Problem 27 - Calculus for Biology and Medicine

Question

(a) Show that $$ f(x)=\sqrt{x-1}, \quad x \geq 1 $$ is continuous from the right at \(x=1\). (b) Graph \(f(x)\). (c) Does it make sense to look at continuity from the left at \(x=1 ?\)

Step-by-Step Solution

Answer

(a) \( f(x) = \sqrt{x-1} \) is continuous from the right at \( x = 1 \). (b) Graph starts at \( (1,0) \) and increases. (c) Left continuity does not apply, as \( f(x) \) is undefined for \( x < 1 \).

Step 1: Define Right-Hand Continuity

To show that a function \( f(x) \) is continuous from the right at \( x = a \), we need to prove that \( \lim_{x \to a^+} f(x) = f(a) \). For the function \( f(x) = \sqrt{x-1} \), this means showing that \( \lim_{x \to 1^+} \sqrt{x-1} = \sqrt{1-1} = 0 \).

Step 2: Evaluate the Right-Hand Limit

Consider \( \lim_{x \to 1^+} \sqrt{x-1} \). For values \( x > 1 \), the expression \( \sqrt{x-1} \) is non-negative and gets closer to zero as \( x \) approaches 1 from the right.Calculate:\[\lim_{x \to 1^+} \sqrt{x-1} = \sqrt{1-1} = 0.\]

Step 3: Verify Continuity from the Right

We established that \( \lim_{x \to 1^+} \sqrt{x-1} = 0 \) and that \( f(1) = \sqrt{1-1} = 0 \). Since the right-hand limit equals \( f(1) \), \( f(x) \) is continuous from the right at \( x = 1 \).

Step 4: Graph the Function

To graph \( f(x) = \sqrt{x-1} \) for \( x \geq 1 \), note that the graph starts at the point \( (1, 0) \) and increases as \( x \) increases. The graph appears as the right half of a horizontally shifted square root graph.

Step 5: Consider Left-Hand Continuity

Since \( f(x) = \sqrt{x-1} \) is defined only for \( x \geq 1 \), it is not meaningful to consider the limit from the left at \( x = 1 \) because \( f(x) \) is not defined for \( x < 1 \).

Key Concepts

Right-Hand Continuity

Right-hand continuity is an important concept in calculus. It refers to a function being continuous when approached from the right side of a point. For a function to be right-hand continuous at a point \(x = a\), the right-hand limit of the function as \(x\) approaches \(a\) must equal the function's value at \(a\). Mathematically, this is expressed as \( \lim_{x \to a^+} f(x) = f(a) \).
Let's consider the function \( f(x) = \sqrt{x-1} \) for \( x \geq 1 \). To check if \( f(x) \) is continuous from the right at \( x = 1 \), we find the limit from the right:

• Calculate \( \lim_{x \to 1^+} \sqrt{x-1} \).
• As \( x \) approaches 1 from the right, \( \sqrt{x-1} \) approaches \( \sqrt{1-1} = 0 \).

Since \( f(1) = 0 \) and the limit as \( x \to 1^+ \) is also \(0\), \( f(x) \) is continuous from the right at \( x = 1 \).

Left-Hand Continuity

Left-hand continuity deals with a function's behavior as it approaches a point from the left. For a function \( f(x) \) to be continuous from the left at \( x = a \), we need \( \lim_{x \to a^-} f(x) = f(a) \).
However, in the context of the function \( f(x) = \sqrt{x-1} \), this concept doesn't apply for \( x = 1 \). The function is only defined for \( x \geq 1 \), meaning there is no domain to consider on the left of \( x = 1 \).
No values exist for \( x < 1 \) in the domain of \( f(x) \), rendering the left-hand limit and thus left-hand continuity meaningless at this point.
In simpler terms, since the function doesn't "exist" to the left of 1, checking for left-hand continuity is not feasible.

Limit Evaluation

Evaluating limits is crucial to understanding continuity. The limit of a function at a point tells us what value the function is approaching as \( x \) gets infinitely close to that point.
In the case of right-hand continuity for \( f(x) = \sqrt{x-1} \) at \( x = 1 \), we need to evaluate the limit \( \lim_{x \to 1^+} \sqrt{x-1} \):

• We examine \( \sqrt{x-1} \) for \( x > 1 \), noticing it gets closer to 0 as \( x \) approaches 1.
• So, \( \lim_{x \to 1^+} \sqrt{x-1} = 0 \).

This sets a strong foundation, demonstrating the function's behavior very close to the boundary of its domain, ensuring the right-hand continuity at that point.

Graph of a Function

Visualizing the graph of a function is a powerful way to grasp its behavior. Graphing \( f(x) = \sqrt{x-1} \) when \( x \geq 1 \) helps us see right away how the function behaves.

• The graph starts at the point \( (1, 0) \) because \( f(1) = 0 \).
• For \( x > 1 \), the function generates the right half of a horizontally shifted square root curve.

The square root function has a distinct shape, appearing to increase steadily as \( x \) increases.
By looking at the graph, we visually confirm the continuous rise starting from \( x = 1 \), supporting our understanding of right-hand continuity. This visual tool becomes invaluable for comprehending the more abstract mathematical concepts.