Problem 6 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 5-10: If \(a\) and \(h\) are real numbers, find (a) \(f(a)\) (b) \(f(-a)\) (c) \(-f(a)\) (d) \(f(a+h)\) (e) \(f(a)+f(h)\) (f) \(\frac{f(a+h)-f(a)}{h}\), if \(h \neq 0\) $$ f(x)=3-4 x $$

Step-by-Step Solution

Answer

(a) 3 - 4a, (b) 3 + 4a, (c) -3 + 4a, (d) 3 - 4a - 4h, (e) 6 - 4a - 4h, (f) -4.

Step 1: Evaluate f(a)

To find \( f(a) \), substitute \( x = a \) into the function \( f(x) = 3 - 4x \). This gives \( f(a) = 3 - 4a \).

Step 2: Evaluate f(-a)

To find \( f(-a) \), substitute \( x = -a \) into the function. Thus, \( f(-a) = 3 - 4(-a) = 3 + 4a \).

Step 3: Evaluate -f(a)

To find \( -f(a) \), we take the negative of \( f(a) \) which is \( 3 - 4a \). Thus, \(-f(a) = -(3 - 4a) = -3 + 4a \).

Step 4: Evaluate f(a+h)

Substitute \( x = a + h \) into the function: \( f(a+h) = 3 - 4(a+h) = 3 - 4a - 4h \).

Step 5: Evaluate f(a) + f(h)

Compute \( f(a) + f(h) \) by adding \( f(a) = 3 - 4a \) and \( f(h) = 3 - 4h \), which gives \( (3 - 4a) + (3 - 4h) = 6 - 4a - 4h \).

Step 6: Evaluate \( \frac{f(a+h) - f(a)}{h} \)

First, find \( f(a+h) - f(a) \) which is \((3 - 4a - 4h) - (3 - 4a) = -4h\). Then, divide by \( h \) to get \( \frac{-4h}{h} = -4 \) for \( h eq 0 \).

Key Concepts

Function Evaluation

Evaluating a function means finding the output of the function based on a specific input value. This is a fundamental concept in understanding how functions operate. Consider the function given in the exercise, \( f(x) = 3 - 4x \). When we evaluate the function for different inputs, such as \( a \) or \( -a \), we substitute these values into the function to find the corresponding outputs.

• For \( f(a) \): Substitute \( a \) into the function, leading to \( f(a) = 3 - 4a \).
• For \( f(-a) \): Substitute \( -a \) and perform the necessary arithmetic, resulting in \( f(-a) = 3 + 4a \).

Function evaluation helps us understand the behavior of the function for various inputs and is essential for graphing or optimizing functions.

Difference Quotient

The difference quotient is a way to measure how a function changes as its input changes. It's a significant concept in calculus because it's directly related to the derivative of a function. Consider the expression \( \frac{f(a+h) - f(a)}{h} \) in the exercise:

• First, find \( f(a+h) \), which involves substituting \( a+h \) into the function to get \( f(a+h) = 3 - 4a - 4h \).
• Subtract \( f(a) \) from \( f(a+h) \), simplifying to \( -4h \).
• The difference quotient is then \( \frac{-4h}{h} = -4 \), provided \( h eq 0 \).

This process demonstrates how the function's rate of change is consistent over small intervals, confirming the linearity of the function. The difference quotient highlights how functions evolve and is foundational for understanding instantaneous rates of change.

Real Numbers

Real numbers are a fundamental part of mathematics, encompassing all the numbers we typically work with in algebra and calculus. They include positive and negative numbers, fractions, and irrational numbers. In this exercise, real numbers are represented by the variables \( a \) and \( h \).

• They allow us to perform operations such as substitution, addition, and substraction with confidence.
• Knowing that \( a \) and \( h \) are real numbers ensures the function \( f(x) = 3 - 4x \) behaves predictably for any real input.

Understanding real numbers provides the groundwork for exploring more complex mathematical concepts and ensures clarity when evaluating functions and their behavior across the entire number line.

Substitution

Substitution is a mathematical technique where numbers or expressions are replaced with variables into a function or equation. It is crucial for evaluating functions, solving equations, and simplifying expressions. In the given solution:

• Substitution is used when evaluating \( f(a) \) by replacing \( x \) with \( a \) to obtain \( f(a) = 3 - 4a \).
• Similarly, for \( f(a+h) \), \( x \) is substituted with \( a+h \), resulting in \( f(a+h) = 3 - 4a - 4h \).

This process makes it possible to analyze how functions respond to different variables, which is fundamental in both algebra and calculus. Substitution simplifies problem-solving by showing intermediate steps that reveal how a function transforms according to its inputs.