4}, \quad|x| \geq 2 $$ is continuous from the right at \(x=2\) and continuous from the left at \(x=-2\). (b) Graph \(f(x)\). (c) Does it make sense to look at continuity from the left at \(x=2\) and at continuity from the right at \(x=-2 ?\)

Step-by-Step Solution

Answer

(a) The function is right continuous at \(x=2\) and left continuous at \(x=-2\). (b) The graph is 'V' shaped. (c) Checking continuity from opposite sides doesn't make sense as the function isn't defined in those regions.

Step 1: Define the Function and Examine the Domain

The function \( f(x) = \sqrt{x^2 - 4} \) is defined for \(|x| \geq 2\). This means that the domain includes all real numbers such that \(x \leq -2\) or \(x \geq 2\). We need to investigate its continuity at the boundaries of this domain.

Step 2: Check Right Continuity at \(x=2\)

Right continuity at \(x=2\) requires that \( \lim_{x \to 2^+} f(x) = f(2) \). Evaluate \(f(2)\):\[ f(2) = \sqrt{2^2 - 4} = \sqrt{0} = 0 \]Evaluate the right-hand limit:\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \sqrt{x^2 - 4} = \sqrt{4 - 4} = 0 \]Both are equal, ensuring right continuity at \(x=2\).

Step 3: Check Left Continuity at \(x=-2\)

Left continuity at \(x=-2\) requires that \( \lim_{x \to -2^-} f(x) = f(-2) \). Evaluate \(f(-2)\):\[ f(-2) = \sqrt{(-2)^2 - 4} = \sqrt{0} = 0 \]Evaluate the left-hand limit:\[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} \sqrt{x^2 - 4} = \sqrt{4 - 4} = 0 \]Both are equal, ensuring left continuity at \(x=-2\).

Step 4: Graph the Function

Graph the function \( f(x) = \sqrt{x^2 - 4} \) over its domain. The graph is a continuous curve for \(x \leq -2\) and \(x \geq 2\) with a minimum at both \(x = -2\) and \(x = 2\) where the function value is 0. Beyond these points, the graph forms a symmetric curve resembling a 'V' shape. It rises away from the x-axis as \(|x|\) increases beyond 2.

Step 5: Discuss Continuity from Opposite Sides

Continuity from the left at \(x = 2\) and from the right at \(x = -2\) has no relevance because the function is not defined for \(x<2\) and \(x>-2\). Therefore, considering continuity from these opposite sides isn't meaningful within the given domain.

Key Concepts

Limits

In calculus, limits are a fundamental concept used to define continuity. A limit helps us understand how a function behaves as it approaches a certain point.
For our function, \( f(x) = \sqrt{x^2 - 4} \), we examine the limits as \( x \) approaches the boundaries of the domain, specifically at \( x = 2 \) from the right and \( x = -2 \) from the left.

• Right-hand limit: This involves checking what happens to \( f(x) \) as \( x \) approaches 2 from values greater than 2. Mathematically, this is expressed as \( \lim_{x \to 2^+} f(x) \).
• Left-hand limit: Here, we analyze what happens to \( f(x) \) as \( x \) nears -2 from values less than -2, written as \( \lim_{x \to -2^-} f(x) \).

In both cases, if the function values (outputs) and the respective limits are equal, it describes continuity at those points in direction specified.

Graphing Functions

Graphing functions is a powerful visual tool that helps to understand behavior across their domains. For \( f(x) = \sqrt{x^2 - 4} \), the graph illustrates where and how the function is continuous.
The domain restriction \( |x| \geq 2 \) dictates that our graph only exists at \( x \leq -2 \) and \( x \geq 2 \). This gives the function an interesting shape:

• Symmetry: Looks like a "V" with its points touching the x-axis at both \( x = 2 \) and \( x = -2 \).
• Behavior: As \( |x| \) gets larger beyond 2, the function moves further away from the x-axis, indicating a rise in value.

Graphing helps identify these points and continuity, showing the limits at work visually by observing the curve's behavior as it reaches known values.

Domain of a Function

Understanding the domain of a function is critical as it defines the set of all permitted inputs. For the function \( f(x) = \sqrt{x^2 - 4} \), the domain is inherently limited due to the square root.

• Domain Definition: Any value inside a square root must be non-negative. Thus, \( x^2 - 4 \geq 0 \), simplifying to \( |x| \geq 2 \). This means \( x \leq -2 \) or \( x \geq 2 \).
• Boundaries in Domain: Makes sense for continuity checks at edges \( x = -2 \) and \( x = 2 \), but not between these values, as \( f(x) \) doesn't exist there.

Proper comprehension of the domain ensures we only explore continuity where the function legitimately exists, avoiding baseless evaluations.