Problem 34 - College Algebra Essentials

Question

Evaluate each function at the given values of the independent variable and simplify. $$ f(r)=\sqrt{25-r}-6 $$ a. \(f(16)\) b. \(f(-24) \) c. \(f(25-2 x)\)

Step-by-Step Solution

Answer

The evaluation of the function at the given values results in: \(f(16) = -3\), \(f(-24) = 1\), and \(f(25-2x) = \sqrt{2x} - 6\).

Step 1: Substitute \(r = 16\) in the function.

To evaluate \(f(16)\), replace \(r\) with 16 in the function. This gives \(f(16) = \sqrt{25-16} - 6\). Now simplify this expression further.

Step 2: Simplify the expression for \(f(16)\)

Simplification gives \(f(16) = \sqrt{9} - 6 = 3 - 6 = -3\)

Step 3: Substitute \(r = -24\) in the function.

To evaluate \(f(-24)\), replace \(r\) with -24 in the function. This gives \(f(-24) = \sqrt{25-(-24)} - 6\). Now simplify this expression further.

Step 4: Simplify the expression for \(f(-24)\)

Simplification gives \(f(-24) = \sqrt{49} - 6 = 7 - 6 = 1\)

Step 5: Substitute \(r = 25-2x\) in the function.

To evaluate \(f(25-2x)\), replace \(r\) with \(25-2x\) in the function. This gives \(f(25-2x) = \sqrt{25-(25-2x)} - 6\). Now simplify this expression further.

Step 6: Simplify the expression for \(f(25-2x)\)

Simplification gives \(f(25-2x) = \sqrt{2x} - 6\)

Key Concepts

Simplifying Algebraic Expressions

The process of simplifying algebraic expressions is fundamental in solving various mathematical problems. It involves reducing an expression to its simplest form by combining like terms, eliminating parentheses, and using arithmetic operations. For example, consider the expression \( f(r)=\sqrt{25-r}-6 \) where \( r \) is the independent variable.

To simplify this expression, you might follow these steps:

• Identify and perform any necessary arithmetic operations inside the square root. For instance, if you're given \( r=16 \) then calculate \( 25-r \) to find \( 25-16 \), which is \( 9 \).
• Evaluate the square root. The square root of \( 9 \) is \( 3 \), which makes the expression become \( 3-6 \).
• Combine like terms or perform the subtraction to finally get \( -3 \).

This procedure not only makes the expressions easier to work with, but it also reveals the underlying structure of the function, making it easier to analyze and apply further operations.

Radical Expressions

Radical expressions contain numbers or variables under the square root, cube root, or higher roots. Simplification of such expressions involves reducing the number under the root to its simplest form and, if possible, eliminating the root entirely if it's a perfect square or cube.

In the context of the provided exercise, for example \( f(-24) = \sqrt{25-(-24)} - 6 \), we focus on the radical \( \sqrt{25-(-24)} \) which simplifies to \( \sqrt{49} \). Since \( 49 \) is a perfect square, its square root is \( 7 \), leading to a simplified expression of \( 7-6 \), which equals \( 1 \). It's crucial to understand the properties of square roots and know the perfect squares to simplify such radical expressions accurately.

Substitution Method in Algebra

Substitution is a powerful tool used to evaluate functions at specific values or to solve equations where one variable can be expressed in terms of another. In the substitution method, you replace a variable with a given number or expression. From our example, to find \( f(25-2x) \), we substitute \( r \) with \( 25-2x \) in the function \( f(r)=\sqrt{25-r}-6 \).

After the substitution, we have \( f(25-2x) = \sqrt{25-(25-2x)} - 6 \) which simplifies to \( \sqrt{2x} - 6 \). This simplification illustrates how substituting a given value into an algebraic function helps us to evaluate and further analyze the function for different variables or expressions. Understanding this method can greatly assist in tackling various mathematical problems across algebra and calculus.