Problem 28 - College Algebra

Question

In Exercises \(21-32,\) evaluate each function at the given values of the independent variable and simplify. $$f(r)=\sqrt{25-r}-6$$ a. \(f(16) \quad\) b. \(f(-24) \quad\) c. \(f(25-2 x)\)

Step-by-Step Solution

Answer

a) \(f(16)=-3\), b) \(f(-24)=1\), c) \(f(25-2x)=\sqrt{2x}-6\)

Step 1: Evaluating \(f(16)\)

To find the value of \(f(16)\), we replace \(r\) in \(f(r)\) with \(16\): \[f(16)=\sqrt{25-16}-6 = \sqrt{9}-6 = 3-6 = -3\]

Step 2: Evaluating \(f(-24)\)

To find the value of \(f(-24)\), we replace \(r\) in \(f(r)\) with \(-24\): \[f(-24)=\sqrt{25 -(-24)}-6 = \sqrt{25+24}-6 = \sqrt{49}-6 = 7-6 = 1\]

Step 3: Evaluating \(f(25-2x)\)

To find the value of \(f(25-2x)\), we replace \(r\) in \(f(r)\) with \(25-2x\): \[f(25-2x)=\sqrt{25-(25-2x)}-6 = \sqrt{2x}-6\] Note: This expression cannot be simplified without knowing the value of \(x\).

Key Concepts

Function Evaluation

Function evaluation is a fundamental concept in algebra. It involves replacing the variable in a function with a given number and performing the necessary calculations to find the result. For example, consider the function provided in the exercise, \(f(r) = \sqrt{25-r} - 6\). To evaluate this function at a specific value such as \(r = 16\), you simply substitute \(16\) for \(r\) and calculate the result, as shown in the solution step for \(f(16)\). Understanding how to correctly replace variables and simplify the resulting expression is key to mastering function evaluation. When evaluating functions, ensure each step of the process is followed carefully: substitution, simplification, and then computation of the final value.

Simplifying Expressions

Simplifying algebraic expressions is crucial to working effectively with algebra. After evaluating a function for a specific value, you often get an expression that can be simplified. Simplification makes expressions easier to understand and work with. In the example of \(f(16)\), the process includes simplifying the radical expression \(\sqrt{9}\). The square root of 9 is 3, and simplifying the expression further by subtracting 6 gives us \(-3\). To simplify expressions proficiently, you should be comfortable with basic algebraic operations and properties. Always combine like terms where possible and perform operations according to the order of operations, also known as PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).

Radical Expressions

Radical expressions, such as \(\sqrt{25-r}\), involve roots of numbers or algebraic expressions. To evaluate and simplify radical expressions, look for perfect squares, cubes, or higher order roots that can be extracted to simplify the expression. For instance, the square root of 25 is 5, and the square root of 49 is 7. When \(r\) is a value such as 16, which makes the radicand (the expression under the square root) a perfect square, you can simplify the expression completely, as seen with \(\sqrt{9} = 3\). It's essential to recognize when a radical can be simplified and to remember that the square root of a negative number is not a real number unless you're working with complex numbers.

Algebraic Functions

An algebraic function, like the one in our exercise \(f(r)\), involves algebraic expressions built using polynomial expressions, radicals, and arithmetic operations. Understanding algebraic functions is about knowing how to manipulate and evaluate expressions for any given value of the variable. The function represents a rule that assigns a specific output for each input value. When given an algebraic function, it is essential to scrutinize the structure of the function to determine its domain (all the possible input values) and to know how to properly evaluate it at different values within that domain. In the example \(f(25-2x)\), the function is evaluated at \(25 - 2x\), which expresses a relationship between two varying quantities, \(r\) and \(x\), showcasing the dynamic nature of algebraic functions.