Problem 15 - Algebra and Trigonometry

Question

Evaluate the function at the indicated values. $$ \begin{array}{l}{g(x)=\frac{1-x}{1+x}} \\ {g(2), g(-2), g\left(\frac{1}{2}\right), g(a), g(a-1), g(-1)}\end{array} $$

Step-by-Step Solution

Answer

g(2) = -\frac{1}{3}, g(-2) = -3, g\left(\frac{1}{2}\right) = \frac{1}{3}, g(a) = \frac{1-a}{1+a}, g(a-1) = \frac{2-a}{a}, g(-1) is undefined.

Step 1: Identify the Function Expression

The function given is \( g(x) = \frac{1-x}{1+x} \). We will substitute different values of \( x \) into this expression to evaluate the function at each of the specified points.

Step 2: Evaluate Function at x = 2

Substitute \( x = 2 \) into the function: \[ g(2) = \frac{1-2}{1+2} = \frac{-1}{3} \].Thus, \( g(2) = -\frac{1}{3} \).

Step 3: Evaluate Function at x = -2

Substitute \( x = -2 \) into the function: \[ g(-2) = \frac{1 - (-2)}{1 + (-2)} = \frac{1 + 2}{1 - 2} = \frac{3}{-1} = -3 \].Thus, \( g(-2) = -3 \).

Step 4: Evaluate Function at x = \( \frac{1}{2} \)

Substitute \( x = \frac{1}{2} \) into the function: \[ g\left(\frac{1}{2}\right) = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} \].Thus, \( g\left(\frac{1}{2}\right) = \frac{1}{3} \).

Step 5: Evaluate Function at x = a

Substitute \( x = a \) into the function: \[ g(a) = \frac{1-a}{1+a} \].Thus, \( g(a) = \frac{1-a}{1+a} \).

Step 6: Evaluate Function at x = a - 1

Substitute \( x = a-1 \) into the function: \[ g(a-1) = \frac{1 - (a-1)}{1 + (a-1)} = \frac{2-a}{a} \].Thus, \( g(a-1) = \frac{2-a}{a} \).

Step 7: Evaluate Function at x = -1

Substitute \( x = -1 \) into the function: \[ g(-1) = \frac{1 - (-1)}{1 + (-1)} = \frac{2}{0} \], which is undefined.Thus, \( g(-1) \) is undefined.

Key Concepts

Rational Functions

Rational functions are fractions where the numerator and the denominator are both polynomials. In our exercise, the function is denoted as \( g(x) = \frac{1-x}{1+x} \). Here, both the numerator, \( 1-x \), and the denominator, \( 1+x \), are linear polynomials. Rational functions can exhibit a variety of behaviors, such as having asymptotes or holes in their graphs. In a rational function, the denominator must never be zero, as division by zero is undefined in mathematics. This makes identifying the values for which the denominator becomes zero crucial when evaluating and plotting these functions.

A rational function will be discontinuous or undefined at points where its denominator is equal to zero. In our function, observe how changing \( x \) affects the function's output. For example, substituting \( x = -1 \) makes the denominator zero, hence rendering the function undefined at this point.

Substitution Method

The substitution method involves replacing the variable in a function with specific values to evaluate it. In the context of our problem, we're substituting different values of \( x \) into the given rational function \( g(x) = \frac{1-x}{1+x} \). The goal is to simplify the expression and find the corresponding output for each given \( x \) value.

Here are general steps for using substitution in function evaluation:

• Identify the function and the specific value of \( x \) for evaluation.
• Replace \( x \) with the specific value in the function expression.
• Simplify the resulting expression to get the output.

For instance, in the given exercise, substituting \( x = 2 \) into the function results in the calculation \( g(2) = \frac{1-2}{1+2} = -\frac{1}{3} \). This method is straightforward but requires careful algebraic manipulation, especially for rational functions.

Function Undefined

A function becomes undefined when certain values lead to an expression that cannot be calculated. With rational functions, this occurs when the denominator equals zero. As highlighted earlier, division by zero is not possible, creating points of discontinuity in the function. Such points can be identified before evaluation by setting the denominator equal to zero and solving for \( x \).

Considering the function \( g(x) = \frac{1-x}{1+x} \), we find that when \( x = -1 \), the denominator becomes zero, making \( g(-1) \) undefined. Identifying and understanding these undefined points is essential if one wants to correctly sketch the graph of a rational function or avoid errors during evaluation.

In practice, finding where a function might be undefined involves:

• Setting the denominator to zero.
• Solving the equation to find the critical values.
• Avoiding these values while performing substitutions to ensure valid calculations.

Being aware of undefined points is crucial for accurate mathematical analysis and interpretation of rational functions.