Finding The Derivative Of F(x)=-x^2+3x-8 Using The Definition

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In the realm of calculus, the derivative of a function unveils its instantaneous rate of change at any given point. This fundamental concept empowers us to analyze the behavior of functions, pinpoint critical points, and construct tangent lines. In this comprehensive exploration, we will embark on a step-by-step journey to determine the derivative of the quadratic function f(x) = -x² + 3x - 8 using the very definition of the derivative. Furthermore, we will evaluate the derivative at specific points, namely x = 1, x = 2, and x = 3, to gain deeper insights into the function's dynamic nature.

Demystifying the Definition of the Derivative

The derivative of a function f(x), denoted as f'(x), is formally defined as the limit:

f'(x) = lim (h->0) [f(x + h) - f(x)] / h

This seemingly intricate expression encapsulates the essence of instantaneous change. Let's dissect it piece by piece to grasp its profound meaning.

  • f(x + h): This term represents the function's value at a point infinitesimally close to x, shifted by a tiny increment h. Think of it as peering into the function's future, just a hair's breadth away from the present.
  • f(x): This is simply the function's value at the point x, our reference point in time.
  • f(x + h) - f(x): The difference between these two values reveals the change in the function's output as we move from x to x + h. It's the function's response to our tiny nudge.
  • h: This represents the infinitesimal increment, the size of our step into the future. It's a crucial element in capturing the instantaneous nature of change.
  • [f(x + h) - f(x)] / h: This ratio, known as the difference quotient, approximates the average rate of change of the function over the interval from x to x + h. It's like calculating the slope of a line connecting two points on the function's graph.
  • lim (h->0): This is the heart of the derivative. As we shrink the increment h towards zero, the difference quotient morphs into the instantaneous rate of change at the point x. It's like zooming in on the function's graph until we see only a single point, and the slope of the line touching that point is the derivative.

The derivative, f'(x), is itself a function that spits out the instantaneous rate of change of f(x) at any input x. It's a powerful tool for understanding how a function behaves, where it's increasing or decreasing, and where it reaches its peaks and valleys.

Applying the Definition to Our Quadratic Function

Now, let's put the definition of the derivative into action and find f'(x) for our specific function, f(x) = -x² + 3x - 8. This will involve a series of algebraic manipulations, each carefully executed to unveil the derivative.

  1. Substitute f(x) into the Definition: We begin by replacing f(x) in the definition with its explicit expression: f'(x) = lim (h->0) [(- (x + h)² + 3(x + h) - 8) - (-x² + 3x - 8)] / h

    This substitution sets the stage for our algebraic dance, transforming the abstract definition into a concrete calculation.

  2. Expand and Simplify: Next, we meticulously expand the terms in the numerator and simplify the expression: f'(x) = lim (h->0) [- (x² + 2xh + h²) + 3x + 3h - 8 + x² - 3x + 8] / h f'(x) = lim (h->0) [-x² - 2xh - h² + 3x + 3h - 8 + x² - 3x + 8] / h f'(x) = lim (h->0) [-2xh - h² + 3h] / h

    Notice how terms artfully cancel out, paving the way for further simplification. This is a common theme in derivative calculations, where algebraic elegance often leads to a more concise expression.

  3. Factor out h: We now factor out h from the numerator: f'(x) = lim (h->0) h(-2x - h + 3) / h

    This factorization is a crucial step, as it allows us to eliminate h from the denominator, resolving the potential division by zero that lurks in the limit.

  4. Cancel h and Evaluate the Limit: We cancel the h terms and evaluate the limit as h approaches zero: f'(x) = lim (h->0) (-2x - h + 3) f'(x) = -2x - 0 + 3 f'(x) = -2x + 3

    And there it is! We have successfully navigated the algebraic labyrinth and arrived at the derivative of our quadratic function: f'(x) = -2x + 3.

Evaluating the Derivative at Specific Points

Now that we have the derivative function, f'(x) = -2x + 3, we can evaluate it at specific points to determine the instantaneous rate of change of f(x) at those points. This provides us with valuable insights into the function's behavior at different locations.

  1. f'(1): To find f'(1), we substitute x = 1 into the derivative function: f'(1) = -2(1) + 3 = -2 + 3 = 1

    This tells us that at x = 1, the function f(x) is increasing at a rate of 1 unit for every 1 unit increase in x. The tangent line to the graph of f(x) at x = 1 has a slope of 1.

  2. f'(2): Similarly, to find f'(2), we substitute x = 2 into the derivative function: f'(2) = -2(2) + 3 = -4 + 3 = -1

    At x = 2, the function f(x) is decreasing at a rate of 1 unit for every 1 unit increase in x. The tangent line to the graph of f(x) at x = 2 has a slope of -1.

  3. f'(3): Finally, to find f'(3), we substitute x = 3 into the derivative function: f'(3) = -2(3) + 3 = -6 + 3 = -3

    At x = 3, the function f(x) is decreasing at a rate of 3 units for every 1 unit increase in x. The tangent line to the graph of f(x) at x = 3 has a slope of -3.

Interpretation and Significance

The values of the derivative at these points offer a glimpse into the function's dynamic behavior. At x = 1, the function is increasing, while at x = 2 and x = 3, it is decreasing. The magnitude of the derivative indicates the steepness of the function's graph – a larger magnitude implies a steeper slope.

The derivative is a cornerstone of calculus, enabling us to analyze the behavior of functions, solve optimization problems, and model real-world phenomena. By understanding the definition of the derivative and mastering the techniques for its computation, we unlock a powerful tool for mathematical exploration and problem-solving.

Conclusion

In this comprehensive journey, we have successfully navigated the definition of the derivative to find f'(x) = -2x + 3 for the quadratic function f(x) = -x² + 3x - 8. We then evaluated the derivative at specific points, x = 1, x = 2, and x = 3, gaining valuable insights into the function's instantaneous rate of change at those locations.

This exercise underscores the power and elegance of the derivative, a fundamental concept in calculus that unlocks a deeper understanding of functions and their dynamic behavior. By mastering the definition of the derivative and its applications, we equip ourselves with a powerful tool for mathematical exploration and problem-solving in various fields of science and engineering.