Finding Relative Extrema And Sketching The Graph Of F(x) = -x³ + 12x - 16
In this article, we will explore how to find the relative extreme points of the function f(x) = -x³ + 12x - 16. We will delve into the process of identifying critical points using the first derivative test and determining whether these points correspond to relative maxima or minima. Furthermore, we will use this information, along with other key features of the function, to sketch an accurate graph. Understanding these concepts is crucial for calculus students and anyone interested in analyzing the behavior of functions.
1. Understanding Relative Extrema
Before we dive into the calculations, it's important to define what we mean by relative extrema. In calculus, relative extrema (also known as local extrema) are the points where a function reaches a local maximum or minimum value within a specific interval. These points are 'relative' because they are the highest or lowest points in their immediate vicinity, but not necessarily the highest or lowest points on the entire function's domain. To find these points, we'll use the powerful tools of differential calculus.
1.1. Importance of Relative Extrema
Relative extrema play a significant role in understanding the overall behavior of a function. They help us identify where the function changes direction (from increasing to decreasing or vice versa) and where it attains its peak and valley values within a certain region. In many real-world applications, finding relative extrema is crucial for optimization problems, where we seek to maximize or minimize a certain quantity.
2. Finding Critical Points: The First Derivative
The cornerstone of finding relative extrema lies in the concept of critical points. These are the points where the first derivative of the function is either equal to zero or undefined. Critical points are potential locations for relative maxima or minima because they indicate where the function's slope is momentarily zero (horizontal tangent) or where the slope is undefined (vertical tangent or a cusp).
2.1. Calculating the First Derivative
To find the critical points of f(x) = -x³ + 12x - 16, our first step is to compute its derivative, f'(x). Using the power rule of differentiation, we get:
f'(x) = -3x² + 12
This derivative represents the slope of the tangent line to the graph of f(x) at any point x. We'll use this information to pinpoint the critical points.
2.2. Setting the Derivative to Zero
To find the critical points, we set the first derivative equal to zero and solve for x:
-3x² + 12 = 0
Divide both sides by -3:
x² - 4 = 0
This is a difference of squares, which we can factor as:
(x - 2)(x + 2) = 0
Therefore, the critical points are x = 2 and x = -2. These are the potential locations of our relative extrema.
3. The First Derivative Test: Identifying Extrema
Now that we've identified the critical points, we need to determine whether they correspond to relative maxima, relative minima, or neither. The first derivative test is a powerful tool for this purpose. It involves analyzing the sign of the first derivative f'(x) in the intervals around the critical points.
3.1. Creating a Sign Chart
To apply the first derivative test, we create a sign chart. This chart divides the number line into intervals based on the critical points and then shows the sign of f'(x) in each interval. Our critical points, x = -2 and x = 2, divide the number line into three intervals: (-∞, -2), (-2, 2), and (2, ∞).
To determine the sign of f'(x) in each interval, we pick a test value within the interval and evaluate f'(x) at that value. For example:
- Interval (-∞, -2): Let's choose x = -3. f'(-3) = -3(-3)² + 12 = -27 + 12 = -15 (negative)
- Interval (-2, 2): Let's choose x = 0. f'(0) = -3(0)² + 12 = 12 (positive)
- Interval (2, ∞): Let's choose x = 3. f'(3) = -3(3)² + 12 = -27 + 12 = -15 (negative)
We can summarize this information in a sign chart:
| Interval | Test Value | f'(x) | f(x) | Behavior |
|---|---|---|---|---|
| (-∞, -2) | x = -3 | - | Decreasing | |
| (-2, 2) | x = 0 | + | Increasing | |
| (2, ∞) | x = 3 | - | Decreasing |
3.2. Interpreting the Sign Chart
The sign chart tells us how the function f(x) is behaving in each interval:
- f'(x) < 0 (negative): The function is decreasing.
- f'(x) > 0 (positive): The function is increasing.
- f'(x) = 0: We have a critical point (potential relative extrema).
Based on our sign chart:
- At x = -2, the function changes from decreasing to increasing. This indicates a relative minimum.
- At x = 2, the function changes from increasing to decreasing. This indicates a relative maximum.
4. Finding the Coordinates of the Extreme Points
To find the exact coordinates of the relative extrema, we need to plug the x-values of the critical points back into the original function, f(x) = -x³ + 12x - 16.
4.1. Relative Minimum at x = -2
f(-2) = -(-2)³ + 12(-2) - 16 = 8 - 24 - 16 = -32
So, the relative minimum point is (-2, -32).
4.2. Relative Maximum at x = 2
f(2) = -(2)³ + 12(2) - 16 = -8 + 24 - 16 = 0
So, the relative maximum point is (2, 0).
5. Sketching the Graph
Now that we have identified the relative extrema and understand the function's increasing/decreasing behavior, we can sketch a graph of f(x) = -x³ + 12x - 16. To make our sketch more accurate, let's also find the y-intercept.
5.1. Finding the y-intercept
The y-intercept is the point where the graph crosses the y-axis, which occurs when x = 0. So, we evaluate f(0):
f(0) = -(0)³ + 12(0) - 16 = -16
The y-intercept is (0, -16).
5.2. Sketching the Graph: A Step-by-Step Approach
- Plot the critical points: Plot the relative minimum (-2, -32) and the relative maximum (2, 0). These points will be the turning points of our graph.
- Plot the y-intercept: Plot the point (0, -16).
- Consider the end behavior: Since the leading term of the function is -x³, the function will approach positive infinity as x approaches negative infinity, and it will approach negative infinity as x approaches positive infinity.
- Connect the points: Starting from the left side of the graph, draw a curve that increases until it reaches the relative maximum at (2, 0). Then, the curve decreases, passing through the y-intercept (0, -16) and reaching the relative minimum at (-2, -32). Finally, continue the curve downwards as x increases.
By following these steps, you can create a reasonable sketch of the graph of f(x) = -x³ + 12x - 16. The graph will be a cubic function with a negative leading coefficient, exhibiting a relative maximum and a relative minimum.
6. Conclusion: Summarizing the Key Findings
In conclusion, we have successfully analyzed the function f(x) = -x³ + 12x - 16 to find its relative extrema and sketch its graph. We used the first derivative test to identify the critical points, which were x = -2 and x = 2. By analyzing the sign of the first derivative around these points, we determined that:
- (-2, -32) is a relative minimum
- (2, 0) is a relative maximum
We also found the y-intercept to be (0, -16). Combining this information with the end behavior of the function, we were able to create a sketch of the graph. This process demonstrates the power of calculus in understanding the behavior of functions and identifying their key features.
By mastering these techniques, you'll be well-equipped to analyze a wide range of functions and solve optimization problems in various contexts. Remember to always carefully calculate the derivatives, construct sign charts, and interpret the results to gain a comprehensive understanding of the function's behavior.
