Problem 58 - Calculus and its applications

Question

Is the function given by $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{3} x+4, & \text { for } x<3 \\ 2 x-1, & \text { for } x \geq 3 \end{array}\right. $$ continuous at \(x=3\) ? Why or why not?

Step-by-Step Solution

Answer

The function is continuous at \(x = 3\) because both limits and \(f(3)\) are equal.

Step 1: Understand the Definition of Continuity

A function is continuous at a point \(x = c\) if it meets three conditions: 1) \(f(c)\) is defined, 2) \(\lim_{x \to c} f(x)\) exists, and 3) \(\lim_{x \to c} f(x) = f(c)\). We will check these conditions for the function at \(x = 3\).

Step 2: Check if f(3) is Defined

For \(x \geq 3\), the function is defined as \(f(x) = 2x - 1\). Replacing \(x\) with 3, we find \(f(3) = 2(3) - 1 = 6 - 1 = 5\). Thus, \(f(3)\) is defined and is equal to 5.

Step 3: Calculate the Left-hand Limit as x Approaches 3

To find the left-hand limit, we use the expression \(\frac{1}{3}x + 4\) since it applies when \(x < 3\). As \(x\) approaches 3 from the left, the limit is \(\lim_{x \to 3^-} \frac{1}{3}x + 4 = \frac{1}{3}(3) + 4 = 1 + 4 = 5\).

Step 4: Calculate the Right-hand Limit as x Approaches 3

For \(x \geq 3\), the function is defined by \(2x - 1\). Therefore, the right-hand limit as \(x\) approaches 3 is \(\lim_{x \to 3^+} 2x - 1 = 2(3) - 1 = 6 - 1 = 5\).

Step 5: Compare the Limits and f(3)

Now that we have both limits, we compare \(\lim_{x \to 3^-} f(x) = 5\), \(\lim_{x \to 3^+} f(x) = 5\), and \(f(3) = 5\). Since all these values are equal, the function is continuous at \(x = 3\).

Key Concepts

Piecewise Functions

A piecewise function is a function that is defined by different expressions for different intervals of its domain. This type of function is particularly useful when dealing with situations that have different outcomes based on the input value. In our exercise, the function is defined by two expressions: one for when the input is less than 3, and another for when the input is greater than or equal to 3.

The piecewise nature of this function can be described as follows:

• For values of \(x < 3\), the function is defined by the expression \(\frac{1}{3}x + 4\).
• For values of \(x \geq 3\), it switches to the expression \(2x - 1\).

It is crucial to understand this distinction to determine if the function is continuous at a point like \(x = 3\).

Left-Hand Limit

The left-hand limit of a function at a certain point is the value that the function approaches as the input approaches that point from the left. When dealing with piecewise functions, we focus on the expression relevant for inputs slightly less than the point of interest.

In our example, we need to determine \(\lim_{x \to 3^-} f(x)\) using the expression \(\frac{1}{3}x + 4\) as this applies to \(x < 3\). By evaluating the limit, we see:

\[\lim_{x \to 3^-} \left(\frac{1}{3}x + 4\right) = \frac{1}{3}(3) + 4 = 1 + 4 = 5\]

This tells us the function approaches 5 as \(x\) comes from values less than 3.

Right-Hand Limit

The right-hand limit is the value that a function approaches as the input comes from the right. For piecewise and step functions, this involves using the expression valid for inputs slightly greater than or exactly at the point of interest.

In our exercise, we calculate \(\lim_{x \to 3^+} f(x)\) using the expression \(2x - 1\), applicable for \(x \geq 3\). Thus, when evaluating the right-hand limit, we get:

\[\lim_{x \to 3^+} \left(2x - 1\right) = 2(3) - 1 = 6 - 1 = 5\]

This result indicates that as \(x\) approaches 3 from the right, the function's value tends towards 5, matching the left-hand limit.

Definition of Continuity

A function is continuous at a specific point if it fulfills the following conditions:

• First, the function must be defined at that point. For our exercise, \(f(3) = 5\) is indeed defined.
• Second, both the left-hand and right-hand limits as \(x\) approaches the point must exist. As we've calculated, both limits were found to be 5.
• Lastly, these limits must equal the function's value at the point itself. Since \(\lim_{x \to 3^-} f(x) = 5\), \(\lim_{x \to 3^+} f(x) = 5\), and \(f(3) = 5\), all conditions are satisfied.

Consequently, the function is continuous at \(x = 3\) as all these requirements are met protecting the solidarity of its graph at this point.