Problem 6 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The simplified form of the difference quotient is \( 2x + h + 1 \) and the table values are 13, 12, 11.1, and 11.01 for each respective \( h \).

Step 1: Understand the Difference Quotient

The difference quotient is given by the formula \( \frac{f(x+h) - f(x)}{h} \). It represents the average rate of change of the function \( f(x) \) over the interval \([x, x+h]\).

Step 2: Compute \( f(x+h) \)

Substitute \( x+h \) into the function \( f(x) = x^2 + x \). This gives: \[ f(x+h) = (x+h)^2 + (x+h) \] Expand this to obtain: \[ f(x+h) = (x^2 + 2xh + h^2) + x + h = x^2 + 2xh + h^2 + x + h \]

Step 3: Find \( f(x+h) - f(x) \)

Subtract \( f(x) = x^2 + x \) from \( f(x+h) \): \[ f(x+h) - f(x) = (x^2 + 2xh + h^2 + x + h) - (x^2 + x) \] Simplifying gives: \[ f(x+h) - f(x) = 2xh + h^2 + h \]

Step 4: Simplify the Difference Quotient

Divide \( f(x+h) - f(x) \) by \( h \): \[ \frac{f(x+h)-f(x)}{h} = \frac{2xh + h^2 + h}{h} \]Simplify the expression: \[ \frac{f(x+h)-f(x)}{h} = \frac{h(2x + h + 1)}{h} = 2x + h + 1 \]

Step 5: Calculate Values for the Table

Substitute \( x = 5 \) into the simplified expression \( 2x + h + 1 \) for different values of \( h \):- For \( h = 2 \): \( 2(5) + 2 + 1 = 13 \)- For \( h = 1 \): \( 2(5) + 1 + 1 = 12 \)- For \( h = 0.1 \): \( 2(5) + 0.1 + 1 = 11.1 \)- For \( h = 0.01 \): \( 2(5) + 0.01 + 1 = 11.01 \)

Key Concepts

Average Rate of Change

The average rate of change is a fundamental concept in calculus, and it's often the stepping stone to understanding derivatives. When we talk about the average rate of change for a function, we're essentially saying: "How much does the function change, on average, over a specific interval?" In mathematical terms, the average rate of change of a function \( f(x) \) between two points \( x \) and \( x+h \) is given by the difference quotient formula:

• \( \frac{f(x+h) - f(x)}{h} \)

This expression captures how the output of the function (let's say it's height) changes as the input (like time) changes from \( x \) to \( x+h \).

Imagine you're driving a car, and your speedometer gives you a different speed every few seconds. The average rate of change would be like checking your speed every minute to see how much it changes. In the context of our function \( f(x) = x^2 + x \), using the differences \( x+h \) and \( x \) gives us a concrete way to calculate how quickly \( f(x) \) climbs as \( x \) increases. This is particularly useful because understanding this change gives a snapshot of the function's behavior over any interval.

Simplifying Expressions

Simplifying expressions is a critical step in making math problems easier to solve and understand. When dealing with the difference quotient, the aim is to simplify the expression \( \frac{2xh + h^2 + h}{h} \).

• First, notice that \( h \) is common in all terms in the numerator. This allows us to factor \( h \) out:
• \( h(2x + h + 1) \)
• Next, simplify by cancelling \( h \) in the numerator and the denominator, resulting in \( 2x + h + 1 \).

This simplification shows how the complex expression depends on both \( x \) and \( h \), but largely decreases complexity since we are reducing the problem to basic arithmetic operations.

Why is this simplification useful? It makes calculations faster and helps us see the relationship between the different elements of the function more clearly. As \( h \) approaches zero (which is where calculus takes over with derivatives), our expression becomes simply \( 2x + 1 \), which provides the instantaneous rate of change at any \( x \). This is why simplifying expressions is emphasized so much in mathematics: it strips problems down to their most fundamental components.

Table Calculation

In exercises like the one posed, after simplifying the difference quotient, the next step is to apply various values into our simplified formula to see how the function behaves closely. This part revolves around table calculations.

With the simplified version \( 2x + h + 1 \), we're ready to fill our table for specific values of \( h \). Let's consider this process:

• At \( x = 5 \), plug in different \( h \) values to get the difference quotient results.
• For \( h = 2 \), we get \( 13 \).
• For \( h = 1 \), the result is \( 12 \).
• As \( h \) gets smaller, like \( 0.1 \) or \( 0.01 \), we observe values closer to the derivative: \( 11.1 \) and \( 11.01 \), respectively.

What table calculation does here is present how the average rate of change converges to a particular value, which is very close to what the derivative at \( x = 5 \) would be as \( h \) approaches zero. This convergence is key: it visually and numerically demonstrates how calculus' core ideas work in practice, transitioning from the average to instantaneous change.