Problem 45 - Intermediate Algebra

Question

If \(f(x)=\sqrt{2 x+1}\), find \(f(3), f(4), f(10)\), and \(f(12)\).

Step-by-Step Solution

Answer

\(f(3) = \sqrt{7}\); \(f(4) = 3\); \(f(10) = \sqrt{21}\); \(f(12) = 5\).

Step 1: Evaluate f(3)

Start by substituting \(x = 3\) into the function \(f(x) = \sqrt{2x + 1}\). You get \(f(3) = \sqrt{2(3) + 1}\). Calculate inside the square root: \(2(3) = 6\), then add \(1\), resulting in \(7\). So, \(f(3) = \sqrt{7}\).

Step 2: Evaluate f(4)

Next, substitute \(x = 4\) into the function \(f(x) = \sqrt{2x + 1}\). This gives you \(f(4) = \sqrt{2(4) + 1}\). Calculate inside the square root: \(2(4) = 8\), then add \(1\), resulting in \(9\). So, \(f(4) = \sqrt{9} = 3\).

Step 3: Evaluate f(10)

Now substitute \(x = 10\) into the function \(f(x) = \sqrt{2x + 1}\). You get \(f(10) = \sqrt{2(10) + 1}\). Calculate inside the square root: \(2(10) = 20\), then add \(1\), resulting in \(21\). So, \(f(10) = \sqrt{21}\).

Step 4: Evaluate f(12)

Finally, substitute \(x = 12\) into the function \(f(x) = \sqrt{2x + 1}\). This gives you \(f(12) = \sqrt{2(12) + 1}\). Calculate inside the square root: \(2(12) = 24\), then add \(1\), resulting in \(25\). So, \(f(12) = \sqrt{25} = 5\).

Key Concepts

Understanding the Square Root Function

The square root function is one of the fundamental components of algebra. It is denoted by the symbol \( \sqrt{} \). In its essence, this function returns a number that, when multiplied by itself, gives the original number under the square root sign. For example, \( \sqrt{9} = 3 \) because \(3 \times 3 = 9 \).
In many algebraic equations, the square root function helps simplify expressions and solve equations. It's widely used in various fields, including geometry and physics, due to its ability to unravel quadratic effects.
In the given problem, the function \( f(x) = \sqrt{2x + 1} \) implies that whatever expression is obtained from \(2x + 1\), its square root will be taken. Understanding this helps with evaluating the function correctly.

Exploring the Substitution Method

The substitution method is a valuable technique in algebra that involves replacing variables with specific values to simplify and solve expressions or equations. This method is particularly useful when evaluating functions at given points. Here’s how it’s applied:

• Identify the expression that contains the variable. In our exercise, that's \(2x + 1\).
• Replace the variable \(x\) with the given number. For example, if \(x = 3\), substitute \(3\) into the expression.
• Calculate the result of the substitution, which simplifies the original function into a more manageable number or expression.

In the example, substituting values into the square root function helps us find specific outputs like \( f(3), f(4), f(10), \) and \( f(12) \).
It effectively transforms potential complexities into simple arithmetic, ensuring clarity and ease of calculation.

Simplifying Algebraic Expressions

Algebraic expressions, such as \(2x + 1\), are combinations of numbers, variables, and arithmetic operations. Simplifying these expressions is crucial when solving equations or evaluating functions. Here’s a breakdown of how this is done:
First, identify each part of the expression. For \(2x + 1\), \(2x\) is a term, meaning \(x\) multiplied by \(2\), and \(1\) is a constant.
When evaluating functions at specific points, like in our exercise, substitution is used first (as previously discussed), and then algebraic simplification follows:

• After substituting the variable, perform the multiplication. For example, in \(2(3)\), multiply \(2\) by \(3\) to get \(6\).
• Next, add or subtract any constants, which further simplifies the expression, resulting in essentially a single number under the square root.

This systematic approach allows one to efficiently manage and resolve mathematical expressions, ensuring accuracy and understanding in function evaluation.