Problem 30 - College Algebra

Question

For the following exercises, evaluate the function \(f\) at the indicated values \(f(-3), f(2), f(-a),-f(a), f(a+h)\). $$ f(x)=\frac{6 x-1}{5 x+2} $$

Step-by-Step Solution

Answer

The evaluations are: \( f(-3) = \frac{19}{13} \), \( f(2) = \frac{11}{12} \), \( f(-a) = \frac{-6a - 1}{-5a + 2} \), \(-f(a) = \frac{-6a + 1}{5a + 2} \), \( f(a+h) = \frac{6a + 6h - 1}{5a + 5h + 2} \).

Step 1 - Evaluate at x = -3

Substitute -3 into the function:\[f(-3) = \frac{6(-3) - 1}{5(-3) + 2} = \frac{-18 - 1}{-15 + 2} = \frac{-19}{-13} = \frac{19}{13}\]

Step 2 - Evaluate at x = 2

Substitute 2 into the function:\[f(2) = \frac{6(2) - 1}{5(2) + 2} = \frac{12 - 1}{10 + 2} = \frac{11}{12}\]

Step 3 - Evaluate at x = -a

Substitute -a into the function:\[f(-a) = \frac{6(-a) - 1}{5(-a) + 2} = \frac{-6a - 1}{-5a + 2}\]

Step 4 - Evaluate at -f(a)

First find \( f(a) \):\[f(a) = \frac{6a - 1}{5a + 2}\]Then, evaluate \(-f(a)\):\[-f(a) = - \left( \frac{6a - 1}{5a + 2} \right) = \frac{-(6a - 1)}{5a + 2} = \frac{-6a + 1}{5a + 2}\]

Step 5 - Evaluate at x = a + h

Substitute \(a + h\) into the function:\[f(a + h) = \frac{6(a + h) - 1}{5(a + h) + 2} = \frac{6a + 6h - 1}{5a + 5h + 2}\]

Key Concepts

Rational Functions

A rational function is essentially a fraction in which the numerator and the denominator are both polynomials. In the function given in this exercise, \( f(x) = \frac{6x-1}{5x+2} \), both parts of the fraction are first-degree polynomials:

• The numerator is \( 6x - 1 \)
• The denominator is \( 5x + 2 \)

Rational functions are important in mathematics because they can model a wide variety of behaviors as \( x \) changes. One key feature is that they are undefined when the denominator is zero, which gives these functions their characteristic behavior near critical points. Understanding this concept helps in analyzing the behavior of the function under different circumstances.

Substitution Method

The substitution method involves replacing a variable in an expression with a specific value to simplify and evaluate the expression. This process is straightforward:

• Take the variable, in this case \( x \).
• Replace or substitute it with a given number or expression.
• Perform the arithmetic operations to solve the expression.

For example, in step 1 of this exercise, the substitution method was used by replacing \( x \) with \(-3\) to find \( f(-3) \). Each time, substitute the given values into the function and simplify, allowing us to understand the behavior of function at different points.

Algebraic Expressions

Algebraic expressions are essentially mathematical phrases that can contain numbers, variables, and arithmetic operations like addition, subtraction, multiplication, and division. In the rational function \( f(x) \), both the numerator and the denominator are algebraic expressions. To evaluate these expressions, one should:

• Understand how variables stand in for numbers.
• Recognize how arithmetic operations interact within the expression.

When expressions include variables, the structure might become complex, but following the basic arithmetic rules makes simplification more systematic.

Function Operations

Function operations refer to the various ways you can manipulate functions, such as through addition, subtraction, multiplication, division, and composition. In this problem, you also see operations like evaluating

• Negative values for functions like \(-f(a)\), which involves negating the expression after it has been evaluated.
• Using addition within brackets as seen in \(f(a+h)\), where the input becomes an expression itself.

These operations help in understanding how functions react to different inputs and modifications, offering insights into broader function behaviors.