Problem 14 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The simplified form is \(2x + h - 3\), and the table values are 9, 8, 7.1, 7.01 for respective \(h\).

Step 1: Find the expression for f(x+h)

Start by substituting \(x + h\) into the function \(f(x)\). Given \(f(x) = x^2 - 3x\), replace \(x\) with \(x + h\). This gives \(f(x+h) = (x+h)^2 - 3(x+h)\).

Step 2: Expand f(x+h)

Simplify \(f(x+h) = (x+h)^2 - 3(x+h)\). Expanding \((x+h)^2\) gives \(x^2 + 2xh + h^2\), and expanding \(-3(x+h)\) gives \(-3x - 3h\). Thus, \(f(x+h) = x^2 + 2xh + h^2 - 3x - 3h\).

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

Subtract \(f(x)\) from \(f(x+h)\). This gives \((x^2 + 2xh + h^2 - 3x - 3h) - (x^2 - 3x)\). Simplify to get \(2xh + h^2 - 3h\).

Step 4: Find the difference quotient

Divide the expression \(2xh + h^2 - 3h\) by \(h\) to find the difference quotient. This simplifies to \(\frac{2xh + h^2 - 3h}{h} = 2x + h - 3\).

Step 5: Evaluate at specific points

Substitute \(x = 5\) into the simplified difference quotient \(2x + h - 3\). For each \(h\) value in the table, compute the result: - For \(h = 2\), it becomes \(2(5) + 2 - 3 = 9\).- For \(h = 1\), it becomes \(2(5) + 1 - 3 = 8\).- For \(h = 0.1\), it becomes \(2(5) + 0.1 - 3 = 7.1\).- For \(h = 0.01\), it becomes \(2(5) + 0.01 - 3 = 7.01\).

Step 6: Complete the table

Fill in the computed values for the difference quotient in the table:- \(h = 2\): 9 - \(h = 1\): 8- \(h = 0.1\): 7.1- \(h = 0.01\): 7.01

Key Concepts

Understanding Calculus

Calculus is an essential branch of mathematics that deals with change. It helps us understand how things evolve over time or space. One of the key ideas in calculus is the concept of a limit, which allows us to determine the behavior of functions as they approach specific points. Calculus is often split into two main areas:

• **Differential Calculus:** This focuses on finding the rate of change of a quantity. It's often concerned with slopes and the concept of derivatives.
• **Integral Calculus:** This deals with the accumulation of quantities, such as areas under curves and the total accumulated change of a quantity.

These concepts are tied together by the Fundamental Theorem of Calculus, which links differentiation with integration. In this exercise, we apply differential calculus to explore how small changes in the input of a function affect its output. The difference quotient is a crucial part of this exploration, representing the average rate of change of the function over an interval. The more we shrink this interval, the more it approximates the instantaneous rate of change, leading us directly into the idea of limits and derivatives.

Exploring Functions

Functions are fundamental building blocks in calculus and mathematics in general. A function, denoted as \(f(x)\), is a relation where each input \(x\) has a unique output \(f(x)\). The function given in the original exercise is \(f(x) = x^2 - 3x\). This is a quadratic function, a specific form of polynomial function characterized by the highest power of \(x\) being 2.When working with functions in calculus, we often want to understand how changes in one variable affect the function's output. Specifically, the difference quotient \[ \frac{f(x+h) - f(x)}{h} \]plays a significant role by measuring the average rate of change over an interval \(h\). In other words, it tells us how the function's values fluctuate as \(x\) is varied by \(h\).Function behavior, such as how they increase, decrease, or inflect, is crucial for understanding real-world problems, from physics to economics. Calculus allows us to delve deeper into the properties of functions, helping us predict and optimize various processes.

The Importance of Limits

Limits are a foundational concept in calculus, allowing us to understand behavior as we approach a particular point. They give us a systematic way to deal with values that seem to approach infinity or become infinitely small.In the context of difference quotients, limits help bridge the gap from average rate of change to instantaneous rate of change. We use limits to define derivatives, an essential tool in calculus for finding slopes and rates of change.As \(h\) approaches 0 in the difference quotient \(\frac{f(x+h)-f(x)}{h}\), we approach the derivative of the function at point \(x\). The calculated values in the original exercise's table show this process: as \(h\) becomes smaller, the difference quotient converges to a particular value, indicating the slope of the tangent to the function at that point.Understanding limits doesn't just help in mathematical theory. It has practical implications in fields that require approximation and precise predictions, such as engineering, physics, and computer science. Limits ensure that we can deal with continuous change efficiently and accurately.