Problem 44 - Intermediate Algebra

Question

If \(f(x)=\sqrt{x-1}\), find \(f(1), f(5), f(13)\), and \(f(26)\).

Step-by-Step Solution

Answer

\(f(1) = 0\), \(f(5) = 2\), \(f(13) = 2\sqrt{3}\), \(f(26) = 5\).

Step 1: Evaluate f(1)

To find \(f(1)\), substitute \(x = 1\) in the function: \(f(x) = \sqrt{x-1}\). We calculate: \[f(1) = \sqrt{1-1} = \sqrt{0} = 0.\] So, \(f(1) = 0\).

Step 2: Evaluate f(5)

To find \(f(5)\), substitute \(x = 5\) in the function: \(f(x) = \sqrt{x-1}\). We calculate: \[f(5) = \sqrt{5-1} = \sqrt{4} = 2.\] So, \(f(5) = 2\).

Step 3: Evaluate f(13)

To find \(f(13)\), substitute \(x = 13\) in the function: \(f(x) = \sqrt{x-1}\). We calculate: \[f(13) = \sqrt{13-1} = \sqrt{12}.\] So, \(f(13) = \sqrt{12}\), which simplifies to \(2\sqrt{3}\).

Step 4: Evaluate f(26)

To find \(f(26)\), substitute \(x = 26\) in the function: \(f(x) = \sqrt{x-1}\). We calculate: \[f(26) = \sqrt{26-1} = \sqrt{25} = 5.\] So, \(f(26) = 5\).

Key Concepts

Square Root Functions

A square root function is a type of algebraic function that involves the square root of a variable or expression. This function is commonly expressed in the form \( f(x) = \sqrt{x} \) or a slight variation such as \( f(x) = \sqrt{x - c} \), where \( c \) is a constant that shifts the graph horizontally. When approaching square root functions, it is essential to remember that the square root symbol indicates the non-negative root of a number. This means that the function will produce values that are zero or positive for non-negative inputs.
Because of the square root, the function \( f(x) = \sqrt{x-1} \), for example, is defined only for values of \( x \) such that \( x-1 \geq 0 \). This implies that the domain of the function includes numbers equal to or greater than 1. The graph of a square root function forms a gentle curve that starts at the point that satisfies the restriction imposed by the square root and stretches towards infinity while moving upwards and to the right.

Substitution Method

The substitution method is a mathematical technique used to evaluate functions like \( f(x) = \sqrt{x-1} \) by replacing the variable \( x \) with a specific number. This technique involves the following steps:

• Identify the function and note what it describes; for example, \( f(x) = \sqrt{x-1} \) indicates a square root function.
• Take the value you want to evaluate (like 1, 5, 13, or 26) and insert it into the function in place of the variable \( x \).
• Perform the arithmetic operations within the function to find the final result. For instance, when evaluating \( f(5) \), you substitute 5 for \( x \), resulting in \( \sqrt{5-1} = \sqrt{4} \), which simplifies to 2.

The substitution method is especially useful because it provides a clear and step-by-step way to approach evaluating expressions, ensuring accuracy and clarity in mathematical problems.

Simplifying Radicals

Simplifying radicals involves rewriting a radical expression like \( \sqrt{12} \) into its simplest form. This process often includes factoring the number inside the radical to identify perfect squares.
To simplify \( \sqrt{12} \), consider the prime factorization of 12: \( 12 = 2^2 \times 3 \). Here, \( 2^2 \) is a perfect square. You can pull the square root of that factor out of the radical to obtain:

• Initially \( \sqrt{12} \) can be broken down into \( \sqrt{4 \times 3} \).
• Since \( \sqrt{4} = 2 \), you can simplify further to \( \sqrt{12} = 2\sqrt{3} \).

This method of simplifying radicals ensures that expressions are presented in their most reduced form, which is helpful for both calculation and interpretation. Simplifying radicals is crucial when working with complicated expressions, in order to make them as concise and understandable as possible.