Increasing and decreasing functions. Find the intervals on whi...

Question

Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.

f(x) = xe⁻(ˣ²/₂)

Accepted Answer

Hello. In this video, we are going to be identifying the intervals of increasing and decreasing of the given function. The function given to us is F of X equal to X2, multiplied by E raise to the power of negative X2 divided by 4. Now, in order to identify the intervals of increasing and decreasing, the first thing we need to do is identify the critical points. In order to obtain the critical points, we are going to need to calculate the derivative of the given function. In order to calculate the derivative, we are going to need to use the product rule. By using the product rule, the derivative of F of X is going to be X2, multiplied by the derivative of Ease the power of negative X2 divided by 4. By the exponential rule that is going to be E, raised the power of negative X2 divided by 4, multiplied by the natural logarithm of E, multiplied by the derivative of negative X2 divided by 4, and that is going to be negative X divided by 2. Then we will add Ease the power of negative X2 divided by 4, and multiply this by the derivative of X2, which is 2 X. Next, what we are going to do is we are going to simplify the function. Notice Well, the first thing here is that the natural logarithm of E will simplify to be 1. Furthermore, notice that each term within the derivative has a common factor of E raised the power of negative X2 divided by 4. What we can do is we can factor route that exponential value. That will give us E raise to the power of negative X2 divided by 4, multiplied by X2 multiplied by X X divided by 2, which is negative X cubed divided by 2, plus 2 X. Now, if we reorder the terms with respect to the positive term, we can rewrite the derivative to be E, raise the power of negative X2 divided by 4, multiplied by the quantity, 2 X minus X cubed divided by 2. This is going to be the simplest form of the derivative. Now what we want to do is we want to obtain the critical numbers. In order to obtain the critical numbers, we will go ahead and set the derivative equal to 0, and solve for the values of X that cause it to be zero. That will be E, raise the power of negative X2 divided by 4, multiplied by 2 X minus X cubed divided by 2, and this will be equal to 0. This will give us two equations to solve. E raised the power of negative X2 divided by 4 equal to 0, and 2 X minus Xcu divided by 2 equal to 0. Now, on the left-hand side, since X is located in the exponential value, there is no possible value of X that is going to give us an exponential value of 0. So we can go ahead and disregard this left-hand equation. But what we want to do now is solve for the right-hand equation. We want to go ahead and eliminate the denominator of 2. So, the first thing we can do is multiply the entire equation by 2. This will give us 4 X minus X cubed equal to 0. The next thing we can do is we can go ahead and factor out an X. This will give us X multiplied by 4 minus X2d is equal to 0. This gives us two equations. X is equal to 0 and 4 minus X2 equal to 0. If we subtract 4, then divide by -1, we get X2d is equal to 4, which further gives us X is equal to positive and -2. So, in total, we have 3 critical points. We have X is equal to 0, and X is equal to positive and negative 2. Now, in order to obtain The intervals of increasing and decreasing, we are going to create a number line. This number line is going to be from negative infinity to positive infinity. Then we will label our critical points at -2, 0, and 2. Now what we want to do is we want to take testing points within each independent interval, plug it into the derivative, and depending on if the derivative is positive or negative, that will tell us the behavior of the graph within that region. Let's go ahead and pick a testing point from negative affinity to negative 2. That testing point is going to be -3. When we plug this into the derivative, F of -3 will give us E raised to the power of -9 divided by 4, multiplied by 15 divided by 2. Now, this exponential value of E is always going to be positive, so we have a positive multiplied by a positive, which gives us a positive derivative. What this means is that since the derivative is positive in this interval, the original function will be increasing within this region. Next, we will pick a testing point between -2 and 0, which is -1. F of -1 will give us E raise the power of -1/4, multiplied by -3 divided by 2. Positive multiplied by negative is a negative value, so that tells us that the region between -2 and 0 is going to be a region where F of X decreases. A testing point between 0 and 2 is 1, F1 is equal to E raise to the power of -14. Multiplied by 3 divided by 2, which is positive. And the testing point between 2 and positive infinity is 3, and F of 3 is equal to E, raises the power of -9 divided by 4, multiplied by -15 divided by 2. This will give us a negative number, so the graph will be decreasing within that region. So based off the number line we just created, The intervals of increasing will be from negative affinity to -2, and from 0 to 2. And the intervals of decreasing will be from -2 to 0 and from 2 to infinity. With that being said, the solution to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem, and I will go ahead and see you all in the next video.

Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.f(x) = xe⁻(ˣ²/₂)

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Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.

f(x) = xe⁻(ˣ²/₂)