Problem 35 - Elementary and Intermediate Algebra

Question

Let \(s(x)=3-x\) and \(t(x)=x^{2}-x-6 .\) Find each function value. $$ (s / t)(1) $$

Step-by-Step Solution

Answer

(s/t)(1) = -1/3.

Step 1: Understand the task

We are asked to evaluate the function value of \((s/t)(1)\), where \(s(x)=3-x\) and \(t(x)=x^2-x-6\). This means we need to substitute \(x=1\) into the expressions for \(s(x)\) and \(t(x)\), then divide the results.

Step 2: Evaluate s(1)

Substitute \(x=1\) into the function \(s(x)=3-x\). This leads to:\[s(1) = 3 - 1 = 2.\] So, \(s(1) = 2\).

Step 3: Evaluate t(1)

Substitute \(x=1\) into the function \(t(x) = x^2 - x - 6\). Calculate as follows:\[t(1) = (1)^2 - 1 - 6 = 1 - 1 - 6 = -6.\] So, \(t(1) = -6\).

Step 4: Compute (s/t)(1)

Now that we have \(s(1)=2\) and \(t(1)=-6\), we find \((s/t)(1)\) by dividing the two results:\[(s/t)(1) = \frac{s(1)}{t(1)} = \frac{2}{-6} = -\frac{1}{3}.\]

Key Concepts

Function Evaluation

When we talk about function evaluation, we're referring to the process of finding the value of a function for a specific input. It’s like finding out the result of a machine when you input a particular value.

You start by taking a function, like in our example, where we have two functions,

• Function s(x) = 3 - x
• Function t(x) = x^2 - x - 6

To evaluate a function, such as s(x) at a specific point, say x = 1, you replace every occurrence of x in the equation with 1.
In our case, this would be:

• s(1) = 3 - 1 = 2

Basically, you're asking, "If x is 1, what does the function s output?" Similarly, you do the same steps for t(x) to get t(1). This concept is fundamental in algebra as it enables you to find specific values that a function assumes for given inputs. This is especially useful in various applications such as physics, economics, and engineering.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials.

In our example, \((s / t)(x) \) is a rational expression because \(s(x)\)and \(t(x)\)are both polynomials.
These expressions are similar to regular fractions, as you can add, subtract, multiply, and divide them. However, because polynomials can have variables, things get a bit more complex.
Rational expressions give you a way to work with ratios of functions, which is very powerful in both algebra and calculus when finding solutions or simplifications.
It’s crucial to ensure the denominator is not zero, as dividing by zero is undefined in mathematics.
This adds an extra step when working with rational expressions: checking that the denominator hasn’t become zero when substituting specific values.
In our problem, we evaluated \((s/t)(1)\), making sure \(t(1)\)did not equal zero.

Substitution Method

The substitution method is a technique in algebra used to replace variables with numbers or other expressions. This approach is particularly handy for simplifying complex expressions or solving equations.

To use substitution effectively:

• Identify the variable you want to substitute.
• Replace the variable with the given number or other expression.

In our original exercise, the task was to evaluate the function values at a specific point, x = 1. This means wherever we saw "x" in the functions \(s(x)\)and \(t(x)\), we swapped it for 1.

By substituting, we turned these algebraic expressions into straightforward arithmetic problems. For instance:

• For \(s(x) = 3-x\), substituting x with 1 gives \(s(1) = 3-1 = 2\).
• For \(t(x) = x^2-x-6\), substituting x with 1 gives \(t(1) = 1^2 - 1 - 6 = -6\).

Substitution is a foundational skill in algebra that simplifies analyzing functions and solving a variety of problems in mathematics.