Problem 15 - Calculus and its applications

Question

(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=x^{2}-3 x+5 $$

Step-by-Step Solution

Answer

The simplified difference quotient is \( 2x + h - 3 \). Calculated values: 9, 8, 7.1, 7.01 for the table rows.

Step 1: Define the Difference Quotient Formula

The difference quotient is defined as \( \frac{f(x+h) - f(x)}{h} \). Here, \( f(x) = x^2 - 3x + 5 \). We will use this formula to find the simplified form for the given function.

Step 2: Find f(x + h)

Substitute \( x+h \) into the function \( f(x) \) to get \( f(x+h) = (x+h)^2 - 3(x+h) + 5 \). Expanding this expression: \( (x+h)^2 = x^2 + 2xh + h^2 \) and \( -3(x+h) = -3x - 3h \). Combine these to get \( f(x+h) = x^2 + 2xh + h^2 - 3x - 3h + 5 \).

Step 3: Substitute and Simplify the Difference Quotient

Substituting \( f(x+h) \) and \( f(x) \) into the difference quotient formula, we get: \( \frac{(x^2 + 2xh + h^2 - 3x - 3h + 5) - (x^2 - 3x + 5)}{h} \). Simplifying this results in \( \frac{2xh + h^2 - 3h}{h} \) which further simplifies to \( 2x + h - 3 \) after canceling \( h \).

Step 4: Apply Difference Quotient to the Table Values

Using the simplified form \( 2x + h - 3 \), compute for each table row by substituting \( x = 5 \) and the given \( h \) values.

Step 5: Calculate for h = 2

Substitute \( x = 5 \) and \( h = 2 \) into \( 2x + h - 3 \) to get \( 2(5) + 2 - 3 = 9 \).

Step 6: Calculate for h = 1

Substitute \( x = 5 \) and \( h = 1 \) into \( 2x + h - 3 \) to get \( 2(5) + 1 - 3 = 8 \).

Step 7: Calculate for h = 0.1

Substitute \( x = 5 \) and \( h = 0.1 \) into \( 2x + h - 3 \) to get \( 2(5) + 0.1 - 3 = 7.1 \).

Step 8: Calculate for h = 0.01

Substitute \( x = 5 \) and \( h = 0.01 \) into \( 2x + h - 3 \) to get \( 2(5) + 0.01 - 3 = 7.01 \).

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with continuous change. It helps us understand how things change at every point rather than over a period of time. A key component of calculus is its focus on deriving quantities that describe change, such as slopes of curves and rates of change. The difference quotient, which we explore in this topic, is one of the initial steps in understanding the derivative, a fundamental concept of calculus. The derivative represents an instantaneous rate of change, and learning the difference quotient is like discovering a secret pathway into it. When you understand the difference quotient, you're beginning to learn how to take the abstract world of functions and bring them into the realm of concrete change and motion.

Simplification

Simplification is the process of reducing expressions to their simplest form. This makes complex problems much easier to handle by stripping them down to their essentials.
In our example, the simplification process takes several terms and streamlines them, which ultimately clarifies the calculation process.
Here's what happens:

• The multiple terms in the difference quotient are combined and reduced.
• Unnecessary parts are eliminated, i.e., like terms cancel each other out.
• The expression is condensed into a form that's easier to interpret and compute.

The key to successfully simplifying mathematical expressions is to keep practicing. Over time, you'll develop an intuition for identifying which parts of an expression can be simplified and how to effectively do it.

Function Evaluation

Function evaluation involves substituting specific values into a function to find its output. This is an essential process that allows us to understand how a function behaves and changes based on different inputs.
For example, when you substitute any number for the variable in a function, you're evaluating that function for that particular number.
In our task:

• We substitute \( x + h \) into the function to learn what the output becomes as we make small changes to \( x \).
• By calculating \( f(x+h) \) and \( f(x) \), we can find how the output of our function shifts with input changes.

This process ultimately provides insights into the characteristics of the function and prepares us for more advanced concepts like limits and derivatives.

Difference Quotient Formula

The difference quotient is a fundamental concept that leads to the derivative in calculus. The formula is generally given by \( \frac{f(x+h) - f(x)}{h} \). It measures the average rate of change of the function \( f(x) \) over a small interval \( h \).
To use the difference quotient effectively:

• Replace \( f(x) \) with the function given.
• Calculate \( f(x+h) \) by substituting \( x+h \) into the function.
• Simplify the expression \( f(x+h) - f(x) \) to find how the output changes.
• Divide by \( h \) to determine the average rate of change.

This formula can appear complex at first, but it becomes quite intuitive with practice, especially when applied to polynomial functions where simplification can significantly streamline the expression. The power of this formula is more evident as \( h \) approaches zero, helping us understand the instantaneous rate of change, or the derivative.