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) The function is right-continuous at \( x=1 \). (b) Graph starts at \( (1, 0) \) and curves upwards for \( x \geq 1 \). (c) Left continuity doesn't apply as \( f(x) \) isn't defined for \( x < 1 \).

Step 1: Define Right-Hand Continuity

A function \( f(x) \) is said to be continuous from the right at a point \( x = c \) if \( \lim_{x \to c^+} f(x) = f(c) \). In this case, we need to evaluate the right-hand limit of \( f(x) = \sqrt{x-1} \) as \( x \to 1^+ \), and compare it to \( f(1) \).

Step 2: Evaluate \( f(1) \)

Substitute \( x = 1 \) into \( f(x) \) to find \( f(1) \). \[ f(1) = \sqrt{1 - 1} = \sqrt{0} = 0. \] So, \( f(1) = 0 \).

Step 3: Evaluate the Right-Hand Limit

Consider the limit as \( x \to 1^+ \). Compute \( \lim_{x \to 1^+} \sqrt{x - 1} \). As \( x \to 1^+ \), \( x - 1 \to 0^+ \). Therefore, \[ \lim_{x \to 1^+} \sqrt{x - 1} = \sqrt{0} = 0. \]

Step 4: Compare Limit with Function Value at \( x=1 \)

We have \( \lim_{x \to 1^+} \sqrt{x - 1} = 0 \) and \( f(1) = 0 \). Since these are equal, \( f(x) \) is continuous from the right at \( x = 1 \).

Step 5: Graph the Function

Graph \( f(x) = \sqrt{x - 1} \) for \( x \geq 1 \). This function starts at the point \( (1, 0) \) and increases, forming a curve similar to the right half of a sideways parabola opening to the right. This demonstrates how \( f(x) \) behaves only for \( x \geq 1 \).

Step 6: Evaluate Continuity from the Left

To consider left-hand continuity at \( x = 1 \), we evaluate \( \lim_{x \to 1^-} f(x) \). However, \( f(x) = \sqrt{x-1} \) is defined only for \( x \geq 1 \). There are no values for \( x < 1 \), so left-hand continuity at \( x=1 \) doesn't make sense.

Key Concepts

Left-Hand Continuity

Left-hand continuity looks at the behavior of a function as we approach a specific point from the left. For a function to be continuous from the left at a point, the limit from the left must equal the function's value at that point. In mathematical terms, a function \( f(x) \) is left continuous at \( x = c \) if \( \lim_{x \to c^-} f(x) = f(c) \). However, if the function is not defined for values less than \( c \), left-hand continuity cannot be evaluated. For example, with the function \( f(x) = \sqrt{x - 1} \) which is defined only for \( x \geq 1 \), there are no values for \( x < 1 \). Therefore, discussing left-hand continuity at \( x=1 \) doesn't make sense, because the function doesn't exist to the left of 1.

Continuous Functions

Continuous functions have no breaks, jumps, or holes at any points in their domain. For a function to be fully continuous at a point \( x = c \), the left-hand limit, right-hand limit, and the function's value at that point must all be equal. In formal terms:

• \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \) which means the left and right limits at \( c \) must exist and be equal.
• \( \lim_{x \to c} f(x) = f(c) \), ensuring the limit matches the function's value at \( c \).

For instance, in our example, \( f(x) = \sqrt{x-1} \) is continuous from the right at \( x=1 \), as the right-hand limit and function value are both zero, but it cannot be left continuous at this point due to the absence of \( x < 1 \) values.

Graphical Representation

Graphical representation helps visualize a function’s behavior and its continuity. By graphing, we can easily observe if a function forms an unbroken or smooth curve. For the function \( f(x) = \sqrt{x - 1} \):- It begins at point \((1, 0)\). - Since \( f(x) \) is only defined for \( x \geq 1 \), the graph starts at \( x=1 \) and rises slowly, curving gently away from the origin. This behavior is similar to the right half of a sideways-opening parabola.Visualizing this graph, we see that there is no discontinuity for \( x \geq 1 \). The continuity from the right is apparent as there is a smooth, continuous curve. However, the graph stops and does not exist for \( x < 1 \), illustrating the absence of left continuity for the function.