Decreasing Intervals Of Y = X / (2ln(x)): Find The Solution

Hey guys! Today, we're diving into a cool calculus problem: figuring out where the function y = x / (2ln(x)) is decreasing. This involves a bit of calculus magic, but don't worry, we'll break it down step by step so it's super easy to follow. Let's get started!

Understanding Decreasing Functions

Before we jump into the math, let's quickly recap what it means for a function to be decreasing. Basically, a function is decreasing over an interval if its value goes down as x increases. Think of it like walking downhill – your altitude decreases as you move forward. In calculus terms, we look at the derivative of the function. If the derivative is negative, the function is decreasing. So, our mission is to find where the derivative of y = x / (2ln(x)) is less than zero. This is a core concept, and understanding it will make the rest of the problem a breeze. Remember, a negative derivative means the function's slope is going downwards, hence, it's decreasing.

Step 1: Finding the Derivative

Okay, first things first, we need to find the derivative of our function, y = x / (2ln(x)). This might look a bit intimidating, but we can totally handle it using the quotient rule. The quotient rule states that if you have a function y = u(x) / v(x), then its derivative y' is given by: y' = (u'v - uv') / v². In our case, u(x) = x and v(x) = 2ln(x). So, let's find u' and v'. The derivative of u(x) = x is simply u'(x) = 1. Now, for v'(x), we need to differentiate 2ln(x). The derivative of ln(x) is 1/x, so the derivative of 2ln(x) is 2 * (1/x) = 2/x. Got it? Great! Now we have all the pieces to plug into the quotient rule. Let's do it! Plugging these into the quotient rule, we get:

y' = (1 * 2ln(x) - x * (2/x)) / (2ln(x))² = (2ln(x) - 2) / (4(ln(x))²). This might look a bit messy, but we're on the right track. We've successfully found the derivative, which is the first big hurdle in solving this problem. Remember, we're trying to find where this derivative is negative, so we're one step closer to our goal.

Step 2: Simplifying the Derivative

Now that we've got the derivative, y' = (2ln(x) - 2) / (4(ln(x))²), let's simplify it a bit to make our lives easier. We can factor out a 2 from the numerator, which gives us: y' = 2(ln(x) - 1) / (4(ln(x))²). Then, we can simplify the fraction by canceling out the 2 in the numerator with the 4 in the denominator: y' = (ln(x) - 1) / (2(ln(x))²). This looks much cleaner, right? Now, to figure out where this derivative is negative, we need to analyze the signs of the numerator and the denominator separately. The denominator, 2(ln(x))², is always non-negative (since it's a square), but we need to be careful about where ln(x) is defined. Remember, ln(x) is only defined for x > 0, and it's equal to zero when x = 1. So, we know that x cannot be 1. This is an important detail to keep in mind. Now, let's focus on the numerator, (ln(x) - 1), which will determine the sign of the derivative.

Step 3: Finding Critical Points

To figure out where the derivative changes sign, we need to find the critical points. These are the points where the derivative is either equal to zero or undefined. We already know that the derivative is undefined when ln(x) = 0, which means x = 1. Now let's find where the derivative is equal to zero. This happens when the numerator, ln(x) - 1, is equal to zero. So, we need to solve the equation ln(x) - 1 = 0. Adding 1 to both sides gives us ln(x) = 1. To solve for x, we take the exponential of both sides: x = e^1 = e. So, we have two critical points: x = 1 and x = e. These points divide the x-axis into intervals, and we need to test each interval to see where the derivative is negative. Remember, critical points are super important in calculus because they often mark where a function changes from increasing to decreasing, or vice versa.

Step 4: Analyzing Intervals

Okay, we've found our critical points: x = 1 and x = e. These points divide the number line into three intervals: (0, 1), (1, e), and (e, ∞). Remember, we're only considering x > 0 because ln(x) is only defined for positive x values. Now, we need to test a value within each interval to see if the derivative y' = (ln(x) - 1) / (2(ln(x))²) is positive or negative.

• Interval (0, 1): Let's pick x = 0.5. Then ln(0.5) is negative, so ln(0.5) - 1 is also negative. The denominator is always positive (except at x=1 where it's zero), so the derivative is negative in this interval. This means the function is decreasing in the interval (0, 1).
• Interval (1, e): Let's pick x = 2. Then ln(2) is between 0 and 1, so ln(2) - 1 is negative. Again, the denominator is positive, so the derivative is negative in this interval. This means the function is also decreasing in the interval (1, e).
• Interval (e, ∞): Let's pick x = e². Then ln(e²) = 2, so ln(e²) - 1 = 1, which is positive. The denominator is still positive, so the derivative is positive in this interval. This means the function is increasing in the interval (e, ∞).

So, by carefully analyzing the sign of the derivative in each interval, we can determine where the function is increasing or decreasing. This is a powerful technique in calculus and helps us understand the behavior of functions.

Step 5: Stating the Answer

Alright, we've done all the hard work! We found the derivative, simplified it, found the critical points, and analyzed the intervals. Now we can confidently state where the function y = x / (2ln(x)) is decreasing. Based on our analysis, the function is decreasing in the intervals (0, 1) and (1, e). Remember that the function is not defined at x = 1, so we exclude it from the intervals. So, there you have it! We've successfully found the intervals where the function is decreasing. This problem involved a few steps, but by breaking it down and tackling each part systematically, we were able to solve it. Great job, guys!

Conclusion

So, to wrap things up, finding the intervals where a function is decreasing involves a few key steps: finding the derivative, simplifying it, finding critical points, and analyzing the intervals. We applied these steps to the function y = x / (2ln(x)) and found that it's decreasing in the intervals (0, 1) and (1, e). This type of problem is a classic example of how calculus can help us understand the behavior of functions. Keep practicing, and you'll become a pro at these in no time! Remember, calculus is all about understanding change, and these types of problems help us visualize and analyze how functions change over different intervals. Keep up the great work!