Problem 45 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

a) -5.5, b) \(\frac{1}{2}b^2 + 3b - 1\), c) \(\frac{1}{2}x^6 + 3x^3 - 1\), d) \(2x^2 - 5.5\).

Step 1: Evaluate g(-3)

Substitute \(-3\) for \(x\) in \(g(x)\). The function is \(g(x) = \frac{1}{2}x^2 + 3x - 1\). Calculating: \[g(-3) = \frac{1}{2}(-3)^2 + 3(-3) - 1\]. First, compute \((-3)^2 = 9\). Then, \(\frac{1}{2} \times 9 = 4.5\). Next, \(3 \times (-3) = -9\). Now, add them together: \[4.5 + (-9) - 1 = -4.5 - 1 = -5.5\]. Thus, \(g(-3) = -5.5\).

Step 2: Evaluate g(b)

Substitute \(b\) for \(x\) in \(g(x)\). The function becomes \(g(b) = \frac{1}{2}b^2 + 3b - 1\). Thus, \(g(b)\) is simply left in terms of \(b\) as \(\frac{1}{2}b^2 + 3b - 1\).

Step 3: Evaluate g(x^3)

Substitute \(x^3\) for \(x\) in \(g(x)\). So, \[g(x^3) = \frac{1}{2}(x^3)^2 + 3(x^3) - 1\]. Next, calculate \((x^3)^2 = x^6\). Therefore, \[g(x^3) = \frac{1}{2}x^6 + 3x^3 - 1\].

Step 4: Evaluate g(2x-3)

Substitute \(2x-3\) for \(x\) in \(g(x)\). Calculate: \[g(2x-3) = \frac{1}{2}(2x-3)^2 + 3(2x-3) - 1\]. First, expand \((2x-3)^2 = 4x^2 - 12x + 9\). Then, substitute back: \[g(2x-3) = \frac{1}{2}(4x^2 - 12x + 9) + 6x - 9 - 1\]. Simplify each term: \[g(2x-3) = 2x^2 - 6x + 4.5 + 6x - 9 - 1\]. Finally, add like terms: \[g(2x-3) = 2x^2 - 5.5\].

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial function distinguished by their degree, which is always 2. These functions have the general form: \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the sign of \( a \).

In our example, the quadratic function \( g(x) = \frac{1}{2}x^2 + 3x - 1 \) includes a leading coefficient of \( \frac{1}{2} \), making the parabola open upwards. Understanding the parts of a quadratic function is crucial: the \( x^2 \) term determines the shape, the \( x \) term affects the slope, and the constant term \( c \) adjusts the position vertically. Quadratics can model numerous real-life situations such as projectile motion and optimizing areas.

Substitution Method

The substitution method is a mathematical technique used to evaluate functions by replacing the variable with a specific number, expression, or another variable. This is particularly useful when you need to simplify the expression or solve for specific values.

In our exercise, we use substitution to find values like \( g(-3) \), where \(-3\) directly replaces \( x \) in \( g(x) = \frac{1}{2}x^2 + 3x - 1 \). Here, substitution allowed us to compute \( g(-3) = -5.5 \).

• Step one is replacing \( x \) with the given value.
• Step two involves following the order of operations (PEMDAS/BODMAS) to simplify.

This method is essential for evaluating and graphing various types of functions, making it an invaluable tool for algebraic problem-solving. The substitution method can also be extended to substitute expressions, as shown in \( g(x^3) \) and \( g(2x-3) \).

Polynomial Functions

Polynomial functions form a broad class of mathematical expressions that encompass quadratic functions and many others. A polynomial function is an expression of the form \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \), where \( n \) is a non-negative integer.

In the context of polynomial functions, each term consists of a coefficient and a power of \( x \). Each polynomial's degree is determined by the highest power of \( x \). The example \( g(x) = \frac{1}{2}x^2 + 3x - 1 \) is a second-degree polynomial (a quadratic) because the highest power is \( x^2 \).

• Polynomial functions can vary greatly in shape and complexity.
• They can have multiple variables, though this exercise focuses on univariate polynomials.

Recognizing and manipulating polynomial functions is a foundational skill in algebra, as they frequently appear in calculus and real-world applications, such as modeling growth rates and financial predictions. Understanding polynomial behavior—roots, shape, and intersections—is critical for deeper mathematical studies.