Problem 63 - Algebra and Trigonometry

Question

Find the domain of the function. $$f(x)=\frac{(x+1)^{2}}{\sqrt{2 x-1}}$$

Step-by-Step Solution

Answer

The domain of the function is \((\frac{1}{2}, \infty)\).

Step 1: Understanding the Function

The function given is \( f(x) = \frac{(x+1)^2}{\sqrt{2x-1}} \). We need to determine for which values of \( x \) the function is defined. This involves looking at both the numerator and the denominator.

Step 2: Analyzing the Numerator

The numerator \((x+1)^2\) is a polynomial which is defined for all real numbers. Therefore, the numerator poses no restrictions on the domain.

Step 3: Analyzing the Denominator

The denominator \(\sqrt{2x-1}\) requires that the expression under the square root be positive, because the square root is only defined for non-negative numbers and must be strictly positive as it's in the denominator. Therefore, \(2x-1 > 0\).

Step 4: Solving the Inequality

To find when \(2x-1 > 0\), solve for \(x\) by adding 1 to each side: \(2x > 1\). Then divide by 2 to isolate \(x\): \(x > \frac{1}{2}\).

Step 5: Conclusion

Combining the requirements, since the numerator is defined for all \(x\) and we have \(x > \frac{1}{2}\) for the denominator, the domain of the function is \((\frac{1}{2}, \infty)\).

Key Concepts

Numerator Analysis

In the study of functions, the numerator provides important information about the domain. If the numerator is a polynomial, it tends to be straightforward because polynomials are defined for all real numbers. Let's look at the specific case of the numerator in the function \(f(x) = \frac{(x+1)^2}{\sqrt{2x-1}}\). Here, the numerator is \((x+1)^2\).

• **Polynomial Nature**: \((x+1)^2\) is a polynomial, as it involves only basic arithmetic operations (addition and multiplication) of the term \(x+1\).
• **Domain of Polynomials**: All polynomials are defined for every real number \(x\), thus this part of the function imposes no restrictions on the domain.

In conclusion, for this specific polynomial numerator, there's no need to exclude any values of \(x\) when determining the domain of the function.

Denominator Analysis

The denominator in a function often brings restrictions to its domain, especially when it includes square roots or divisions by zero. In our function \(f(x) = \frac{(x+1)^2}{\sqrt{2x-1}}\), the denominator is \(\sqrt{2x-1}\). When dealing with square roots, consider the following:

• **Non-Negative Condition**: The expression inside a square root, \(2x-1\), must be non-negative for the root to be defined. However, since it appears in the denominator, it must be strictly positive to prevent division by zero.
• **Strict Inequality**: Thus, solve \(2x-1 > 0\) to find the relevant \(x\) values where the denominator stays positive.

This approach ensures that both the square root and division restrictions are respected. Non-positive values would either make the denominator undefined or cause division by zero, which are both mathematically impossible.

Solving Inequalities

Inequalities are solved in a manner similar to equations, but with special attention when multiplying or dividing by negative numbers. For the inequality \(2x-1 > 0\), the steps are straightforward:

• **Isolate the variable**: Add 1 to both sides to get \(2x > 1\).
• **Solve for \(x\)**: Divide both sides by 2, giving \(x > \frac{1}{2}\).

These steps ensure that we correctly identify the range of \(x\) values that satisfy the inequality. Solving inequalities allows us to determine valid \(x\) values where all parts of a function are defined and behave appropriately. For the function \(f(x)\), it means the real number values from \((\frac{1}{2}, \infty)\) are included in the domain, where the entire expression is valid.