Simplified Difference Quotient For F(x) = 10x² + 5x - 1

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This article delves into the process of determining the simplified form of the difference quotient for the quadratic function f(x) = 10x² + 5x - 1. The difference quotient is a fundamental concept in calculus, serving as the foundation for understanding derivatives, which measure the instantaneous rate of change of a function. Mastering the simplification of the difference quotient is crucial for grasping the core principles of differential calculus. This article will provide a step-by-step guide, ensuring clarity and comprehension for learners of all levels.

Understanding the Difference Quotient

At its core, the difference quotient represents the average rate of change of a function f(x) over a small interval. It is defined by the following formula:

DQ = [f(x + h) - f(x)] / h

Where:

  • f(x) is the function we are analyzing.
  • h is a small change in the input x.
  • f(x + h) is the function's value when the input is x + h.

Essentially, the difference quotient calculates the slope of the secant line that passes through two points on the function's graph: (x, f(x)) and (x + h, f(x + h)). As h approaches zero, this secant line approaches the tangent line, and the difference quotient approaches the derivative, which represents the instantaneous rate of change at the point x. For the function at hand, f(x) = 10x² + 5x - 1, we will methodically substitute and simplify to arrive at the difference quotient's simplified form.

Step 1: Calculate f(x + h)

The first step in finding the difference quotient is to determine f(x + h). This involves substituting (x + h) for x in the original function:

f(x + h) = 10(x + h)² + 5(x + h) - 1

Now, we need to expand and simplify this expression. First, expand the squared term:

(x + h)² = x² + 2xh + h²

Substitute this back into the expression for f(x + h):

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

Next, distribute the constants:

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

This expression represents the value of the function at the point x + h. This expansion is a critical step because it sets the stage for subtracting f(x) and ultimately simplifying the difference quotient. Pay close attention to the distribution of constants and the expansion of the squared term to avoid errors. Careful execution at this stage ensures the accuracy of the subsequent steps.

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

Now that we have f(x + h), the next step is to subtract the original function, f(x), from it. This will give us the numerator of the difference quotient:

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

Distribute the negative sign to the terms inside the second parenthesis:

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

Now, we can simplify by combining like terms. Notice that several terms cancel out:

  • 10x² and -10x² cancel each other.
  • 5x and -5x cancel each other.
  • -1 and +1 cancel each other.

This leaves us with:

f(x + h) - f(x) = 20xh + 10h² + 5h

This expression represents the change in the function's value over the interval h. This simplification is crucial because it isolates the terms that contribute to the rate of change. The cancellation of terms involving only x is a common pattern in difference quotient calculations, reflecting the focus on the change in the function's value with respect to h.

Step 3: Divide by h

The next step is to divide the expression we obtained in the previous step by h. This completes the difference quotient formula:

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

Now, we can factor out an h from the numerator:

(20xh + 10h² + 5h) / h = h(20x + 10h + 5) / h

Since h is assumed to be non-zero in the difference quotient (otherwise, we would be dividing by zero), we can cancel the h in the numerator and denominator:

h(20x + 10h + 5) / h = 20x + 10h + 5

This simplification is a critical step in finding the simplified difference quotient. Dividing by h and canceling the common factor allows us to eliminate the indeterminate form that would arise when h approaches zero. The resulting expression represents the average rate of change of the function over the interval h, and it's now in a form that allows us to easily take the limit as h approaches zero to find the derivative.

Step 4: The Simplified Difference Quotient

After dividing by h and simplifying, we arrive at the simplified form of the difference quotient:

DQ = 20x + 10h + 5

This expression represents the average rate of change of the function f(x) = 10x² + 5x - 1 over the interval h. It is a crucial result because it allows us to understand how the function's value changes as x changes. Furthermore, this expression is a stepping stone to finding the derivative of the function. By taking the limit of this expression as h approaches zero, we can find the instantaneous rate of change of the function at a specific point. This simplified form is the key to understanding the function's behavior and its derivative.

Conclusion

In summary, we have successfully found the simplified form of the difference quotient for the function f(x) = 10x² + 5x - 1. The steps involved are:

  1. Calculate f(x + h) by substituting (x + h) into the function.
  2. Subtract f(x) from f(x + h).
  3. Divide the result by h.
  4. Simplify the expression.

The simplified difference quotient, 20x + 10h + 5, provides valuable insights into the function's rate of change. It is a fundamental concept in calculus, paving the way for understanding derivatives and other advanced topics. Mastering the process of finding and simplifying the difference quotient is essential for anyone studying calculus. This skill allows for a deeper understanding of how functions change and provides a solid foundation for further exploration of calculus concepts. By understanding the difference quotient, one can better appreciate the core principles that underpin differential calculus.