Analyzing F(x) = 2x³ + 3x² - 36x - 54 Derivatives Critical Values And Intervals

In this detailed exploration, we will delve into the intricacies of the function f(x) = 2x³ + 3x² - 36x - 54. Our primary goal is to dissect its behavior by finding its first derivative, identifying critical values, and pinpointing intervals of increase and decrease. This comprehensive analysis will provide a thorough understanding of the function's characteristics and graphical representation. Understanding the behavior of functions is crucial in calculus and has numerous applications in various fields, including physics, engineering, and economics. By examining the function's derivative, we can determine its slope at any given point, which helps us identify where the function is increasing, decreasing, or has a local maximum or minimum. The critical values, where the derivative is zero or undefined, are particularly important as they often correspond to these turning points. Let's embark on this analytical journey to unravel the properties of this cubic function.

a. Finding the First Derivative: f'(x)

The first step in our analysis is to find the first derivative of the function. The first derivative, denoted as f'(x), provides invaluable information about the function's rate of change. Specifically, it tells us the slope of the tangent line at any point on the function's graph. This slope is crucial in determining whether the function is increasing, decreasing, or has a stationary point (where the slope is zero). To compute the first derivative, we will apply the power rule, which states that if f(x) = axⁿ, then f'(x) = naxⁿ⁻¹. Applying this rule to each term of our function, f(x) = 2x³ + 3x² - 36x - 54, we get: The derivative of 2x³ is 6x² (3 * 2 * x^(3-1)). The derivative of 3x² is 6x (2 * 3 * x^(2-1)). The derivative of -36x is -36 (1 * -36 * x^(1-1)). The derivative of -54, a constant, is 0. Combining these results, we obtain the first derivative: f'(x) = 6x² + 6x - 36. This quadratic function, f'(x), is the key to unlocking the behavior of our original cubic function, f(x). The first derivative, f'(x) = 6x² + 6x - 36, is a quadratic equation that represents the slope of the original function, f(x) = 2x³ + 3x² - 36x - 54, at any given point. By analyzing this derivative, we can determine critical points, intervals of increase, and intervals of decrease. This is because the sign of f'(x) tells us whether the original function is increasing (f'(x) > 0), decreasing (f'(x) < 0), or has a horizontal tangent (f'(x) = 0). Therefore, finding and understanding the first derivative is a crucial step in analyzing the behavior of a function. In the next sections, we will use this derivative to find the critical values and the intervals where the function is increasing or decreasing.

b. Identifying Critical Values

Having computed the first derivative, our next task is to list any critical values. Critical values are the x-values where the derivative, f'(x), is either equal to zero or undefined. These points are pivotal because they often correspond to local maxima, local minima, or saddle points of the original function. In other words, critical values are the x-coordinates where the function's slope is either zero (horizontal tangent) or undefined (vertical tangent or a cusp). To find these critical values, we need to solve the equation f'(x) = 0. Recall that our first derivative is f'(x) = 6x² + 6x - 36. Setting this equal to zero gives us the quadratic equation: 6x² + 6x - 36 = 0. To simplify the equation, we can divide all terms by 6, resulting in: x² + x - 6 = 0. Now, we can solve this quadratic equation by factoring. We are looking for two numbers that multiply to -6 and add to 1. These numbers are 3 and -2. Therefore, we can factor the quadratic as: (x + 3)(x - 2) = 0. Setting each factor equal to zero gives us the solutions: x + 3 = 0 or x - 2 = 0. Solving for x, we find the critical values: x = -3 and x = 2. These are the x-coordinates where the function f(x) may have local maxima, local minima, or points of inflection. The critical values x = -3 and x = 2 are essential for understanding the local behavior of the function f(x). At these points, the tangent line to the curve is horizontal, indicating a potential change in the function's direction (from increasing to decreasing or vice versa). To determine whether these critical points are local maxima, local minima, or neither, we can use the first derivative test or the second derivative test. The first derivative test involves examining the sign of f'(x) on either side of the critical points. If f'(x) changes from positive to negative at a critical point, then it is a local maximum. If f'(x) changes from negative to positive, then it is a local minimum. If f'(x) does not change sign, then it is neither a local maximum nor a local minimum. The second derivative test involves finding the second derivative, f''(x), and evaluating it at the critical points. If f''(-3) > 0, then x = -3 is a local minimum. If f''(2) < 0, then x = 2 is a local maximum. In the following sections, we will use these critical values to identify the intervals of increase and decrease.

c. Identifying Intervals of Increase

Now that we have the critical values, we can identify intervals of increase. Intervals of increase are those sections of the x-axis where the function's value is increasing as x increases. In other words, the function is rising as we move from left to right on the graph. This corresponds to the intervals where the first derivative, f'(x), is positive (f'(x) > 0). To find these intervals, we will use the critical values we found earlier (x = -3 and x = 2) to divide the number line into intervals. These critical values are the potential turning points of the function, where it may switch from increasing to decreasing or vice versa. Our critical values, x = -3 and x = 2, divide the number line into three intervals: (-∞, -3), (-3, 2), and (2, ∞). To determine where f(x) is increasing, we need to test a value from each interval in the first derivative, f'(x) = 6x² + 6x - 36. If f'(x) > 0 for a test value in an interval, then the function is increasing on that interval. Let's choose test values: For the interval (-∞, -3), let's choose x = -4. Then, f'(-4) = 6(-4)² + 6(-4) - 36 = 96 - 24 - 36 = 36, which is positive. So, f(x) is increasing on the interval (-∞, -3). For the interval (-3, 2), let's choose x = 0. Then, f'(0) = 6(0)² + 6(0) - 36 = -36, which is negative. So, f(x) is decreasing on the interval (-3, 2). For the interval (2, ∞), let's choose x = 3. Then, f'(3) = 6(3)² + 6(3) - 36 = 54 + 18 - 36 = 36, which is positive. So, f(x) is increasing on the interval (2, ∞). Therefore, the intervals where f(x) is increasing are (-∞, -3) and (2, ∞). These intervals of increase tell us where the function's graph is rising as we move from left to right. Understanding these intervals, along with the intervals of decrease, helps us sketch the overall shape of the function's graph. The intervals of increase are crucial for understanding the function's behavior. A function is increasing when its first derivative is positive, indicating that the slope of the tangent line is positive. By identifying these intervals, we gain insight into where the function is climbing, which is essential for sketching the function's graph and understanding its overall trend. In this case, the function f(x) increases as x approaches negative infinity, reaches a local maximum at x = -3, decreases in the interval between the critical points, and then increases again as x approaches positive infinity.

d. Identifying Intervals of Decrease

Finally, we will identify intervals of decrease. Intervals of decrease are the sections of the x-axis where the function's value is decreasing as x increases. In other words, the function is falling as we move from left to right on the graph. This corresponds to the intervals where the first derivative, f'(x), is negative (f'(x) < 0). As we did with the intervals of increase, we will use the critical values (x = -3 and x = 2) to guide our analysis. We have already divided the number line into three intervals: (-∞, -3), (-3, 2), and (2, ∞). When we analyzed the intervals of increase, we tested a value from each interval in the first derivative, f'(x) = 6x² + 6x - 36. We found that: For the interval (-∞, -3), f'(-4) = 36, which is positive. So, f(x) is increasing on this interval. For the interval (-3, 2), f'(0) = -36, which is negative. So, f(x) is decreasing on this interval. For the interval (2, ∞), f'(3) = 36, which is positive. So, f(x) is increasing on this interval. From this analysis, we can conclude that the interval where f(x) is decreasing is (-3, 2). This means that between the critical points x = -3 and x = 2, the function's graph is falling. The interval of decrease provides further insight into the function's behavior and shape. A function is decreasing when its first derivative is negative, indicating that the slope of the tangent line is negative. In this case, the function f(x) decreases between the critical points x = -3 and x = 2, which means that the function reaches a local maximum at x = -3 and then decreases until it reaches a local minimum at x = 2. By identifying the intervals of increase and decrease, we can create a rough sketch of the function's graph. In this case, we know that the function increases until x = -3, then decreases until x = 2, and finally increases again for x > 2. This information, combined with the critical values, helps us understand the function's local and global behavior.

In summary, by finding the first derivative, identifying critical values, and analyzing the intervals of increase and decrease, we have gained a comprehensive understanding of the behavior of the function f(x) = 2x³ + 3x² - 36x - 54. This process is fundamental in calculus and provides valuable insights into the properties of functions. This comprehensive analysis has provided a clear picture of the function's behavior, allowing us to understand its shape and characteristics. This approach can be applied to analyze a wide range of functions and is a cornerstone of calculus. Understanding the critical points, intervals of increase, and decrease helps in sketching the function's graph and solving optimization problems. By understanding these concepts, we can better analyze and interpret the behavior of various functions, which is a fundamental skill in mathematics and its applications.