Problem 48 - Precalculus with Limits

Question

In Exercises 43-50, evaluate the function for the indicated values. \(k(x) = [[\frac{1}{2}x + 6]]\) (a) \(k(5)\) (b) \(k(-6.1)\) (c) \(k(0.1)\) (d) \(k(15)\)

Step-by-Step Solution

Answer

The answers for the given exercise are (a) k(5) = 8.5, (b) k(-6.1) = 2.95, (c) k(0.1) = 6.05, and (d) k(15) = 13.5.

Step 1: Evaluate k(5)

Substitute x = 5 in the function k(x) = 1/2x + 6. Thus, k(5) = 1/2 * 5 + 6 = 2.5 + 6 = 8.5.

Step 2: Evaluate k(-6.1)

Substitute x = -6.1 in the function k(x) = 1/2x + 6. Thus, k(-6.1) = 1/2 * -6.1 + 6 = -3.05 + 6 = 2.95.

Step 3: Evaluate k(0.1)

Substitute x = 0.1 in the function k(x) = 1/2x + 6. Thus, k(0.1) = 1/2 * 0.1 + 6 = 0.05 + 6 = 6.05.

Step 4: Evaluate k(15)

Substitute x = 15 in the function k(x) = 1/2x + 6. Thus, k(15) = 1/2 * 15 + 6 = 7.5 + 6 = 13.5.

Key Concepts

Understanding Linear Functions

A linear function is one of the most fundamental concepts in algebra. It is a function whose graph is a straight line, which means that there is a constant rate of change. Typically, it is expressed in the form of:

• \( y = mx + c \)

where \( m \) represents the slope of the line and \( c \) denotes the y-intercept—that is, where the line crosses the y-axis.
Linear functions are very predictable because as \( x \) increases by one unit, \( y \) changes by the slope \( m \) consistently. For example, in the function given in the problem, \( k(x) = \frac{1}{2}x + 6 \), the slope \( m \) is \( \frac{1}{2} \) and the y-intercept \( c \) is 6.
This means every increase of 1 in \( x \) will increase \( k(x) \) by \( \frac{1}{2} \). Understanding the role of \( m \) and \( c \) helps in predicting how the line behaves and assists in plotting the function on a graph.

The Substitution Method in Function Evaluation

The substitution method is a practical way to evaluate functions, especially when you have specific values you need to compute. This method involves replacing the variable \( x \) in the function with the given numerical value.
In the function \( k(x) = \frac{1}{2}x + 6 \), to find \( k(5) \), you substitute \( x \) with 5. So it becomes:

• \( k(5) = \frac{1}{2} \times 5 + 6 \)
• This simplifies to \( 2.5 + 6 = 8.5 \)

By solving the expression, you evaluate the function at that specific point. This process is repeated for every substitution required, as demonstrated for values like \( k(-6.1) \), \( k(0.1) \), and \( k(15) \).
This method helps to determine the specific output or function value for given input values, providing a concrete way to see how changes in \( x \) affect \( k(x) \).

Creating and Simplifying Algebraic Expressions

Algebraic expressions are fundamental in representing mathematical relationships and functions. They consist of numbers, variables, and the arithmetic operations that connect them, such as addition, subtraction, multiplication, and division.
In the context of the given problem, the expression \( \frac{1}{2}x + 6 \) is algebraic. To evaluate this expression, you perform operations like multiplication of fractions and addition:

• First, multiply the given \( x \) value by \( \frac{1}{2} \)
• Then, add 6 to the result

These expressions can be simplified to find concrete answers.
For instance, to simplify \( \frac{1}{2} \times 5 + 6 \), you multiply \( 5 \) by \( \frac{1}{2} \) giving \( 2.5 \), and then add 6 to get a final answer of 8.5.
Mastering algebraic expressions helps in translating everyday problems into mathematical language, making them easier to solve.