Problem 33 - Elementary and Intermediate Algebra

Question

Let \(s(x)=3-x\) and \(t(x)=x^{2}-x-6 .\) Find each function value. $$ (s \cdot t)(-2) $$

Step-by-Step Solution

Answer

\((s \cdot t)(-2) = 0\)

Step 1: Identify the Composition

The problem asks for the value of \(s \cdot t\)(-2). This means we will multiply the outputs of the functions \s(x)\ and \t(x)\ for the input \x = -2\. We need to evaluate each function separately at \x = -2\.

Step 2: Evaluate \(s(-2)\)

Substitute \x = -2\ into \s(x) = 3 - x\.\(s(-2) = 3 - (-2)\)Calculate:\(s(-2) = 3 + 2 = 5\)

Step 3: Evaluate \(t(-2)\)

Substitute \x = -2\ into \t(x) = x^2 - x - 6\.\(t(-2) = (-2)^2 - (-2) - 6\)Calculate:\((-2)^2 = 4\)\(-(-2) = 2\)So,\(t(-2) = 4 + 2 - 6 = 0\)

Step 4: Compute \((s \, \cdot \, t)(-2)\)

After evaluating \(s(-2) = 5\) and \(t(-2) = 0\), we calculate:\[(s \, \cdot \, t)(-2) = s(-2) \, \times \, t(-2) = 5 \, \times \, 0 \]The result is:\[(s \, \cdot \, t)(-2) = 0\]

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions that consist of variables raised to whole number powers, coefficients, and constants, which are added or subtracted together. An example of a polynomial is the function \( t(x) = x^2 - x - 6 \). This function is a quadratic polynomial because its highest degree term is \( x^2 \). Quadratic polynomials are characterized by a parabolic graph shape.

Polynomials are very versatile and widely used in mathematical modeling. They can model physical phenomena and relationships in a variety of fields, such as physics and economics. As a polynomial function's degree increases, so does the curve's number of turns and level of complexity. Higher-degree polynomials may exhibit more complex behaviors, but they still follow predictable patterns. With practice, understanding how to handle polynomial functions becomes a valuable skill in problem-solving.

Function Evaluation

Function evaluation involves substituting a specific value for the variable in a function's expression. In the original exercise, both \( s(x) = 3 - x \) and \( t(x) = x^2 - x - 6 \) needed to be evaluated for \( x = -2 \).

To evaluate a function, simply replace the variable with the given value and perform arithmetic operations as needed. For example, to evaluate \( s(-2) \), substitute \(-2\) into \(s(x)\):

• Substitute to get: \(s(-2) = 3 - (-2)\)
• Calculate: \(s(-2) = 3 + 2 = 5\)

Similarly, for \( t(-2) \):

• Substitute to get: \(t(-2) = (-2)^2 - (-2) - 6\)
• Simplify: \(t(-2) = 4 + 2 - 6 = 0\)

Function evaluation is a fundamental skill, facilitating the calculation of function values for specific arguments, which is crucial in graphing and further analysis.

Algebraic Expressions

Algebraic expressions are combinations of variables, coefficients, and constants. They form the building blocks of more complex mathematical functions like polynomials. Each part of an algebraic expression needs to be accurately computed to ensure correct results.

In the exercise, the algebraic expressions \( 3 - x \) and \( x^2 - x - 6 \) were key. Within them, operations like addition, subtraction, multiplication, and exponentiation are used. By applying rules of arithmetic, algebraic expressions can be manipulated and simplified.

Clear understanding of algebraic expressions facilitates solving equations, finding function values, and further composing functions, such as in our example with \((s \cdot t)(-2)\). Each step—substituting values, then calculating output—traces back to correctly handling algebraic expressions. Grasping the interaction among components of an expression ensures your calculated results are reliable.