Analyzing The Function F(x) = (x^4 - 50x^2)/9 A Comprehensive Guide
This article provides a comprehensive analysis of the function f(x) = (x^4 - 50x^2)/9. We will explore its properties, including finding its derivative, critical points, intervals of increase and decrease, concavity, and inflection points. We will also utilize a graphing tool to visualize the function and its key features. This detailed exploration will provide a deep understanding of the function's behavior and characteristics. Understanding polynomial functions like this is crucial in various fields, including physics, engineering, and economics, where they are used to model real-world phenomena. By mastering the techniques presented here, you will be well-equipped to analyze and interpret similar functions in various contexts.
(a) Finding the Derivative f'(x)
To determine the derivative of the function f(x) = (x^4 - 50x^2)/9, we can apply the power rule of differentiation. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1). We will apply this rule to each term in the function. Before differentiating, let's rewrite the function as f(x) = (1/9)x^4 - (50/9)x^2. This makes it easier to apply the power rule to each term separately. Remember, the derivative of a constant times a function is the constant times the derivative of the function.
Applying the power rule to the first term, (1/9)x^4, we multiply the coefficient (1/9) by the exponent 4 and reduce the exponent by 1, resulting in (4/9)x^3. For the second term, -(50/9)x^2, we multiply the coefficient (-50/9) by the exponent 2 and reduce the exponent by 1, resulting in -(100/9)x. Therefore, the derivative of f(x), denoted as f'(x), is the sum of these two results. Mastering the power rule is fundamental to differential calculus and is used extensively in finding derivatives of polynomial functions. This skill is essential for understanding rates of change and optimization problems.
Thus, the derivative f'(x) is given by:
f'(x) = (4/9)x^3 - (100/9)x
We can further simplify this expression by factoring out a common factor of (4/9)x:
f'(x) = (4/9)x(x^2 - 25)
And further factor using the difference of squares:
f'(x) = (4/9)x(x - 5)(x + 5)
This factored form of the derivative is particularly useful for finding the critical points of the function, which are the points where the derivative is either zero or undefined. Critical points are crucial for determining the intervals of increase and decrease of the function, as well as locating local maxima and minima. The ability to find and analyze critical points is a key skill in calculus and has wide applications in optimization problems.
(b) Finding Critical Points
To find the critical points of f(x), we need to determine where the derivative f'(x) = (4/9)x(x - 5)(x + 5) is equal to zero or undefined. Since f'(x) is a polynomial, it is defined for all real numbers. Therefore, we only need to find the values of x for which f'(x) = 0. Setting f'(x) = 0, we have:
(4/9)x(x - 5)(x + 5) = 0
This equation is satisfied when any of the factors are equal to zero. Thus, we have three critical points:
- x = 0
- x = 5
- x = -5
These critical points divide the real number line into four intervals: (-∞, -5), (-5, 0), (0, 5), and (5, ∞). We will use these intervals to determine the intervals of increase and decrease of the function. The critical points are like turning points on a road, indicating a change in the direction of the slope. Understanding how to find and interpret critical points is essential for sketching the graph of a function and understanding its behavior.
(c) Intervals of Increase and Decrease
To determine the intervals where f(x) is increasing or decreasing, we analyze the sign of the derivative f'(x) in each of the intervals defined by the critical points. If f'(x) > 0 in an interval, then f(x) is increasing in that interval. Conversely, if f'(x) < 0 in an interval, then f(x) is decreasing in that interval. We can create a sign chart to organize our analysis.
| Interval | Test Value (x) | f'(x) = (4/9)x(x - 5)(x + 5) | Sign of f'(x) | Increasing/Decreasing | |
|---|---|---|---|---|---|
| (-∞, -5) | -6 | (4/9)(-6)(-11)(-1) | - | Decreasing | |
| (-5, 0) | -1 | (4/9)(-1)(-6)(4) | + | Increasing | |
| (0, 5) | 1 | (4/9)(1)(-4)(6) | - | Decreasing | |
| (5, ∞) | 6 | (4/9)(6)(1)(11) | + | Increasing |
From the sign chart, we can conclude the following:
- f(x) is decreasing on the intervals (-∞, -5) and (0, 5).
- f(x) is increasing on the intervals (-5, 0) and (5, ∞).
This information allows us to identify local extrema. Since f(x) changes from decreasing to increasing at x = -5, there is a local minimum at x = -5. Similarly, since f(x) changes from increasing to decreasing at x = 0, there is a local maximum at x = 0. And, since f(x) changes from decreasing to increasing at x = 5, there is a local minimum at x = 5. Understanding intervals of increase and decrease is crucial for sketching the graph of a function and identifying its local extrema. These concepts are widely used in optimization problems to find the maximum or minimum values of a function.
(d) Finding Local Maxima and Minima
Based on the intervals of increase and decrease, we can identify the local maxima and minima of f(x). A local maximum occurs where the function changes from increasing to decreasing, and a local minimum occurs where the function changes from decreasing to increasing. From our analysis in part (c), we found that:
- f(x) has a local minimum at x = -5.
- f(x) has a local maximum at x = 0.
- f(x) has a local minimum at x = 5.
To find the y-coordinates of these extrema, we substitute these x-values back into the original function f(x) = (x^4 - 50x^2)/9:
-
Local minimum at x = -5:
f(-5) = ((-5)^4 - 50(-5)^2)/9 = (625 - 1250)/9 = -625/9 ≈ -69.44 -
Local maximum at x = 0:
f(0) = (0^4 - 50(0)^2)/9 = 0 -
Local minimum at x = 5:
f(5) = ((5)^4 - 50(5)^2)/9 = (625 - 1250)/9 = -625/9 ≈ -69.44
Therefore, the local extrema are:
- Local minimum: (-5, -625/9)
- Local maximum: (0, 0)
- Local minimum: (5, -625/9)
Identifying local maxima and minima is a fundamental application of calculus and is essential for solving optimization problems. These points represent the peaks and valleys of the function's graph, providing valuable information about its behavior.
(e) Finding the Second Derivative f''(x)
To analyze the concavity of f(x), we need to find the second derivative, f''(x). We start with the first derivative, f'(x) = (4/9)x^3 - (100/9)x, and differentiate it again using the power rule.
Applying the power rule to (4/9)x^3, we multiply the coefficient (4/9) by the exponent 3 and reduce the exponent by 1, resulting in (4/3)x^2. For the term -(100/9)x, we multiply the coefficient (-100/9) by the exponent 1 and reduce the exponent by 1, resulting in -(100/9). Therefore, the second derivative f''(x) is given by:
f''(x) = (4/3)x^2 - (100/9)
We can simplify this expression by factoring out a common factor of (4/9):
f''(x) = (4/9)(3x^2 - 25)
Finding the second derivative is a crucial step in understanding the concavity of a function. The second derivative tells us whether the function is curving upwards (concave up) or downwards (concave down). This information is essential for sketching an accurate graph of the function and identifying inflection points.
(f) Finding Points of Inflection
Points of inflection occur where the concavity of the function changes. To find these points, we need to determine where the second derivative f''(x) = (4/9)(3x^2 - 25) is equal to zero or undefined. Since f''(x) is a polynomial, it is defined for all real numbers. Therefore, we only need to find the values of x for which f''(x) = 0. Setting f''(x) = 0, we have:
(4/9)(3x^2 - 25) = 0
This implies that:
3x^2 - 25 = 0
Solving for x, we get:
3x^2 = 25
x^2 = 25/3
x = ±√(25/3) = ±(5/√3) = ±(5√3)/3
Thus, we have two potential points of inflection:
- x = (5√3)/3 ≈ 2.887
- x = -(5√3)/3 ≈ -2.887
To confirm that these are indeed points of inflection, we need to check the sign of f''(x) in the intervals determined by these points. These points of inflection mark a change in the curvature of the graph, transitioning from concave up to concave down or vice versa. Identifying inflection points helps to provide a complete picture of the function's shape.
(g) Determining Concavity
To determine the intervals where f(x) is concave up or concave down, we analyze the sign of the second derivative f''(x) in the intervals defined by the potential points of inflection. If f''(x) > 0 in an interval, then f(x) is concave up in that interval. Conversely, if f''(x) < 0 in an interval, then f(x) is concave down in that interval. We can create a sign chart to organize our analysis.
| Interval | Test Value (x) | f''(x) = (4/9)(3x^2 - 25) | Sign of f''(x) | Concavity |
|---|---|---|---|---|
| (-∞, -(5√3)/3) | -3 | (4/9)(3(9) - 25) | + | Concave Up |
| (-(5√3)/3, (5√3)/3) | 0 | (4/9)(-25) | - | Concave Down |
| ((5√3)/3, ∞) | 3 | (4/9)(3(9) - 25) | + | Concave Up |
From the sign chart, we can conclude the following:
- f(x) is concave up on the intervals (-∞, -(5√3)/3) and ((5√3)/3, ∞).
- f(x) is concave down on the interval (-(5√3)/3, (5√3)/3).
Since the concavity changes at x = ±(5√3)/3, these are indeed points of inflection. Understanding the concavity of a function helps to refine the sketch of its graph and provides additional insight into its behavior. Concavity is particularly important in optimization problems, as it can help to determine whether a critical point corresponds to a local maximum or a local minimum.
To find the y-coordinates of the inflection points, we substitute x = ±(5√3)/3 into the original function f(x) = (x^4 - 50x^2)/9:
f((5√3)/3) = (((5√3)/3)^4 - 50((5√3)/3)^2)/9 = (625/9 - 50(25/3))/9 = (625/9 - 1250/3)/9 = (625 - 3750)/81 = -3125/81 ≈ -38.58
f(-(5√3)/3) = (((-5√3)/3)^4 - 50((-5√3)/3)^2)/9 = (625/9 - 50(25/3))/9 = (625/9 - 1250/3)/9 = (625 - 3750)/81 = -3125/81 ≈ -38.58
Therefore, the points of inflection are approximately:
- ((5√3)/3, -3125/81) ≈ (2.887, -38.58)
- (-(5√3)/3, -3125/81) ≈ (-2.887, -38.58)
(h) Sketching the Graph
Now that we have gathered all the necessary information, we can sketch the graph of f(x) = (x^4 - 50x^2)/9. We know the following:
- Zeros: x = 0, x = ±√50 ≈ ±7.07
- Local minimum: (-5, -625/9) ≈ (-5, -69.44)
- Local maximum: (0, 0)
- Local minimum: (5, -625/9) ≈ (5, -69.44)
- Points of inflection: ±(5√3)/3, -3125/81) ≈ ±(2.887, -38.58)
- Concave up on (-∞, -(5√3)/3) and ((5√3)/3, ∞)
- Concave down on (-(5√3)/3, (5√3)/3)
- Decreasing on (-∞, -5) and (0, 5)
- Increasing on (-5, 0) and (5, ∞)
Using this information, we can sketch a graph that captures the key features of the function. The graph will have a W-shape, with local minima at x = ±5 and a local maximum at x = 0. The graph will be concave down between the inflection points and concave up outside of them. A graphing tool can be used to verify the sketch and provide a more precise representation of the function.
The process of sketching the graph of a function is a powerful way to visualize its behavior and understand its properties. By combining analytical techniques with graphical tools, we can gain a deep understanding of functions and their applications in various fields.
In this comprehensive analysis, we have thoroughly explored the function f(x) = (x^4 - 50x^2)/9. We successfully found its derivative, identified critical points, determined intervals of increase and decrease, located local maxima and minima, calculated the second derivative, found points of inflection, and analyzed concavity. By combining these analytical techniques, we were able to sketch an accurate representation of the function's graph. This detailed exploration demonstrates the power of calculus in understanding the behavior of functions and its applications in various fields. The skills and concepts discussed in this article are essential for anyone working with mathematical models and optimization problems.
