First Derivative Test

a. Locate the critical ...

Question

First Derivative Test

a. Locate the critical points of f.

b. Use the First Derivative Test to locate the local maximum and minimum values.

c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

f(x) = x²/(x² - 1) on [-4,4]

Accepted Answer

Hello. In this video, we are going to be identifying critical points using the first derivative test and identifying absolute maximum and minimum values. So, the problem asks us to consider the function F of X equal to 3 X2, divided by X2 minus 9 on the interval from -5 to 5. We want to find the critical points of F and use the first derivative test to classify these points. Then we want to determine the absolute maximum and minimum values of F on the specified interval. Let's go ahead and start this problem by approaching the first half of the problem. We will start by identifying the critical points of the given function. Now, in order to find the critical points of the function, we want to find the values of the function that causes its derivative to be zero. So in order to find critical points, we first need to take the derivative of the given function and set it equal to zero. So, the function given to us is F of X equal to 3 X2 divided by the quantity X2 minus 9. We will take the derivative with respect to X by using the quotient rule. That will give us F X equal to X2 minus 9 multiplied by the derivative of the numerator, which is 6 X minus 3X2 multiplied by the derivative of the denominator, which will be 2 X, and this will all be divided by the quantity X2 minus 9 quantity squared. Next, what we want to do is we want to simplify the derivative. We will distribute 6x into the quantity X2 minus 9. That will allow us to simplify the numerator to be 6 x cubed minus 54 X. Then we will subtract the product of 3 x 2 and 2 X, which will be 6 x cubed. And again, this will all be divided by X2 minus 9 quantity squared. We will then combine like terms together in the numerator, eliminating the 6 X cube terms, and that will give us a final derivative of negative 54 X divided by the quantity X2 minus 9 squad. Now, this is a derivative of a rational function. So, in order to determine the critical points, what we want to do is you want to find the values of X that cause the numerator to be zero. If the numerator is 0, then the entirety of the function becomes zero. So in order to find the critical points, we will take the numerator -54 X and set it equal to 0. If we divide both sides of this equation by -54, we get the value X is equal to 0. What this means is that we have one critical point of the function. The critical point is going to be located at the value X is equal to 0, which is the first solution to this problem. Now, what we want to do is we want to use the first derivative test to identify if this critical point is a local maximum or local minimum. In order to do this, we will go ahead and use a number line. Now, on this number line, we are going to label the boundaries of our interval. Now, recall that we are finding or working on the interval from -5 to 5. So we will go ahead and label that on our number line. Next, we will go ahead and label the critical point on our number line. That is going to be at 0. Now, there is one more thing we need to label on the number line. We need to label the restrictions of the domain of the function. Now, keep in mind that the original function was 3X2d divided by X2 minus 9. In order to find the restrictions of the domain, we will take the denominator, set it equal to zero, and solve for X. And when we do that, we are going to get two values of X. We get X is equal to positive and -3, and these two values represent vertical asymptotes. So we will go ahead and label the vertical asymptotes on our number line. Now, in order to use the first derivative test, what we are going to do is we are going to take testing points on the interval from -3 to 0, and from 0 to 3. We will then plug those testing points into the derivative to determine the behaviors of the graph on that interval. So what is a testing point that we can take on the interval from -3 to 0? Well, the easiest testing point we can work with is -1. If we take -1 and plug it into the derivative, F of -1 will give us -54 multiplied by -1 divided by -1 2 minus 9 squad. This will simplify to give us the value of 27 divided by 32. Now, since this testing point produced a positive value, this tells us that the graph on the interval from -3 to 0, will be increasing. We will go ahead and repeat this process for the interval from 0 to 3. An easy testing point we can pick in this region is 1, and if we plug in 1 into the derivative, we will get the value negative 27 divided by 32. And since this is a negative number, this tells us that the graph will be decreasing on the interval from 0 to 3. Now, by the first derivative test, since the behaviors of the graph are switching from increasing to decreasing as it passes through the critical 0.0 this tells us that we have a local maximum at the value X is equal to 0, and this is how you would classify the critical point by using the first derivative test. Now what we want to do is you want to identify the absolute maximums and minimums. Now, there is a problem with this type of function though. Recall that earlier we stated that we're vertical asymptotes of the function. Now, if you were to approach or use the limit tests, as X approaches the value of -3 from the right hand side, and as X approaches the value of 3, From the right hand, uh, from the left hand side. Both of these values of X will approach negative infinity. And vice versa if we approach -3 from the left hand side. And we approach. Positive 3 from the right hand side, the graph will increase infinitely. That means that the highest value that the graph reaches will be positive infinity, and the lowest value the graph reaches is negative infinity. But since these are both non-determined numbers or these are not finite numbers, that means that there is not going to be any local, any absolute maximum or minimum value. So there are no absolute extrema. And that is going to be the final part of this problem. So, I hope this video helps you in understanding how to use the first derivative test and to find local extrema, or absolute extrema, and I will go ahead and see you all in the next video.

First Derivative Testa. Locate the critical points of f.b. Use the First Derivative Test to locate the local maximum and minimum values.c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).f(x) = x²/(x² - 1) on [-4,4]

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First Derivative Test

a. Locate the critical points of f.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

f(x) = x²/(x² - 1) on [-4,4]