Problem 33 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to find the value of the product of two functions, \((s \cdot t)(-2)\), which means we need to evaluate both functions \(s(x)\) and \(t(x)\) at \(x = -2\) and then multiply the results together.

Step 2: Evaluate the Function \(s(x)\)

The function given is \(s(x) = 3 - x\). Substitute \(x = -2\) into the function:\[ s(-2) = 3 - (-2) = 3 + 2 = 5. \]

Step 3: Evaluate the Function \(t(x)\)

The function given is \(t(x) = x^2 - x - 6\). Substitute \(x = -2\) into the function:\[t(-2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0.\]

Step 4: Multiply the Results

Now multiply the results from Step 2 and Step 3 to find \((s \cdot t)(-2)\):\[(s \cdot t)(-2) = s(-2) \cdot t(-2) = 5 \cdot 0 = 0.\]

Step 5: Final Result

The value of \((s \cdot t)(-2)\) is 0.

Key Concepts

Function Evaluation

Function evaluation is a vital concept in mathematics that involves substituting a specific value of the variable into a given function. This process is crucial for understanding how the function behaves at certain points. Suppose we have a function like \( s(x) = 3 - x \). If we're asked to evaluate \( s(x) \) at \( x = -2 \), we replace every occurrence of \( x \) in the equation with \( -2 \).
The calculation would be:

• \( s(-2) = 3 - (-2) \)
• This simplifies to \( 3 + 2 \)
• So, \( s(-2) = 5 \).

Once the specific value is substituted and calculated, you have the function's value at \( x = -2 \). This simple substitution technique is foundational for more complex evaluations involving other functions or polynomials.

Polynomial Functions

A polynomial function is an expression made up of constants and variables, combined using only addition, subtraction, multiplication, and non-negative power laws of variables. Polynomial functions are named based on their degree, the highest power of the variable present in the expression.
For example, the function \( t(x) = x^2 - x - 6 \) is a quadratic polynomial because the highest power of \( x \) is 2.

• The term \( x^2 \) defines it as a quadratic.
• The function also includes linear terms like \( -x \), and constant terms like \( -6 \).

These polynomials are widely used as they allow us to easily model various real-world phenomena and solve algebraic problems. Evaluating a polynomial function involves substituting a value into the polynomial and simplifying the result, just like in function evaluation, as shown in the steps for \( t(-2) \).

Algebraic Expressions

An algebraic expression is a mathematical phrase composed of numbers, variables, and operational symbols. Unlike equations, expressions don't have an equality sign and are often simplified or evaluated to find their values.
Consider an expression like \( 3 - x \), which is part of the function \( s(x) = 3 - x \). Each term in the expression can include:

• Constants: numbers not multiplied by variables (like \( 3 \))
• Variables: symbols that can take various values (like \( x \))
• Coefficients: numbers multiplied by variables (though not explicitly shown here)

Algebraic expressions can become complex, with multiple terms combined using addition or subtraction, like in \( x^2 - x - 6 \). Understanding these components is essential for evaluating expressions, simplifying them, and solving algebraic equations.