Increasing And Decreasing Intervals And Local Extrema Of H(x) = -x³ + X²

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In the realm of calculus, understanding the behavior of functions is paramount. This article delves into the analysis of the function h(x) = -x³ + x², focusing on identifying the intervals where it increases or decreases and pinpointing its local extreme values. This exploration involves leveraging the power of derivatives to unveil the function's dynamic characteristics. By understanding these concepts, we gain valuable insights into the function's graph and its overall behavior.

Decoding Increasing and Decreasing Intervals

To determine the intervals where the function h(x) = -x³ + x² is increasing or decreasing, we first need to find its first derivative, denoted as h'(x). The first derivative provides crucial information about the function's slope at any given point. A positive derivative indicates an increasing function, while a negative derivative signifies a decreasing function. To find the first derivative, we apply the power rule of differentiation to each term of the function:

h'(x) = d/dx (-x³ + x²) h'(x) = -3x² + 2x

Now that we have the first derivative, we need to find its critical points. Critical points are the points where the derivative is either equal to zero or undefined. These points are crucial because they often mark the transition between increasing and decreasing intervals. To find the critical points, we set h'(x) equal to zero and solve for x:

-3x² + 2x = 0 x(-3x + 2) = 0

This equation has two solutions:

x = 0 -3x + 2 = 0 => x = 2/3

Thus, the critical points are x = 0 and x = 2/3. These points divide the x-axis into three intervals: (-∞, 0), (0, 2/3), and (2/3, ∞). To determine whether the function is increasing or decreasing in each interval, we can test a point within each interval by plugging it into the first derivative h'(x).

  • Interval (-∞, 0): Let's test x = -1: h'(-1) = -3(-1)² + 2(-1) = -3 - 2 = -5 Since h'(-1) is negative, the function is decreasing in the interval (-∞, 0).

  • Interval (0, 2/3): Let's test x = 1/3: h'(1/3) = -3(1/3)² + 2(1/3) = -3(1/9) + 2/3 = -1/3 + 2/3 = 1/3 Since h'(1/3) is positive, the function is increasing in the interval (0, 2/3).

  • Interval (2/3, ∞): Let's test x = 1: h'(1) = -3(1)² + 2(1) = -3 + 2 = -1 Since h'(1) is negative, the function is decreasing in the interval (2/3, ∞).

Therefore, we can conclude that the function h(x) = -x³ + x² is increasing on the open interval (0, 2/3) and decreasing on the open intervals (-∞, 0) and (2/3, ∞).

Identifying Local Extreme Values

Local extreme values, also known as local maxima and minima, represent the points where the function reaches its highest or lowest values within a specific neighborhood. These points are crucial in understanding the function's overall shape and behavior. To find the local extreme values of h(x) = -x³ + x², we utilize the first derivative test, which leverages the information about increasing and decreasing intervals.

We already found the critical points x = 0 and x = 2/3, and we know the intervals where the function is increasing and decreasing. Now, we can apply the first derivative test:

  • At x = 0: The function changes from decreasing to increasing. This indicates a local minimum at x = 0. To find the value of the function at this point, we plug x = 0 into the original function: h(0) = -(0)³ + (0)² = 0 So, there is a local minimum at (0, 0).

  • At x = 2/3: The function changes from increasing to decreasing. This indicates a local maximum at x = 2/3. To find the value of the function at this point, we plug x = 2/3 into the original function: h(2/3) = -(2/3)³ + (2/3)² = -8/27 + 4/9 = -8/27 + 12/27 = 4/27 So, there is a local maximum at (2/3, 4/27).

Therefore, the function h(x) = -x³ + x² has a local minimum at (0, 0) and a local maximum at (2/3, 4/27). These points represent the function's valleys and peaks within its local neighborhood.

Visualizing the Function's Behavior

To further solidify our understanding, let's visualize the function h(x) = -x³ + x² and its key features. The graph of the function is a cubic curve that exhibits both increasing and decreasing behavior. The local minimum at (0, 0) represents a valley in the curve, while the local maximum at (2/3, 4/27) represents a peak. The intervals where the function is increasing and decreasing are visually evident in the graph's upward and downward slopes.

By plotting the function, its critical points, and its local extrema, we gain a comprehensive understanding of its behavior. The graph serves as a visual confirmation of our analytical findings, reinforcing the concepts of increasing and decreasing intervals and local extreme values.

Conclusion: Mastering Function Analysis

This exploration of the function h(x) = -x³ + x² has demonstrated the power of calculus in understanding a function's behavior. By finding the first derivative, identifying critical points, and applying the first derivative test, we successfully determined the intervals where the function increases and decreases and located its local extreme values. These techniques are fundamental tools in calculus and are applicable to a wide range of functions.

Understanding increasing and decreasing intervals and local extrema is crucial for sketching function graphs, optimizing mathematical models, and solving real-world problems. By mastering these concepts, we unlock a deeper understanding of the mathematical world and its applications.

In summary, we have meticulously analyzed the function h(x) = -x³ + x², identifying its increasing interval as (0, 2/3) and its decreasing intervals as (-∞, 0) and (2/3, ∞). Furthermore, we pinpointed a local minimum at (0, 0) and a local maximum at (2/3, 4/27). This comprehensive analysis showcases the power of calculus in dissecting and understanding the behavior of functions. Understanding these concepts are critical for succeeding in more advanced calculus topics. Therefore, practice is key to master function analysis and its applications. By continually practicing and refining your understanding, you will strengthen your problem-solving skills and gain confidence in tackling complex mathematical challenges.