Problem 14 - Calculus: Early Transcendentals

Question

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number \( a \). \( f(x) = 3x^4 - 5x + \sqrt[3]{x^2 + 4}, \hspace{5mm} a = 2 \)

Step-by-Step Solution

Answer

The function is continuous at \( x = 2 \).

Step 1: Definition of Continuity

A function is defined as continuous at a number \( a \) if \( \lim_{x \to a} f(x) = f(a) \). Therefore, to show that the function \( f(x) = 3x^4 - 5x + \sqrt[3]{x^2 + 4} \) is continuous at \( a = 2 \), we need to show that \( \lim_{x \to 2} f(x) = f(2) \).

Step 2: Calculate \( f(a) \)

Substitute \( a = 2 \) directly into the function: \[ f(2) = 3(2)^4 - 5(2) + \sqrt[3]{(2)^2 + 4} \]Calculate each term:\( 3(2)^4 = 48 \)\( 5(2) = 10 \)\( (2)^2 + 4 = 8 \) and \( \sqrt[3]{8} = 2 \)Combine the results:\[ f(2) = 48 - 10 + 2 = 40 \]

Step 3: Calculate \( \lim_{x \to 2} 3x^4 \)

For the polynomial part \( 3x^4 \), use the direct substitution to find the limit:\[ \lim_{x \to 2} 3x^4 = 3(2)^4 = 48 \]

Step 4: Calculate \( \lim_{x \to 2} (-5x) \)

Similarly, for the linear term \(-5x\), find the limit directly:\[ \lim_{x \to 2} (-5x) = -5(2) = -10 \]

Step 5: Calculate \( \lim_{x \to 2} \sqrt[3]{x^2 + 4} \)

Use substitution for the cube root function:\[ \lim_{x \to 2} \sqrt[3]{x^2 + 4} = \sqrt[3]{(2)^2 + 4} = \sqrt[3]{8} = 2 \]

Step 6: Combine the Limits

Combine the limits found in Steps 3, 4, and 5:\[ \lim_{x \to 2} (3x^4 - 5x + \sqrt[3]{x^2 + 4}) = 48 - 10 + 2 = 40 \]

Step 7: Compare Limits with Function Value

Since \( \lim_{x \to 2} f(x) = 40 \) and \( f(2) = 40 \), by the definition of continuity, the function is continuous at \( x = 2 \).

Key Concepts

Limits

Limits are a fundamental concept in calculus. They describe the value that a function approaches as the input approaches some point. Mathematically, if we want to find the limit of a function \( f(x) \) as \( x \) approaches a number \( a \), we use the notation \( \lim_{x \to a} f(x) \). Understanding limits is crucial for analyzing the behavior of functions near particular points.

In the original exercise, limits help us determine if a function is continuous at a specific point. We examined the limits for each part of the function separately, which simplifies calculating the overall limit.

• First, we found the limit of the polynomial term \( 3x^4 \) as \( x \rightarrow 2 \).
• Next, we looked at the linear component \( -5x \).
• Finally, we assessed the limit of the cube root \( \sqrt[3]{x^2 + 4} \).

Each part was evaluated separately before the results were combined to find the full limit of the function as \( x \) approaches 2.

Polynomial Functions

Polynomial functions consist of terms that are non-negative integer powers of \( x \). These functions look like \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \) where \( n \) is a non-negative integer, and the coefficients \( a_i \) are constants. In the exercise, one of the components of our function is a polynomial, specifically \( 3x^4 - 5x \).

Polynomials are continuous everywhere, meaning you can find their limits by direct substitution. This is because polynomial functions do not have breaks, holes or asymptotes. When checking continuity at \( x = 2 \), simply plug in the value to evaluate the polynomial term.

• The term \( 3x^4 \) becomes \( 3(2)^4 = 48 \).
• The term \( -5x \) becomes \( -5(2) = -10 \).

This is why they are often one of the simpler elements of a function to deal with when exploring limits and continuity.

Cube Roots

The cube root of a number \( x \) is represented as \( \sqrt[3]{x} \). This operation asks what number, when multiplied by itself twice, results in \( x \). Cube roots are an important concept to understand because they behave differently from square roots, especially since they are defined for all real numbers, including negatives.

In the problem, one part of the function is a cube root expression \( \sqrt[3]{x^2 + 4} \). To find the limit as \( x \to 2 \), substitute \( x = 2 \) directly:

• Calculate \( (2)^2 + 4 = 8 \).
• Find \( \sqrt[3]{8} = 2 \).

The cube root function is continuous because it produces smooth curves without interruptions.
This smoothness ensures that limits can be calculated directly by substitution for real numbers, just like using direct substitution with polynomials.

Continuous Functions

A function is continuous at a point \( a \) if \( \lim_{x \to a} f(x) = f(a) \). Continuity means no breaks or holes in the graph of the function at that point. It gives us the ability to work smoothly with the function without encountering undefined behaviors.

To prove continuity at a particular point, like in our exercise, we follow these steps:

• First, calculate \( f(a) \) by substituting \( a \) into the function.
• Then, find \( \lim_{x \to a} f(x) \) by evaluating the function’s individual components and combining them.
• If these two values are equal, the function is continuous at \( a \).

In our exercise, the function value and limit at \( x = 2 \) both equaled 40, confirming its continuity there.