Problem 21 - 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(\frac{1}{2}) = \frac{1}{3}, g(a) = \frac{1-a}{1+a}, g(a-1) = \frac{2-a}{a}, g(-1) \text{ is undefined}."

Step 1: Evaluate g(2)

Plug in the value 2 into the function.\[ g(2) = \frac{1-2}{1+2} = \frac{-1}{3} \] Thus, \( g(2) = -\frac{1}{3} \).

Step 2: Evaluate g(-2)

Plug in the value -2 into the function.\[ g(-2) = \frac{1-(-2)}{1+(-2)} = \frac{1+2}{1-2} = \frac{3}{-1} \] Thus, \( g(-2) = -3 \).

Step 3: Evaluate g(\frac{1}{2})

Plug in the value \( \frac{1}{2} \) 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 4: Evaluate g(a)

Plug in \( a \) into the function without altering the expression.\[ g(a) = \frac{1-a}{1+a} \] Therefore, \( g(a) \) remains \( \frac{1-a}{1+a} \).

Step 5: Evaluate g(a-1)

Plug in \( a-1 \) 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 6: Evaluate g(-1)

Plug in \(-1\) into the function. Notice that the denominator becomes zero.\[ g(-1) = \frac{1-(-1)}{1+(-1)} = \frac{2}{0} \] The function is undefined at \( g(-1) \).

Key Concepts

Function Evaluation

Function evaluation is the process of finding the output of a function for specific input values. In this context, the given function is \( g(x) = \frac{1-x}{1+x} \), and the goal is to find \( g(2) \), \( g(-2) \), \( g\left( \frac{1}{2} \right) \), \( g(a) \), \( g(a-1) \), and \( g(-1) \).To evaluate a function:

• Select the specific input value.
• Substitute this value into the function in place of the variable.
• Simplify the resulting expression to find the output.

This helps in understanding how the function behaves at different points. It is crucial to simplify carefully to avoid mistakes, especially when dealing with fractions or more complex expressions.

Variable Substitution

Variable substitution is replacing the variable in an expression with a specific value or another expression. In the context of the function \( g(x) = \frac{1-x}{1+x} \), it involves replacing \( x \) with given values like \( 2 \) or expressions like \( a-1 \).Each substitution follows a straightforward approach:

• Identify the variable in the function (here \( x \)).
• Replace \( x \) with the given value or expression wherever it appears.
• Re-calculate to get the result.

Substitution helps in converting a function into a form where it can be directly evaluated or further analyzed.

Undefined Expressions

An undefined expression occurs when the mathematical process results in an operation that isn't allowed, such as division by zero. This concept is crucial when dealing with rational functions like \( g(x) = \frac{1-x}{1+x} \).When evaluating \( g(-1) \) for instance:

• The denominator becomes \( 1 + (-1) = 0 \).
• Division by zero is undefined in real numbers.

In such cases, the function does not provide a valid output, indicating a point of discontinuity or a break in the graph of the function. Knowing where a function is undefined helps in graphing and understanding its complete behavior.

Rational Functions

Rational functions are expressions created by dividing two polynomials. In the given exercise, \( g(x) = \frac{1-x}{1+x} \) is a rational function where both the numerator and the denominator are polynomials of degree one.Key characteristics of rational functions include:

• They may have vertical asymptotes where the denominator is zero but the numerator is non-zero, indicating undefined points.
• The value of the function approaches infinity or minus infinity near vertical asymptotes.
• They can include holes, removable discontinuities, if the same factor cancels out in both the numerator and denominator, leaving behind a reduced form.

Understanding rational functions involves considering the domain (all permissible values of \( x \)) and identifying any points of discontinuity where the function is not defined.