Problem 52 - Contemporary Precalculus

Question

Determine the domain of the function according to the usual convention. $$h(x)=\frac{\sqrt{x-1}}{x^{2}-1}$$

Step-by-Step Solution

Answer

Question: What is the domain of the given function: $$f(x)=\frac{\sqrt{x-1}}{x^{2}-1}$$ Answer: The domain of the given function is $$x>1$$.

Step 1: Finding the values of x for which the square root is defined

The square root of a number is defined for all non-negative numbers. In this case, we must ensure that the expression inside the square root is non-negative, so we have: $$x-1\geq0$$ Solving this inequality, we get: $$x\geq1$$

Step 2: Finding the values of x for which the denominator is not zero

The denominator of our function is defined as long as it is not equal to zero. To find the values of x where the denominator equals zero, we set the denominator equal to zero and solve for x: $$x^{2}-1=0$$ This is a difference of squares, and it can be factored as: $$(x-1)(x+1)=0$$ So, the values of x where the denominator is zero are x=1 and x=-1.

Step 3: Finding the intersection of the two intervals

We now have two sets of values to consider in order to find the domain of the function. The first set is x≥1, and the second set consists of all x-values except x=1 and x=-1. To find the intersection of these two sets, we note that all values of x≥1 are acceptable, except for x=1 itself. Therefore, the domain of the function is: $$x>1$$

Key Concepts

Inequality Solving

Inequality solving is essential in determining the domain of functions as it identifies the acceptable range of values for the variable. In our example, the function is \( h(x)=\frac{\sqrt{x-1}}{x^2-1} \). To ensure the square root is defined, the inequality \( x-1 \geq 0 \) is set up because square roots are real for non-negative numbers only.

Solving this inequality is straightforward: simply add 1 to both sides, yielding \( x \geq 1 \). This ensures that the value inside the square root is non-negative. However, the solution to the inequality isn't complete without considering other restrictions from the function, such as the denominator's condition. The process is an example of using inequality solving to piece together the domain of a complex function.

Square Roots in Functions

Understanding square roots in functions is crucial when examining their domain. The square root is a function itself and only outputs real numbers when the input is zero or positive. This property restricts the domain of the outer function, like \( h(x) \) in our case.

Whenever a function's formula involves a square root, like \( \sqrt{x-1} \) here, an inequality is formed to determine where the expression under the square root is non-negative. Keep in mind that the equality part of \( \geq \) is only applicable if the square root is not in the denominator of a fraction. Otherwise, that particular value would make the denominator zero, which is not permissible in mathematics as it would be undefined.

Denominator Restrictions

When determining the domain of a function with a rational expression, identifying denominator restrictions is key. A function is undefined if its denominator equals zero because division by zero is not allowed. In the provided function \( h(x) \), we have \( x^2 - 1 \) as the denominator which can be factored into \( (x - 1)(x + 1) \). This factoring reveals the roots of the equation, x = 1 and x = -1, which we exclude from the domain.

The intersection of intervals from solving the inequality and respecting the denominator restrictions leads to the conclusion that while x can be equal to or greater than 1 from the square root constraint, x must not be 1 due to the denominator's constraint. Therefore, only values greater than 1 are included in the domain, leading us to a final domain for \( h(x) \) of \( x > 1 \).