Determining Intervals Of Increase And Decrease For H(x) = √x(x - 12)

Jul 13, 2025 by ADMIN 69 views

Finding Intervals of Increase and Decrease for h(x) = √x(x - 12)

<p>In this article, we will delve into the process of determining the intervals where the function h(x) = √x(x - 12), defined for x > 0, is either increasing or decreasing. This involves calculus techniques, specifically analyzing the first derivative of the function. By understanding the sign of the derivative, we can effectively map out the behavior of the function across its domain.</p>

1. Understanding Increasing and Decreasing Functions

To effectively determine the intervals where a function is increasing or decreasing, it's essential to understand the underlying concepts. A function is said to be increasing on an interval if its values increase as the input (x) increases. Conversely, a function is decreasing if its values decrease as the input increases. Calculus provides a powerful tool to analyze this behavior using the first derivative. If the first derivative of a function is positive on an interval, the function is increasing on that interval. If the first derivative is negative, the function is decreasing. The points where the first derivative is zero or undefined are critical points, which are potential locations where the function changes its behavior from increasing to decreasing or vice versa. These critical points are vital in identifying the intervals of interest.

When analyzing a function, it's important to first consider its domain, which in this case is defined as x > 0 due to the presence of the square root. This restriction impacts the analysis, as we only focus on the positive x-axis. The function h(x) = √x(x - 12) combines a square root and a linear term, making it a good example to illustrate the application of calculus in understanding function behavior. By finding the first derivative and analyzing its sign, we can precisely identify the intervals where the function rises or falls, providing a comprehensive understanding of its graphical representation and behavior.

The first step in this analysis is to find the derivative of the function. This often involves applying the product rule, chain rule, and other differentiation techniques, depending on the complexity of the function. Once the derivative is found, we set it equal to zero and solve for x to find the critical points. These points divide the domain into intervals, and we then test a value within each interval in the derivative to determine its sign. A positive sign indicates an increasing function, a negative sign indicates a decreasing function, and a zero value suggests a potential local maximum or minimum. Understanding these steps is crucial for mastering the analysis of function behavior and is a fundamental concept in calculus.

2. Computing the First Derivative

To begin our analysis, we need to compute the first derivative of the function h(x) = √x(x - 12). This will involve the application of the product rule, which states that the derivative of a product of two functions, u(x) and v(x), is given by (uv)' = u'v + uv'. In our case, we can identify u(x) = √x and v(x) = (x - 12). First, we find the derivatives of u(x) and v(x) separately. The derivative of u(x) = √x, which can also be written as x^(1/2), is u'(x) = (1/2)x^(-1/2), which simplifies to 1/(2√x). The derivative of v(x) = (x - 12) is simply v'(x) = 1.

Now, applying the product rule, we get h'(x) = u'(x)v(x) + u(x)v'(x) = (1/(2√x))(x - 12) + √x(1). To simplify this expression, we need to combine the terms. We can do this by finding a common denominator, which in this case is 2√x. Thus, we rewrite the equation as h'(x) = (x - 12)/(2√x) + (2x)/(2√x). Combining the numerators, we get h'(x) = (x - 12 + 2x)/(2√x), which further simplifies to h'(x) = (3x - 12)/(2√x). This is the simplified form of the first derivative of h(x). The derivative is a critical component in determining the intervals of increase and decrease, as it tells us the rate of change of the function at any given point.

The next step is to find the critical points of the function by setting h'(x) equal to zero and solving for x. These critical points are where the function may change direction, from increasing to decreasing or vice versa. Additionally, we need to consider where the derivative is undefined, which can also indicate a change in the function's behavior. In this case, the derivative is undefined when the denominator 2√x is equal to zero, which occurs at x = 0. However, since the original function is defined for x > 0, we only need to consider the values greater than zero. The simplified form of the derivative makes it easier to identify these critical points and proceed with the analysis of the intervals of increase and decrease.

3. Finding Critical Points

Critical points are crucial in identifying where a function's behavior changes from increasing to decreasing, or vice versa. To find the critical points of the function h(x) = √x(x - 12), we need to analyze the first derivative, h'(x) = (3x - 12)/(2√x). Critical points occur where the derivative is either equal to zero or undefined. First, let's find the points where the derivative is equal to zero. This happens when the numerator of the derivative is zero, i.e., 3x - 12 = 0. Solving this equation for x, we get 3x = 12, which gives us x = 4. So, x = 4 is one critical point.

Next, we need to consider where the derivative is undefined. The derivative h'(x) = (3x - 12)/(2√x) is undefined when the denominator is zero. The denominator, 2√x, is zero when x = 0. However, since the original function h(x) = √x(x - 12) is defined only for x > 0, we do not include x = 0 as a critical point in our analysis. It's important to remember that the domain of the original function plays a significant role in determining which points are relevant for analysis. Therefore, the only critical point we need to consider is x = 4.

This single critical point, x = 4, divides the domain of the function (x > 0) into two intervals: (0, 4) and (4, ∞). We will use these intervals to determine where the function is increasing and decreasing. The critical point acts as a potential turning point for the function, where it may switch from increasing to decreasing or vice versa. The next step is to test a value within each of these intervals in the first derivative to determine its sign. The sign of the derivative will tell us whether the function is increasing or decreasing in that interval. By carefully analyzing the critical points and the intervals they create, we can gain a comprehensive understanding of the function's behavior.

4. Determining Intervals of Increase and Decrease

Now that we have identified the critical point at x = 4, we can determine the intervals where the function h(x) = √x(x - 12) is increasing or decreasing. The critical point divides the domain (x > 0) into two intervals: (0, 4) and (4, ∞). To determine the behavior of the function in each interval, we need to test a value within each interval in the first derivative, h'(x) = (3x - 12)/(2√x). The sign of the derivative in each interval will tell us whether the function is increasing or decreasing.

Let's start with the interval (0, 4). We can choose any value within this interval, such as x = 1. Plugging x = 1 into the derivative, we get h'(1) = (3(1) - 12)/(2√1) = (3 - 12)/2 = -9/2. Since h'(1) is negative, the function is decreasing on the interval (0, 4). This means that as x increases from 0 to 4, the value of the function h(x) decreases. The negative sign of the derivative indicates a downward slope of the function's graph in this interval.

Next, we consider the interval (4, ∞). We can choose any value within this interval, such as x = 9. Plugging x = 9 into the derivative, we get h'(9) = (3(9) - 12)/(2√9) = (27 - 12)/(2*3) = 15/6 = 5/2. Since h'(9) is positive, the function is increasing on the interval (4, ∞). This means that as x increases beyond 4, the value of the function h(x) increases. The positive sign of the derivative indicates an upward slope of the function's graph in this interval. By analyzing the sign of the derivative in each interval, we have successfully determined where the function is increasing and decreasing.

5. Final Intervals and Conclusion

In conclusion, by analyzing the first derivative of the function h(x) = √x(x - 12), we have successfully identified the intervals where the function is increasing and decreasing. We found that the critical point x = 4 divides the domain (x > 0) into two intervals: (0, 4) and (4, ∞). By testing values within these intervals in the first derivative h'(x) = (3x - 12)/(2√x), we determined that:

The function is decreasing on the interval (0, 4).
The function is increasing on the interval (4, ∞).

These findings provide a comprehensive understanding of the behavior of the function h(x). In the interval (0, 4), the function's values decrease as x increases, indicating a downward trend. At x = 4, the function reaches a critical point, which is a local minimum. Beyond x = 4, in the interval (4, ∞), the function's values increase as x increases, indicating an upward trend. This analysis is crucial for sketching the graph of the function and understanding its overall characteristics.

The application of calculus, specifically the use of derivatives, allows us to precisely determine the intervals of increase and decrease for a given function. This information is essential in various fields, including optimization problems, where finding the maximum or minimum values of a function is required. The ability to analyze function behavior using calculus is a fundamental skill in mathematics and has wide-ranging applications in science, engineering, and economics. Understanding these concepts not only helps in solving mathematical problems but also provides a deeper insight into the behavior of real-world phenomena that can be modeled using functions.