Problem 41 - Thinking Mathematically

Question

Evaluate each function at the given value of the variable. \(h(r)=3 r^{2}+5\) a. \(h(4)\) b. \(h(-1)\)

Step-by-Step Solution

Answer

The values of the function for given \(r\) are \(h(4)=53\) and \(h(-1)=8\).

Step 1: Evaluate function at \(r=4\)

To evaluate this function at \(r=4\), replace every occurrence of \(r\) with \(4\) in the equation. This results in \(h(4)=3*(4)^{2}+5\). Calculate inside the parenthesis first, then execute the exponent, followed by the multiplication, and then finally the addition. Solve this to get the value of \(h(4)\).

Step 2: Results of \(h(4)\)

To calculate the value of \(h(4)\), we follow the order of operations. First, we square the 4 within the parenthesis to get 16. Then we multiply 3 with 16 to get 48. Lastly, we add 5 to 48 to get the final result. Therefore, \(h(4)=48+5=53\).

Step 3: Evaluate function at \(r=-1\)

To evaluate this function at \(r=-1\), replace every occurrence of \(r\) with \(-1\) in the equation. This results in \(h(-1)=3*(-1)^{2}+5\). Execute the exponent first, followed by the multiplication, then finally the addition, to get the value of \(h(-1)\).

Step 4: Result of \(h(-1)\)

Following the rules of operations, first square the values in the brackets, which yields \(1\). Multiply 3 by this result to get 3. Lastly, add 5 to 3 to get the result. Therefore, \(h(-1)=3+5=8\).

Key Concepts

Function Evaluation

Understanding function evaluation is crucial in algebra. When you're tasked with evaluating a function at a given value, like in the problem involving h(r) = 3r^2 + 5, what you're really doing is substituting a number for the variable in the function and simplifying the expression. For example, for h(4), we replace every instance of r with 4 to get h(4) = 3(4)^2 + 5.

Function evaluation often involves following specific algebraic rules to simplify the function, and in most cases, it includes a combination of basic arithmetic operations such as addition, multiplication, and exponentiation. Through practice and reinforcement, students can become proficient in this fundamental algebraic skill.

Order of Operations

The key to correctly evaluating functions is the order of operations, commonly remembered using the acronym PEMDAS, which stands for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). This rule tells us which operations to perform first to accurately simplify an expression.

When evaluating h(4), we start with the exponent, square 4 to get 16, then carry out the multiplication 3*16, and finally, we add 5 to get 53. When followed consistently, the order of operations ensures that everyone can arrive at the same correct result.

Exponentiation

Exponentiation is the process of raising a number to a power, which signifies how many times to multiply the number by itself. It's important when simplifying expressions during function evaluation. In the expression h(r), the term r^2 indicates that whatever value r holds should be squared.

For instance, for h(4), we evaluate (4)^2 by multiplying 4 by itself to get 16. Exponentiation takes precedence over multiplication and addition in the order of operations. When dealing with negative bases, like in h(-1), remember that a negative number raised to an even exponent results in a positive number; hence, (-1)^2 becomes 1.

Substitution in Algebra

Substitution is a method used in algebra to replace variables with numbers or other expressions. It's particularly useful in function evaluation, allowing us to find the value of functions at specific points. For example, in evaluating h(r) for r = 4 and r = -1, we substitute 4 and -1 for r respectively.

This technique not only helps us solve for specific cases but also illustrates how functions behave under different inputs. It's essential to substitute correctly and to carefully follow through with the order of operations to arrive accurately at solutions such as h(4) = 53 and h(-1) = 8.