Problem 20 - APEX Calculus

Question

Determine if \(f\) is continuous at the indicated values. If not, explain why. \(f(x)=\left\\{\begin{array}{ll}x^{3}-x & x<1 \\ x-2 & x \geq 1\end{array}\right.\) (a) \(x=0\) (b) \(x=1\)

Step-by-Step Solution

Answer

(a) Continuous at \(x=0\). (b) Not continuous at \(x=1\).

Step 1: Understanding Continuity

A function is continuous at a point if three conditions are satisfied: 1) The function is defined at that point. 2) The limit of the function exists as it approaches that point. 3) The limit of the function equals the function's value at that point.

Step 2: Determine Continuity at x=0

For the function at \(x = 0\), since \(0 < 1\), use \(f(x) = x^3 - x\). Find \(f(0)\): \(f(0) = 0^3 - 0 = 0\). Next, determine \(\lim_{{x \to 0}} f(x)\): \(\lim_{{x \to 0}} (x^3 - x) = 0^3 - 0 = 0\). Since \(f(0)\) and \(\lim_{{x \to 0}} f(x)\) exist and are equal, \(f\) is continuous at \(x = 0\).

Step 3: Determine Continuity at x=1

For the function at \(x = 1\), find \(f(1)\): \(f(1) = 1 - 2 = -1\). Check the left-hand limit: \(\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} (x^3 - x) = 1^3 - 1 = 0\). Check the right-hand limit: \(\lim_{{x \to 1^+}} f(x) = \lim_{{x \to 1^+}} (x - 2) = 1 - 2 = -1\). Since \(\lim_{{x \to 1^-}} f(x) eq \lim_{{x \to 1^+}} f(x)\), limit as \( x\to1 \) does not exist. Thus, \(f\) is not continuous at \(x = 1\).

Key Concepts

Piecewise Functions

Piecewise functions are types of functions that have different expressions for different intervals of the input variable. They are defined in 'pieces', which means the function behaves differently in distinct parts of its domain. This can oftentimes be seen with functions that use different formulas based on whether the input value is less than, greater than, or equal to a certain number. For example, consider the given function:\[f(x) = \begin{cases} x^3 - x & \text{if } x < 1 \ x - 2 & \text{if } x \geq 1 \end{cases}\]This tells us that for values of \(x\) less than 1, the function is represented as \(x^3 - x\). For values greater than or equal to 1, it adopts the form \(x - 2\). Having multiple expressions allows for complex behavior, so investigating continuity at certain points can sometimes be more involved than with a simple one-expression function.

Limit of a Function

The limit of a function is a fundamental concept in calculus, describing the behavior of a function as the input approaches a certain value. It indicates what the output (or 'y' value) of the function is getting close to. Limits are essential when analyzing the continuity of a function at a point.For example, when examining a point such as \(x = 0\) or \(x = 1\) in our piecewise function, we calculate the left-hand and right-hand limits to see how the function behaves from both sides of the point. Limits can indicate whether a function is heading toward a particular value consistently from both directions. If they do so from both the left and the right direction consistently, then the standard limit is said to exist at that point, contributing to the function's continuity.

Continuity at a Point

A function is said to be continuous at a particular point if there is no interruption in its graph at that point. To determine this mathematically, we check three criteria:

• The function must be defined at the point.
• The limit of the function as it approaches the point must exist.
• The limit must be equal to the function's value at that point.

Take our function at \(x = 0\) as an example. We have:- The function is defined at \(x = 0\).- The limit as \(x\) approaches 0 for \(x < 1\) is \(0\).- Since both the function's value and the limit are \(0\), the function is continuous at this point.Similarly, analyzing these conditions at \(x = 1\) reveals that because the left-hand and right-hand limits do not match (0 and -1, respectively), the function is not continuous.

Function Definition at a Point

A function's definition at a specific point entails knowing exactly what value the function takes at that location. In terms of piecewise functions, it's crucial to identify which piece of the function applies to a given point.For example, examine the piecewise function at \(x = 1\):- At \(x = 1\), we use the second part of the piecewise function, \(x - 2\), thus providing the function value \(f(1) = 1 - 2 = -1\).The definition at a point is a straightforward but fundamental part of determining continuity. If a function is not defined at a point, it cannot be continuous. However, in our example, while \(f(1)\) is defined, the diverging limits from the left and right (0 and -1 respectively) contribute to the function not being continuous there.