Problem 31 - College Algebra

Question

v$$ h(x)=\sqrt[3]{\frac{1}{2 x-3}} $$

Step-by-Step Solution

Answer

The domain of \( h(x) \) is all real numbers except \( x = \frac{3}{2} \).

Step 1: Identify the Inner Function

The given function is a composite function. Identify the inner function first. Here, the inner function is the expression inside the cube root: \( f(x) = \frac{1}{2x-3} \).

Step 2: Determine the Domain of the Inner Function

The function \( f(x) = \frac{1}{2x-3} \) is a rational function. For this function to be valid, its denominator cannot be zero. Thus, solve \( 2x - 3 eq 0 \) to find the domain restriction: \( x eq \frac{3}{2} \).

Step 3: Apply the Cube Root Function

The outer function is the cube root, which is defined for all real numbers. Therefore, the domain restriction from the inner function \( x eq \frac{3}{2} \) governs the entire function.

Step 4: Conclusion of Domain

Considering all the restrictions identified, the domain of \( h(x) = \sqrt[3]{\frac{1}{2x-3}} \) is all real numbers except \( x = \frac{3}{2} \), which can be written as: \( x \in \mathbb{R} \setminus \left\{ \frac{3}{2} \right\} \).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values, or 'x' values, that you can use without encountering any mathematical errors, like division by zero or taking the square root of a negative number. When finding the domain, you should always check for restrictions.

In our case with the composite function, comprising both a rational function and a cube root, we need to ensure each component is valid. While cube root functions are versatile, accepting any real number, the rational inner function has its limitations. With rational functions, the denominator can't be zero, or else it becomes undefined. Thus, for the function \( f(x) = \frac{1}{2x-3} \), it's crucial to solve \(2x - 3 eq 0\) to establish the domain. Solving this gives \(x eq \frac{3}{2}\).

This finding determines the domain for the whole composite function, \( h(x) = \sqrt[3]{\frac{1}{2x-3}} \), keeping it to all real numbers except \( x = \frac{3}{2} \).

Rational Functions

Rational functions are expressions of the form \( \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials. These functions are only invalid when their denominator \( q(x) \) equals zero, causing undefined values, which results in restrictions on their domain.

It's fundamental in mathematics to solve \( q(x) = 0 \) for rational functions to find where the denominator becomes zero. For example, in the function \( f(x) = \frac{1}{2x-3} \), the denominator \( 2x-3 \) should not be zero. Solving \( 2x-3 = 0 \) gives \( x = \frac{3}{2} \).

• This means \( f(x) \) is undefined at \( x = \frac{3}{2} \).
• Therefore, the domain of \( f(x) \) excludes \( x = \frac{3}{2} \).

Understanding these restrictions helps when building or analyzing composite functions where rational expressions are part of the mix.

Cube Root Functions

Cube root functions, represented as \( \sqrt[3]{x} \), differ from square root functions in their domain qualities. They accept all real numbers, meaning you can plug in any real number, positive, negative, or zero, without issue.

This unrestricted feature stems from the fact that you can take the cube root of any number without encountering mathematical anomalies like undefined or imaginary numbers. So, when using cube roots, there’s no need for additional restrictions beyond what's pre-established in composite functions.

• This wide applicability pertains to all real numbers: \( x \in \mathbb{R} \).
• In our given function, \( h(x) = \sqrt[3]{\frac{1}{2x-3}} \), the cube root aspect doesn’t impose limits. Any restrictions arise solely from the inner rational function, such as \( x eq \frac{3}{2} \).

Appreciating this allows you to differentiate where limitations on a composite function’s domain truly originate.