Problem 11 - Calculus for Biology and Medicine

Question

In Problems 9-12, determine at which points \(f(x)\) is discontinuous. $$ f(x)=\left\\{\begin{array}{cl} x^{2}-1 & \text { if }|x| \geq 1 \\ x-1 & \text { if }|x|<1 \end{array}\right. $$

Step-by-Step Solution

Answer

The function is discontinuous at \(x = -1\).

Step 1: Understand the pieces of the function

The function is piecewise and changes its definition based on the value of \(x\). Specifically, for \(|x| \geq 1\), it is defined as \(f(x) = x^2 - 1\), and for \(|x| < 1\), it is \(f(x) = x - 1\).

Step 2: Identify the points where the definition of the function changes

The transition in the function's pieces occurs at the boundaries of \(|x| < 1\), specifically at \(x = -1\) and \(x = 1\). Thus, these are the points that require checking for discontinuity.

Step 3: Check continuity at \(x = -1\)

For \(x = -1\), the function definition is \(f(x) = x^2 - 1\) when \(x = -1\). Calculating the value gives \(f(-1) = (-1)^2 - 1 = 0\).To check for continuity, find the limit from both sides. The left side (\(x < -1\)) is evaluated using \(f(x) = x^2 - 1\):\[\lim_{x \to -1^-} f(x) = (-1)^2 - 1 = 0\].The right side (\(x > -1\)) is evaluated using \(f(x) = x-1\):\[\lim_{x \to -1^+} f(x) = -1 - 1 = -2\].Since the left and right limits are not equal, \(f(x)\) is discontinuous at \(x = -1\).

Step 4: Check continuity at \(x = 1\)

For \(x = 1\), the function definition is \(f(x) = x^2 - 1\) at \(x = 1\). Calculating the value gives \(f(1) = 1^2 - 1 = 0\).To check for continuity, find the limit from both sides. The left side (\(x < 1\)) is evaluated using \(f(x) = x-1\):\[\lim_{x \to 1^-} f(x) = 1 - 1 = 0\].The right side (\(x > 1\)) is evaluated using \(f(x) = x^2 - 1\):\[\lim_{x \to 1^+} f(x) = 1^2 - 1 = 0\].Since both the left and right limits and the actual function value are equal, \(f(x)\) is continuous at \(x = 1\).

Key Concepts

Limit Analysis

Limit analysis helps us understand how a function behaves as the input approaches a certain point. In piecewise functions, calculating limits is crucial when we suspect discontinuities at boundary points where the function's formula changes.

For the function given in our problem, at points where \(|x|=1\), the definition changes, making it important to check the limits from both sides:

• Left-hand limit: Calculated by approaching the point from values less than it.
• Right-hand limit: Calculated by approaching from values greater than the concerned point.

If the left-hand limit and right-hand limit are equal, the limit at that point exists. For instance, at \(x=1\), the limits approach zero on both sides, showing consistency. However, at \(x=-1\), differing limits suggest a break or gap in the function's graph, indicating discontinuity.

Continuity at a Point

Continuity at a point means that the function doesn't have any gaps, jumps, or interruptions at that specific input. For a function to be continuous at a point \(c\), three conditions must hold:

• The function \(f(x)\) must be defined at \(x=c\).
• The limit of \(f(x)\) must exist as \(x\) approaches \(c\).
• The limit of \(f(x)\) as \(x\) approaches \(c\) must equal \(f(c)\).

In the exercise, we check continuity at \(x=-1\) and \(x=1\).At \(x=1\), all conditions are satisfied, so it is continuous there. But at \(x=-1\), the second condition fails, as the left-hand and right-hand limits are not equal, revealing a discontinuity. Understanding these conditions allows us to spot where a piecewise function breaks or continues smoothly.

Piecewise Functions

Piecewise functions are defined by different expressions depending on the value of \(x\). They often look strange because they switch rules abruptly at certain points.

For example, our function has two distinct parts:

• \(|x| \geq 1\): Uses the expression \(x^2 - 1\).
• \(|x| < 1\): Uses the expression \(x - 1\).

These expressions change based on intervals of \(x\), and this can introduce discontinuities at certain points, making it necessary to handle them cautiously. The boundaries between these expressions, such as \(x=-1\) and \(x=1\) in our problem, require special attention to ensure that the piecewise function is well-behaved. This kind of function challenges students to think about behavior at edges, making them an excellent case for studying continuity and limits.