Problem 34 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Find \(f(x)\) for the indicated values of \(x\), if possible. (b) Find the domain of \(f\). $$ f(x)=\frac{1}{\sqrt{x-1}} \text { for } x=0, a^{2}-a+1 $$

Step-by-Step Solution

Answer

(a) f(0) is undefined; f(a²-a+1) is undefined for a=0, 1; defined otherwise. (b) Domain: (1, ∞).

Step 1: Evaluate f(x) for x = 0

We start by substituting \(x = 0\) into the function \(f(x) = \frac{1}{\sqrt{x-1}}\). The expression becomes \(f(0) = \frac{1}{\sqrt{0-1}} = \frac{1}{\sqrt{-1}}\), which is undefined because the square root of a negative number is not a real number. Therefore, \(f(0)\) does not exist.

Step 2: Evaluate f(x) for x = a² - a + 1

Next, substitute \(x = a^2 - a + 1\) into the function \(f(x) = \frac{1}{\sqrt{x-1}}\). This gives us \(f(a^2 - a + 1) = \frac{1}{\sqrt{a^2 - a + 1 - 1}} = \frac{1}{\sqrt{a^2 - a}}\). For \(f(x)\) to be defined, \(a^2 - a\) must be positive and non-zero. If \(a^2 - a = 0\) (i.e., \(a(a-1) = 0\)), then \(f(x)\) becomes undefined. Thus, \(a\) cannot be \(0\) or \(1\). Otherwise, \(f(a^2 - a + 1)\) is defined.

Step 3: Find the Domain of f(x)

To find the domain of \(f(x) = \frac{1}{\sqrt{x-1}}\), we need the expression under the square root to be positive: \(x - 1 > 0\). Solving this inequality gives \(x > 1\). Therefore, the domain of \(f(x)\) is \((1, \infty)\).

Key Concepts

Undefined Functions

Functions can become undefined for several reasons. It often happens when the operations in the function are mathematically impossible, such as dividing by zero or taking the square root of a negative number.
For instance, in the function \( f(x) = \frac{1}{\sqrt{x-1}} \), the function becomes undefined if the denominator equals zero or if our square root has a negative number.

• If \( x = 0 \), substituting into the function gives us \( f(0) = \frac{1}{\sqrt{-1}} \), which is undefined because the square root of a negative number doesn't give a real number.
• If \( x = a^2 - a + 1 \), substituting leads to \( f(x) = \frac{1}{\sqrt{a^2 - a}} \). Here, the function also becomes undefined if \( a^2 - a = 0 \).

Understanding when a function is undefined helps us identify where a function cannot have any real number outputs.

Square Root Properties

The properties of square roots are essential in figuring out where a function will be defined.
Square roots can only represent the roots of non-negative quantities in terms of real numbers. For example, \( \sqrt{-1} \) isn't a real number, causing the function that contains it to be undefined unless we are considering complex numbers.

• Consider \( \sqrt{x-1} \) in the function \( f(x) = \frac{1}{\sqrt{x-1}} \). Here, \( x-1 \) must be greater than zero as we need to take the square root of a positive number. If \( x-1 > 0 \), only then can you calculate \( \sqrt{x-1} \).

Knowing these properties allows us to define the constraints for where functions are valid or usable within the set of real numbers.

Inequalities

Inequalities are vital tools used to determine the domain of a function, especially when square roots are involved.

• For the function \( f(x) = \frac{1}{\sqrt{x-1}} \), solving \( x-1 > 0 \) ensures that the expression inside the square root remains positive, thus permitting a real number solution.
• This leads us to solve the inequality \( x > 1 \), so the domain from which this function can accept inputs is any value greater than 1, denoted as the interval \( (1, \infty) \).

Inequalities help ensure functions remain defined by keeping significant expressions valid under mathematical operations. In this context, they show where the square root will yield a real number rather than an undefined expression.