Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = x³ / (3x² + 1)

Accepted Answer

Hello, in this video, we are going to be considering the function F of X is equal to X2 divided by X + 2. We want to describe the open intervals on which F of X is increasing and decreasing, and identify its local extreme values. So, since we are asked to find the intervals of increasing and decreasing, and the local extreme values, the first thing we want to do is we want to take the derivative of the given function. Now, the function given to us is F of X is equal to X2 divided by X + 2. Let's just go ahead and quickly discuss the domain. Now, the only value of X that causes the domain to be undefined is X's equal to -2. So that means that the domain is going to be from negative infinity to -2, union -2 to positive infinity. Now that we've discussed the domain, let's go ahead and take the first derivative of this function. By using the quotient rule, we will have F of X equal to X + 2 multiplied by 2 X minus X2 multiplied by 1, and divided by X + 2 quantity squared. By simplifying the numerator, we will have 2 X2 + 4 X minus X2 divided by the quantity X plus 2 quantity squared, and by combining like terms, we will get X2 + 4 X divided by X + 2 quantity squared. Now, in order to determine the critical points to find the increasing and decreasing intervals, what we want to do is we want to find the values of X that cause the numerator of this function to be zero. In order to do this, we will take the numerator X2 + 4 X, set it equal to 0, and solve for X. We will go ahead and factor out an X from the left hand side of this equation, that will give us X multiplied by the quantity X + 4 equal to 0. And since we have two factors of X equal to 0, we can set each independent factor of X equal to 0. That will give us X equal to 0, and X is equal to -4. And these are going to be two critical points for the number line. One more thing to consider too is the restriction of the domain. Now, originally, we stated that the only restriction we have is negative 2. So when we use the first derivative test to find the intervals of increasing and decreasing, we also need to take the domain restriction into consideration. So that is going to be a critical point as well. Now, in order to use the first derivative test, we will go ahead and create a number line and label all of our critical values on this number line. We have -4. 2 And 0 Now what we want to do is we want to take testing points within these neighborhoods of the critical points to plug back into the derivative and determine the behaviors of the graph. Some testing points we can pick are -5, -3. -1 and 1. Now F of -5 is going to give us an output of 0.6. Since this is a positive value, the function will be increasing from negative infinity to -4. If we take -3 and plug it into the derivative, we get the output of -3, which tells us that the function is decreasing within this region. If we take -1 and plug it into the derivative, we get -3, and that tells us that the derivative, or the overall function is decreasing within this region, and plugging in positive 1 into the derivative will give us the value of 0.6, which again tells us the the overall graph is increasing in this region. So let's just go ahead and label the regions of increasing and decreasing. Now, the graph will be increasing from negative infinity to -4, and it is also increasing from 0 to positive infinity. We also determined that the graph will be decreasing. On the intervals from -4 to -2, union -2. To 0 So these are going to be our intervals of increasing and decreasing, which is the first part or the first solution that we want to this problem. Now what we want to do, recall, is that we want to find the local extreme values. Now, in order to do this, we first need to identify our local maximum and local minimum values. Notice that as we pass through -4, the graph shifts from increasing to decreasing. This tells us that we have a local maximum at X is equal to 4. There is no shift in increase or decrease as we passed -2, but that is a vertical asymptote, so that's understandable. And as we pass through zero, the graph shifts from decreasing to increasing, which means that we have a local minimum at X is equal to 0. Now, in order to find the extreme values, we have to take these values of X is equal to 4 and X is equal to 0, and plug them back into the original equation. So at the local maximum, which is located at X is equal to 4, we get F of 4, and that is going to give us the output of -8. And for the local minimum. Which is located at X is equal to 0, plugging in 0 into the original equation is going to give us the output of 0. And these are going to be the local extreme values to the function, which is the second solution to this problem. So I hope this video helps you in understanding on how to approach this type of problem, and I will go ahead and see you all in the next video.

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = x³ / (3x² + 1)

Key Concepts

Derivative and Critical Points

First Derivative Test

Interval Testing

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