Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 24-35, at what points, if any, are the functions discontinuous? $$ g(u)=\frac{u^{2}+|u-1|}{\sqrt[3]{u+1}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify Domains

First, we need to identify the domain of each component of the function. For \(g(u)\), we have a square root and modulus function in the numerator, and a cube root in the denominator. The numerator \(u^2 + |u-1|\) is defined for all real numbers. The cube root \(\sqrt[3]{u+1}\) is also defined for all real numbers, but we must check if it introduces any discontinuities when set in the denominator as it cannot be zero.

Step 2: Find Points of Discontinuity

To find discontinuities, analyze where \(\sqrt[3]{u+1} = 0\), as division by zero is undefined. Solve \(u+1=0\), which gives us \(u = -1\). This value makes the denominator zero, indicating a discontinuity at \(u = -1\).

Step 3: Check Other Discontinuities

Since the numerator \(u^2 + |u-1|\) is always defined and the cube root in the denominator is otherwise continuous, no other discontinuities exist.

Key Concepts

Domain of a Function

The domain of a function specifies the set of all possible input values (usually represented by 'x' or 'u') for which the function is defined. In simpler terms, it tells us what numbers we can plug into the function without causing any mathematical errors. For the function given in the original problem, we have a numerator that includes a modulus operation and a denominator involving a cube root.

Here's how to find the domain of a function:

• Identify any values that make the function undefined. Common reasons include division by zero and taking the square root or logarithm of a negative number.
• Check each element of the function separately to determine its domain.
• Combine the results to identify the domain for the entire function.

In the problem, the numerator, consisting of a square and modulus function, is defined for all real numbers. The cube root in the denominator, \(rac{1}{\sqrt[3]{u+1}}\), is also defined for all real numbers but cannot equal zero, so we need to specifically check for this by setting the cube root equal to zero and solving for 'u'. When \((u + 1) = 0\), we find \(u = -1\), which is where discontinuity occurs.

Cube Root Function

A cube root function is a type of radical function where the root is cube root (also known as the third root). The general form is \(\sqrt[3]{x}\), and it is considered more forgiving than a square root function because it is defined for all real numbers.

Key properties of the cube root function:

• It can accept negative numbers, unlike the square root.
• The cube root of zero is zero.
• It maintains continuity across all real numbers, meaning it does not jump or break at any point.

In the provided function, the cube root component \(\sqrt[3]{u+1}\) is in the denominator. While it normally remains continuous, its occurrence in the denominator means we must check where it equals zero, because dividing by zero is undefined. Therefore, \(u+1=0\) leads to \(u=-1\), indicating a discontinuity despite the usual continuous nature of cube root functions.
It's important to grasp that the issue arises here not from the cube root function itself, but from its placement in the denominator of a fraction.

Modulus Function

The modulus function, often represented by \(|x|\), outputs the absolute value of a given number. Absolute value reflects the distance of a number from zero on the number line, ensuring it's always non-negative. In our function, it's written as \(|u-1|\).

Key insights into modulus functions:

• Deals with positive and negative numbers, turning negatives into their positive counterparts.
• It never leads to a negative result.
• The graph of a modulus function forms a 'V' shape, introducing a sharp turn at the point where the expression inside equals zero.

In the given exercise function, the modulus function does not introduce any discontinuities. This is because it is defined and continuous for all real numbers, smoothly transitioning across the entire number line. This characteristic helps to keep the numerator \(u^2 + |u-1|\) always defined and free from any breaks, contributing to the understanding of where the actual discontinuity (at \(u=-1\)) is caused by the cube root in the denominator.