Difference Quotient Of F(x) = 10x² + 5x - 1: Step-by-Step Solution

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Introduction

In the realm of calculus, the difference quotient serves as a cornerstone for understanding the concept of the derivative, which in turn forms the foundation for analyzing rates of change and slopes of curves. Mastering the simplification of the difference quotient is therefore crucial for anyone venturing into calculus and related fields. This article delves into the process of simplifying the difference quotient for the specific function f(x) = 10x² + 5x - 1, providing a step-by-step guide and insightful explanations to enhance your understanding. The difference quotient, often denoted as DQ, represents the average rate of change of a function over a small interval. It's the slope of the secant line through two points on the function's graph. As the interval shrinks, this secant line approaches the tangent line, and the difference quotient approaches the derivative. Therefore, understanding the difference quotient is like getting a sneak peek into the world of derivatives, which are essential for solving many real-world problems. From physics and engineering to economics and finance, derivatives help us model and optimize systems that change over time. In physics, they describe velocity and acceleration. In economics, they help maximize profits and minimize costs. And in engineering, they are used to design everything from bridges to airplanes. So, even if you're not a mathematician, understanding the difference quotient can open doors to a deeper understanding of the world around you. The function we're going to work with today, f(x) = 10x² + 5x - 1, is a simple quadratic function. But the principles we learn here can be applied to more complex functions as well. The key is to break down the problem into smaller, manageable steps and to understand the algebraic manipulations involved. We'll start by defining the difference quotient and then systematically work through the process of simplifying it for our given function. Along the way, we'll highlight common pitfalls and offer strategies for avoiding them. By the end of this article, you'll not only be able to simplify the difference quotient for this specific function, but you'll also have a solid understanding of the underlying concepts and techniques that you can apply to other functions as well.

Defining the Difference Quotient

The difference quotient is formally defined as:

DQ = [f(x + h) - f(x)] / h, where h ≠ 0

Here, h represents a small change in x, and f(x + h) represents the function's value at the point x + h. The difference quotient, therefore, calculates the change in the function's value (f(x + h) - f(x)) divided by the change in the input (h). This gives us the average rate of change of the function over the interval from x to x + h. Let's break down this definition further. The numerator, f(x + h) - f(x), represents the change in the function's output as we move from x to x + h. This is the "rise" in the slope calculation. The denominator, h, represents the change in the input, which is the "run" in the slope calculation. So, the difference quotient is essentially a slope formula, calculating the slope of the secant line connecting the points (x, f(x)) and (x + h, f(x + h)) on the graph of the function. The condition h ≠ 0 is crucial. If h were zero, we would be dividing by zero, which is undefined. This condition highlights the fact that we're looking at the average rate of change over a non-zero interval. As h gets smaller and smaller, the secant line gets closer and closer to the tangent line at the point (x, f(x)). This is the core idea behind the derivative, which is the instantaneous rate of change of the function at a single point. The difference quotient is a powerful tool because it allows us to approximate the derivative without actually taking a limit. This is particularly useful in situations where we don't have a closed-form expression for the derivative or when we're working with numerical data. By calculating the difference quotient for small values of h, we can get a good estimate of the derivative at a given point. Now that we have a clear understanding of the definition of the difference quotient, let's move on to the next step: applying this definition to our specific function, f(x) = 10x² + 5x - 1.

Applying the Definition to f(x) = 10x² + 5x - 1

To find the difference quotient for f(x) = 10x² + 5x - 1, we need to substitute this function into the DQ formula. This involves several steps, each requiring careful algebraic manipulation. First, we need to find f(x + h). This means replacing every instance of x in the function's expression with (x + h). This is a crucial step, and it's where many mistakes can happen if you're not careful with the order of operations and the distribution of terms. Remember to use parentheses correctly and to expand any squared terms completely. Once we have f(x + h), we can then subtract f(x) from it, which is the second step in calculating the numerator of the difference quotient. This step often involves simplifying the expression by combining like terms and canceling out terms that appear with opposite signs. Finally, we divide the entire result by h, which is the denominator of the difference quotient. This division step is often the key to simplifying the expression further, as it can lead to the cancellation of h from the numerator and denominator. Let's go through each of these steps in detail:

  1. Find f(x + h):

    Substitute (x + h) for x in the function:

    • f(x + h) = 10(x + h)² + 5(x + h) - 1

    Now, expand and simplify:

    • f(x + h) = 10(x² + 2xh + h²) + 5x + 5h - 1
    • f(x + h) = 10x² + 20xh + 10h² + 5x + 5h - 1

    This step involves careful expansion of the squared term and distribution of the constants. Make sure to double-check your work to avoid any arithmetic errors.

  2. Calculate f(x + h) - f(x):

    Subtract the original function, f(x) = 10x² + 5x - 1, from f(x + h):

    • f(x + h) - f(x) = (10x² + 20xh + 10h² + 5x + 5h - 1) - (10x² + 5x - 1)

    Now, simplify by distributing the negative sign and combining like terms:

    • f(x + h) - f(x) = 10x² + 20xh + 10h² + 5x + 5h - 1 - 10x² - 5x + 1
    • f(x + h) - f(x) = 20xh + 10h² + 5h

    Notice how several terms cancel out in this step. This is a common occurrence when calculating the difference quotient, and it simplifies the expression significantly.

  3. Divide by h:

    Divide the result by h:

    • DQ = [f(x + h) - f(x)] / h = (20xh + 10h² + 5h) / h

    Now, factor out an h from the numerator:

    • DQ = h(20x + 10h + 5) / h

    Finally, cancel the common factor of h:

    • DQ = 20x + 10h + 5

    This is the simplified form of the difference quotient for the given function. The h in the expression tells us that this is still an average rate of change over an interval of size h. To find the instantaneous rate of change (the derivative), we would need to take the limit as h approaches zero. But for now, we have successfully simplified the difference quotient.

Simplified Form of the Difference Quotient

Therefore, the simplified form of the difference quotient for the function f(x) = 10x² + 5x - 1 is:

DQ = 20x + 10h + 5

This expression represents the average rate of change of the function over an interval of size h around the point x. It's a crucial stepping stone in understanding the concept of the derivative, which gives the instantaneous rate of change at a single point. Let's delve a bit deeper into what this simplified form tells us. The expression 20x + 10h + 5 is a linear function of h, with the slope being 10 and the y-intercept being 20x + 5. This means that the average rate of change of the function changes linearly as the interval size h changes. The term 20x represents the part of the rate of change that depends on the value of x. This makes sense because the rate of change of a quadratic function is not constant; it varies depending on where you are on the curve. The term 10h represents the adjustment to the rate of change due to the size of the interval h. As h gets smaller, this term becomes less significant, and the difference quotient approaches the derivative. The term 5 is a constant term that contributes to the overall rate of change. It's worth noting that if we were to take the limit of this expression as h approaches zero, we would get 20x + 5, which is the derivative of the function f(x) = 10x² + 5x - 1. This is a powerful connection between the difference quotient and the derivative, and it highlights the importance of understanding the difference quotient as a precursor to understanding derivatives. In summary, the simplified form of the difference quotient gives us a clear picture of how the average rate of change of the function varies with both the input x and the interval size h. It's a fundamental concept in calculus that allows us to approximate the derivative and understand the behavior of functions.

Conclusion

In conclusion, we have successfully navigated the process of finding the simplified form of the difference quotient for the function f(x) = 10x² + 5x - 1. By meticulously applying the definition and performing the necessary algebraic manipulations, we arrived at the result:

DQ = 20x + 10h + 5

This journey has not only provided us with the solution but has also illuminated the significance of the difference quotient as a foundational concept in calculus. Understanding the difference quotient is paramount for grasping the derivative, a powerful tool for analyzing rates of change and slopes of curves. The steps involved in simplifying the difference quotient – substituting (x + h) into the function, subtracting the original function, and dividing by h – are fundamental techniques that are applicable to a wide range of functions. By mastering these techniques, you'll be well-equipped to tackle more complex problems in calculus and related fields. Furthermore, the insights gained from this exercise extend beyond the specific function we analyzed. The difference quotient provides a crucial link between the average rate of change and the instantaneous rate of change, which is the essence of the derivative. By understanding how the difference quotient approximates the derivative, you gain a deeper appreciation for the power and elegance of calculus. The simplified form of the difference quotient, 20x + 10h + 5, gives us a clear understanding of how the average rate of change varies with both the input x and the interval size h. This understanding is crucial for interpreting the behavior of the function and for making predictions about its future values. As we've seen, the difference quotient is not just a mathematical formula; it's a powerful tool for understanding the world around us. From physics and engineering to economics and finance, the concepts of rate of change and derivatives are essential for modeling and optimizing systems that change over time. So, whether you're a student just starting your calculus journey or a seasoned professional applying these concepts in your work, a solid understanding of the difference quotient will serve you well. This article has provided a comprehensive guide to simplifying the difference quotient for a specific function, but the principles and techniques discussed here can be applied to a much broader range of problems. Continue practicing and exploring, and you'll find that the difference quotient becomes an invaluable tool in your mathematical toolkit.