Hey guys! Let's dive into analyzing the function f(x) = 5(x-4)^(2/3). This function presents some interesting challenges, especially when we're trying to figure out where it's increasing or decreasing. We'll break it down step by step, so you can understand the concepts thoroughly. The key here is identifying the critical numbers and then testing intervals around those numbers. Let's get started!
Finding the Critical Number A
The first thing we need to do is find the critical numbers of the function. Critical numbers are those points where the derivative of the function is either zero or undefined. These points are crucial because they often mark transitions between increasing and decreasing intervals. Understanding critical numbers is paramount in analyzing function behavior, as they pinpoint potential maxima, minima, and inflection points. By identifying these critical junctures, we can effectively map out the function's trajectory and grasp its overall dynamics.
Calculating the Derivative
To find the critical numbers, we need to calculate the derivative of f(x). Remember the power rule? We'll be using that here. The power rule states that if f(x) = ax^n, then f'(x) = nax^(n-1). Applying this to our function, we get:
f(x) = 5(x-4)^(2/3)
f'(x) = 5 * (2/3) * (x-4)^((2/3) - 1)
f'(x) = (10/3) * (x-4)^(-1/3)
f'(x) = 10 / (3 * (x-4)^(1/3))
So, there’s our derivative. Take your time to go over the steps if you need to. Derivatives are the backbone of calculus, and mastering these calculations sets a strong foundation for more advanced concepts. The derivative we found will help us identify critical points, which are essential for understanding the function’s behavior.
Identifying Points Where the Derivative is Zero or Undefined
Now, we need to find the values of x where f'(x) is either zero or undefined. A fraction is undefined when its denominator is zero. So, we need to find when:
3 * (x-4)^(1/3) = 0
(x-4)^(1/3) = 0
x - 4 = 0
x = 4
The derivative is undefined at x = 4. Now, let's see if the derivative can ever be zero. The numerator of our derivative is 10, which is never zero. Therefore, the derivative f'(x) is never zero.
Thus, we have only one critical number: A = 4. This critical number, derived from the function's derivative, is pivotal in determining intervals of increase and decrease. The derivative, a cornerstone of calculus, provides invaluable insights into a function's behavior, revealing critical points where the function's direction shifts. By pinpointing these critical values, we gain a deeper understanding of the function's dynamics and overall characteristics.
Determining Intervals of Increase and Decrease
Now that we've found our critical number A = 4, we can determine the intervals where the function is increasing or decreasing. We’ll use a sign chart for this, which is a super helpful tool for visualizing the behavior of the derivative across different intervals. Sign charts allow us to efficiently track whether the derivative is positive or negative in various regions, giving us direct insight into the function’s increasing or decreasing nature.
Creating a Sign Chart
We have two intervals to consider: (-∞, 4) and (4, ∞). We'll pick a test value in each interval and plug it into f'(x) to see if the derivative is positive or negative.
Interval (-∞, 4):
Let's pick x = 0.
f'(0) = 10 / (3 * (0-4)^(1/3))
Since (0-4)^(1/3) is a negative number, f'(0) will be negative. This indicates that the function is decreasing on this interval. The negative derivative signifies a downward slope, a key characteristic of a decreasing function.
Interval (4, ∞):
Let's pick x = 5.
f'(5) = 10 / (3 * (5-4)^(1/3))
Since (5-4)^(1/3) is a positive number, f'(5) will be positive. This indicates that the function is increasing on this interval. A positive derivative is a telltale sign of an increasing function, where the slope moves upward.
Intervals of Increase and Decrease
Based on our sign chart, we can conclude:
- f(x) is decreasing on the interval (-∞, 4).
- f(x) is increasing on the interval (4, ∞).
So, there you have it! We’ve successfully identified the critical number and the intervals where the function is increasing and decreasing. Understanding these behaviors is crucial for graphing the function and understanding its properties. By analyzing the derivative, we’ve uncovered the function's trend, revealing where it climbs and where it descends.
Detailed Analysis of the Intervals
Let’s break down why these intervals behave the way they do. By understanding the underlying principles, you can apply these concepts to any function you encounter. It’s not just about getting the right answer, but understanding the ‘why’ behind the math.
Interval (-∞, 4): The Decreasing Trend
In the interval (-∞, 4), the function f(x) = 5(x-4)^(2/3) is decreasing. This means that as x moves from negative infinity towards 4, the value of the function f(x) gets smaller. The negative derivative in this interval confirms this behavior, indicating a downward slope. This decreasing trend is a result of the fractional exponent and the shift caused by (x-4). The function's behavior on this interval is dictated by the interplay between the fractional exponent and the linear term, creating a consistent downward trajectory.
To really get a sense of this, imagine picking some values in this interval and plugging them into the function. For instance:
- If x = 0, f(0) = 5(-4)^(2/3) ≈ 12.599
- If x = 3, f(3) = 5(-1)^(2/3) = 5
You can see that as x increases from 0 to 3, the function value decreases from approximately 12.599 to 5. This further illustrates the decreasing nature of the function in this interval. The numerical decrease highlights the function's downward slide, showcasing the impact of the negative derivative and fractional exponent in this region.
Interval (4, ∞): The Increasing Trend
In the interval (4, ∞), the function f(x) = 5(x-4)^(2/3) is increasing. This means that as x moves from 4 towards infinity, the value of the function f(x) gets larger. The positive derivative in this interval confirms this behavior, indicating an upward slope. The function's increasing nature is also a result of the fractional exponent, but this time the (x-4) term is positive, leading to a different outcome.
Again, let’s plug in some values to see this in action:
- If x = 5, f(5) = 5(1)^(2/3) = 5
- If x = 8, f(8) = 5(4)^(2/3) ≈ 12.599
As x increases from 5 to 8, the function value increases from 5 to approximately 12.599. This demonstrates the increasing nature of the function in this interval. The numerical increase underscores the function's upward climb, highlighting the influence of the positive derivative and fractional exponent in this segment.
The Critical Point A = 4: A Change in Direction
The critical point A = 4 is where the function changes direction. To the left of x = 4, the function is decreasing, and to the right of x = 4, the function is increasing. This point represents a local minimum for the function. At this point, the function transitions from a downward trajectory to an upward climb, marking a crucial shift in its behavior.
Graphically, this point would be the bottom of a