Problem 70 - College Algebra with Modeling and Visualization

Question

Complete the following for the given \(f(x)\) (a) Find \(f(x+h)\) (b) Find the difference quotient of \(f\) and simplify. $$ f(x)=-5 $$

Step-by-Step Solution

Answer

(a) \(f(x+h) = -5\), (b) The difference quotient is 0.

Step 1: Find Expression for f(x+h)

Given that \( f(x) = -5 \), we need to find \( f(x+h) \). For the function \( f(x) = -5 \), this means for any input, it always returns \(-5\). Therefore, \( f(x+h) = -5 \).

Step 2: Calculate the Difference Quotient

The difference quotient is given by \( \frac{f(x+h) - f(x)}{h} \). Substituting the values, the expression becomes: \[ \frac{-5 - (-5)}{h} \] which simplifies to \[ \frac{0}{h} \].

Step 3: Simplify the Difference Quotient

The expression \( \frac{0}{h} \) simplifies to \( 0 \) for all \( h eq 0 \). Therefore, the simplified difference quotient is \( 0 \).

Key Concepts

Function Evaluation

When you evaluate a function, you're essentially replacing the variable in the function with a specific value. This helps in understanding how the function behaves for different inputs.

• What is a constant function? A constant function is one where the output doesn't change, no matter what the input is. For example, for the function \( f(x) = -5 \), regardless of the value you substitute for \( x \), the function always returns \(-5\).
• How to evaluate \( f(x+h) \)? Since the given function is \( f(x) = -5 \), evaluating \( f(x+h) \) doesn't change its value. It simply means placing \( x+h \) into the function. But since it's a constant function, the output remains \(-5\), because the value of \( x \) or \( x+h \) doesn't affect the function's outcome.

Keep this in mind: Evaluating a constant function is straightforward as they always give the same result, irrespective of the input.

Simplification

Simplification is the process of making a math expression easier to understand or solve. It involves removing any unnecessary complexity.

• First Simplification Step: In the context of the difference quotient, we start with the expression \( \frac{f(x+h) - f(x)}{h} \). For a constant function like \( f(x) = -5 \), this becomes: \[ \frac{-5 - (-5)}{h} \]
• Simplification Process: Calculate the expression inside the numerator first. Here, \(-5 - (-5)\) simplifies to \(0\), so your difference quotient simplifies to \( \frac{0}{h} \).
• Final Simplification: Since \( \frac{0}{h} \) equals \(0\) for any \(h eq 0\), the difference quotient result is \(0\). This means no matter the non-zero value of \(h\), the simplification yields \(0\).

Simplification helps eliminate unnecessary steps and clarifies the result, especially in simple constant functions.

Constant Function

A constant function plays a unique role in mathematics. It is defined as a function that returns a fixed output for any input given.

• Definition: A constant function is written generally as \( f(x) = c \), where \(c\) is a constant real number.
• Example: For \( f(x) = -5 \), no matter what value of \(x\) you use, the output is always \(-5\).
• Impact on Difference Quotient: For a constant function, the difference between \( f(x+h) \) and \( f(x) \) is always zero, leading to a simplified difference quotient of \(0\).

Understanding constant functions is important because they make calculations easier. They illustrate that some mathematical phenomena do not change with differing inputs.