Problem 70 - Calculus Early Transcendentals

Question

Use the continuity of the absolute value function (Exercise 66 ) to determine the interval(s) on which the following functions are continuous. $$h(x)=\left|x^{2}+2 x+5\right|+\sqrt{x}$$

Step-by-Step Solution

Answer

Answer: The function h(x) is continuous on the interval \([0, \infty)\).

Step 1: Identify the domains of the individual functions

The absolute value function, which is defined as \(|x^2 + 2x + 5|\), is continuous for all x-values because the quadratic function in the absolute value is also continuous for all x-values. For the square root function, defined as \(\sqrt{x}\), it is only continuous where x is non-negative, so \(x\geq 0\).

Step 2: Combine the domains

Now we need to combine both domains to determine the interval where the entire function h(x) is continuous. Since the absolute value function has no restrictions, we only need to consider the restrictions of the square root function, which is continuous for \(x\geq 0\). Therefore, combining both domains, the function h(x) is continuous on the interval \([0, \infty)\).

Key Concepts

Absolute Value Function

The absolute value function is represented by the notation \(|x|\) and it measures the distance of a number, \(x\), from zero on the number line. Regardless of whether \(x\) is positive or negative, \( |x| \) will always be non-negative.
In mathematical terms:

• If \( x \geq 0\), then \( |x| = x\).
• If \( x < 0\), then \( |x| = -x\).

The property of absolute value functions being continuous is important. It implies that the function \( f(x) = |x| \) doesn’t have any breaks or holes when plotted graphically.
This continuity extends to any expression inside the absolute value, such as \( |x^2 + 2x + 5| \), despite the quadratic expression itself having roots. Because absolute value is continuous for all real numbers, \( |x^2 + 2x + 5| \) too is defined for every real \( x \). Thus, its own continuity does not impose any restriction on the domain of the overall function.

Square Root Function

The square root function, denoted as \( \sqrt{x}\), provides the positive number that, when multiplied by itself, equals \( x \).
This function is unique because it is only defined for non-negative values of \( x \.\) This comes from the fact that you cannot take the square root of a negative number within the real numbers without running into complex numbers.

• When \( x \geq 0\), \( \sqrt{x} \) is defined and provides continuous and smooth values.
• When \( x < 0\), \( \sqrt{x} \) is not real-valued, and hence is not considered continuous in the real number system.

Thus, the domain of the square root function is \( x \geq 0\). In practical terms, for any function that includes a square root component, this restriction on the domain must be carefully considered to ensure continuity in the context of real numbers.

Domains of Functions

Understanding the domain of a function is foundational in determining where a function is continuous and defined. The domain is essentially the set of all possible input values (or \( x \)-values) for which the function is valid.
In our example, the function \( h(x) = |x^2 + 2x + 5| + \sqrt{x} \) incorporates both an absolute value and a square root function.

• The absolute value component, \( |x^2 + 2x + 5| \), is defined for all \( x \)-values, leading it to have a domain spread across all real numbers.
• The square root component, \( \sqrt{x} \), restricts the domain to non-negative \( x \)-values, i.e., \( x \geq 0\).

To find the domain of the composite function \( h(x) \,\), you take the intersection where both components are valid, leading to \( [0, \infty) \).
Understanding and finding the domain is critical as it helps identify the interval over which a function is continuous and ensures that all its parts work together seamlessly without any breaks or undefined points.