Problem 30 - College Algebra with Corequisite Support

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

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

Step 1: Evaluate f at -3

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

Step 2: Evaluate f at 2

Substitute \(2\) for \(x\) in the function \(f(x) = \frac{6x - 1}{5x + 2}\).\[f(2) = \frac{6(2) - 1}{5(2) + 2} = \frac{12 - 1}{10 + 2} = \frac{11}{12}\]

Step 3: Evaluate f at -a

Substitute \(-a\) for \(x\) in the function \(f(x) = \frac{6x - 1}{5x + 2}\).\[f(-a) = \frac{6(-a) - 1}{5(-a) + 2} = \frac{-6a - 1}{-5a + 2}\]

Step 4: Evaluate -f(a)

First, find \(f(a)\) by substituting \(a\) for \(x\):\[f(a) = \frac{6a - 1}{5a + 2}\]Then take the negative of this result:\[-f(a) = -\frac{6a - 1}{5a + 2}\]

Step 5: Evaluate f(a+h)

Substitute \(a + h\) for \(x\) in the function \(f(x) = \frac{6x - 1}{5x + 2}\).\[f(a+h) = \frac{6(a + h) - 1}{5(a + h) + 2} = \frac{6a + 6h - 1}{5a + 5h + 2}\]

Key Concepts

Algebraic Functions

Algebraic functions are a pivotal concept in algebra that involves the use of variables and constants. These functions are expressed in the form of algebraic expressions, which may include operators like addition, subtraction, multiplication, division, and even exponents. In our function, \(f(x) = \frac{6x - 1}{5x + 2}\), we observe an algebraic function involving both a numerator and a denominator.
The variables, in this case, are subject to operations as defined by the function. Constants play a crucial role, often determining the slope or the y-intercept of linear functions. In more complex expressions, they serve as coefficients that modify the variable's contribution to the overall expression.
With algebraic functions, one of the primary goals is to understand how substituting different values for the variable changes the outcome of the function. This is crucial for solving real-world problems and understanding relationships between varying quantities.

Substitution Method

The substitution method is a standard procedure used to evaluate functions like the one in our exercise, \(f(x) = \frac{6x - 1}{5x + 2}\). To substitute means to replace the variable in the function with a given number or expression.
For instance, if we need to determine \(f(-3)\), we substitute \(-3\) in place of \(x\) throughout the function. This substitution process allows us to find the value of the function at specific points or for specific expressions.
This method is not only limited to numeric values but can also be extended to include expressions, as seen with \(f(a+h)\). Here, the expression \(a+h\) replaces \(x\), necessitating a careful simplification to ensure accuracy. Substitution is a critical strategy in problem-solving, enabling students to determine the practical significance of functions in different contexts.

Rational Expressions

Rational expressions, such as \(f(x) = \frac{6x - 1}{5x + 2}\), play a vital role in algebra. These are fractions where the numerator and/or the denominator are polynomials. They offer a versatile means to represent real-world situations where relationships between quantities need to be expressed as ratios.
Handling rational expressions involves understanding the domain restrictions imposed by the denominator. Specifically, the denominator cannot be zero as this would render the expression undefined. In our exercise, the denominator is \(5x + 2\), which means \(x\) cannot take a value that causes this expression to become zero. Solving \(5x + 2 = 0\) gives a restricted value for \(x\), ensuring the rational expression remains valid.
These expressions often require additional algebraic manipulation, including factoring and simplifying, which helps in comparing and combining rational expressions effectively.

Problem-Solving in Algebra

Problem-solving in algebra is about applying a series of logical steps to work through an equation or expression to reach a solution. The exercise involving \(f(x) = \frac{6x - 1}{5x + 2}\) illustrates a typical algebraic problem-solving scenario.
Each step requires precise calculations and substitutions to evaluate the function at various values like \(-3\), \(2\), and \(-a\). It's crucial to execute these steps systematically to avoid errors. Checking each calculation step ensures accuracy, which is paramount for achieving the correct solution.
Moreover, with expressions like \(f(a+h)\), problem-solving extends to include algebraic simplifications that integrate multiple variables and terms. These steps enhance skills like critical thinking and demonstrate the interplay of various algebraic concepts like rational expressions and function evaluation, proving invaluable in both academic and real-life problem scenarios.