Problem 42 - College Algebra with Modeling and Visualization

Question

Exercises \(41-52:\) For the given \(g(x),\) evaluate each of the following. $$ \begin{array}{lllll} \text { (a) } g(-3) & \text { (b) } g(b) & \text { (c) } g\left(x^{3}\right) & \text { (d) } g(2 x-3) \end{array} $$ $$ g(x)=5-\frac{1}{2} x $$

Step-by-Step Solution

Answer

(a) 6.5; (b) \(5 - \frac{1}{2}b\); (c) \(5 - \frac{1}{2}x^3\); (d) \(6.5 - x\)."

Step 1: Evaluate g(-3)

To find \( g(-3) \), substitute \( -3 \) into the function \( g(x) = 5 - \frac{1}{2}x \). This gives us: \[ g(-3) = 5 - \frac{1}{2}(-3) \] Simplify by calculating: \[ g(-3) = 5 + \frac{3}{2} = 5 + 1.5 = 6.5\] So, \( g(-3) = 6.5 \).

Step 2: Evaluate g(b)

For \( g(b) \), substitute \( b \) for \( x \) in the function: \[ g(b) = 5 - \frac{1}{2}b \] Since there's no further simplification possible without a specific value for \( b \), we state the result as \( g(b) = 5 - \frac{1}{2}b \).

Step 3: Evaluate g(x^3)

To find \( g(x^3) \), substitute \( x^3 \) into \( g(x) \): \[ g(x^3) = 5 - \frac{1}{2}x^3 \] Thus, the expression for \( g(x^3) \) is \( 5 - \frac{1}{2}x^3 \).

Step 4: Evaluate g(2x-3)

Substitute \(2x - 3\) into the function: \[ g(2x-3) = 5 - \frac{1}{2}(2x - 3) \] Distribute \( -\frac{1}{2} \) to both terms in the parentheses: \[ g(2x-3) = 5 - x + \frac{3}{2} \] Combine like terms: \[ g(2x-3) = 5 + \frac{3}{2} - x = 6.5 - x \] So, \( g(2x-3) = 6.5 - x \).

Key Concepts

Substitution

Substitution is a technique in algebra where you replace a variable in an expression with a given value. This is essential for evaluating functions like linear ones. Imagine you have a function, such as \( g(x) = 5 - \frac{1}{2}x \). When you're asked to find \( g(-3) \), you're expected to substitute \(-3\) in place of every \(x\) in the function.

Steps to Substitute:

• Identify the function and the value to substitute.
• Replace every instance of the variable in the function with the given number.
• Calculate the expression's value by following proper arithmetic operations.

For instance, by substituting \(-3\) into our function, you get \( g(-3) = 5 - \frac{1}{2}(-3) \), leading to \( g(-3) = 6.5 \). Substitution helps in determining specific outputs of a function, giving insight into how changes in the input affect the output.

Linear Functions

Linear functions are a central concept in algebra and are characterized by straight-line graphs. The general form of a linear function is \( g(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In the specific function you're exploring, \( g(x) = 5 - \frac{1}{2}x \), it's presented in terms of subtraction, which can be restructured to the common form \( -\frac{1}{2}x + 5 \).

Understanding Linear Functions:

• They represent relationships with constant rates of change.
• The slope indicates the rate of change of the function.
• The y-intercept shows where the line crosses the y-axis.

Linear functions are fundamental in predicting outcomes and analyzing data since they describe direct proportional relationships between variables. For example, as \( x \) increases, the value of \( g(x) \) will decrease if the slope is negative, just like in this function.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operations. Evaluating these expressions involves understanding how to handle elements like terms and coefficients. In functions like \( g(x) = 5 - \frac{1}{2}x \), it's crucial to be able to manipulate and simplify such expressions.

Breaking Down Algebraic Expressions:

• Observe each term in the expression separately.
• Combine like terms for simplification.
• Apply multiplication, division, addition, or subtraction as per order of operations.

An example is simplifying \( g(2x-3) = 5 - \frac{1}{2}(2x-3) \). You distribute \(-\frac{1}{2}\), which modifies the expression to \( g(2x-3) = 6.5 - x \). Mastery over algebraic expressions allows you to carry out operations effectively and solve more complex mathematical problems.