Problem 10 - Calculus and its applications

Question

(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=-2 x+5 $$

Step-by-Step Solution

Answer

The simplified form of the difference quotient is \(-2\); the table entries are all \(-2\).

Step 1: Understand the Function

The function given is a linear function: \( f(x) = -2x + 5 \). This is a straight line with a slope of -2.

Step 2: Write the Difference Quotient

The difference quotient is given by \( \frac{f(x+h) - f(x)}{h} \). We need to substitute the expression for \( f(x) \) into this formula.

Step 3: Calculate \( f(x+h) \)

Substitute \( x+h \) into the function: \( f(x+h) = -2(x+h) + 5 = -2x - 2h + 5 \).

Step 4: Substitute into the Difference Quotient

Substitute \( f(x+h) \) and \( f(x) \) into the difference quotient: \\( \frac{f(x+h)-f(x)}{h} = \frac{(-2x - 2h + 5) - (-2x + 5)}{h} \).

Step 5: Simplify the Expression

Simplify the expression: \\( \frac{(-2x - 2h + 5) + 2x - 5}{h} = \frac{-2h}{h} \). \This reduces to \( -2 \) when \( h eq 0 \).

Step 6: Fill in the Table

For each value of \( h \), the difference quotient is constant: \\( \frac{f(x+h)-f(x)}{h} = -2 \) (since the expression simplifies to \( -2 \) for all \( h eq 0 \)). \Therefore, the table entries for all \( h \) are \(-2\).

Key Concepts

Linear Functions

A linear function is a simple yet powerful mathematical tool that can describe a straight line on a graph. The general formula for a linear function is \( f(x) = mx + b \), where \( m \) represents the slope and \( b \) represents the y-intercept. For the given function \( f(x) = -2x + 5 \), we can see that it is a linear function where:

• The slope \( m \) is \(-2\).
• The y-intercept \( b \) is \(5\).

The slope indicates how steep the line is, and in this case, a slope of \(-2\) means the line falls two units downwards for every one unit it moves to the right.
The y-intercept tells us that the line crosses the y-axis at the point \((0, 5)\). These two components create a specific line on the coordinate plane, making it easy to predict and understand linear changes.

Calculus Problem Solving

Calculus problem solving often requires applying formulas and concepts like the difference quotient to analyze the behavior of functions. The difference quotient itself, \( \frac{f(x+h) - f(x)}{h} \), serves as a foundational concept for understanding rates of change and the behavior of functions.
In solving calculus problems, especially involving linear functions, the first step is to understand the function's structure. Recognizing that \( f(x) = -2x + 5 \) is linear allows us to simplify our problem-solving approach.

• Substituting \( x+h \) into the function gives us \( f(x+h) = -2(x+h) + 5 \).
• This becomes \( -2x - 2h + 5 \) after distributing the \(-2\).
• Substituting back into the difference quotient, we simplify to find consistent results across different increments \( h \).

By applying these steps methodically, calculus problems become manageable and less daunting.

Derivative Calculation

The difference quotient is a stepping stone to understanding derivatives, key tools in calculus for measuring how functions change. The derivative of a function gives us the slope or rate of change at any point. For a linear function like \( f(x) = -2x + 5 \), the derivative turns out to be constant, because the slope of a straight line never changes.
When we simplified the difference quotient:

• \( \frac{f(x+h) - f(x)}{h} = \frac{-2h}{h} \) became \(-2\).

This implies that the derivative of \( f(x) \) with respect to \( x \) is \(-2\). This matches the slope of the linear function, confirming our calculations. In calculus, this constant value of the derivative indicates a constant rate of change, typical for linear functions. Thus, understanding derivatives begins with mastering these basic concepts, leading to advanced mathematical insights.