Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

g(x) = (x − 2) / (x²−1), 0 ≤ x < 1

Accepted Answer

Hello. In this video, we are going to be identifying the local extrema for the given function F X on the interval from -5 to 5. The function F of X is given to us as 2 X minus 5 divided by X2 minus 4. Now, in order to solve for the local extreme values, what we want to do is we want to first take the derivative of our given function. In order to take the derivative, we will use the quotient rule. That will be the original denominator, X2 minus 4 multiplied by the derivative of the numerator, which will be 2 minus the original numerator to X minus 5 multiplied by the derivative of the denominator, which will be 2 X, all divided by the original denominator squared, which is X2 minus 4 squad. In the numerator, when we simplify the terms, we will distribute 2 into X2 minus 4, leaving us with 2 X2 minus 8 and 2 X distributed into 2 X minus 5, and then multiplying by -1 will leave us with negative 4 X2 + 10 X, and this will all be divided by X2 minus 4 quantity squared. And by combining like terms, we can simplify the derivative F of X to be negative to X2 plus 10 X minus 8 divided by X2 minus 4 quantity squared. So, in order to identify the local extreme values, we need to first identify the restrictions of the domain of the of the derivative and the values of X that cause the derivative to be zero. For the domain. We don't want X2 minus 42 to be zero. So to solve for the restrictions, we will take our we will take our denominator, X2 minus 4 quantity square, set it equal to 0, and solve for X. By taking the square root of both sides of this equation, we get X2 minus 4 equal to 0. By adding 4, we have X2 equal to 4, and by taking the square root, we get X's equal to positive or negative 2. What this means is that we do not want X to be positive or negative too, because it will cause our derivative to be undefined. Now what we want to do is we want to solve for the critical points, which are the values of X that cause the derivative to be zero. In order to do this, we will take the numerator of our function, -2 X2 plus 10 X minus 8, set it equal to 0, and solve for X. In order to simplify this equation, since every coefficient has a multiple of two, and the leading coefficient is negative, we can divide the entire equation by negative 2. That will allow us to simplify the equation to be X2 minus 5 X plus 4 equal to 0. The left hand side will factor to be X minus 4 multiplied by X minus 1, and this will give us the values of X. X is equal to 4 and 1. So, these are going to be the critical points. And now what we want to do is you want to determine the behaviors around the critical points. Now recall that in the original problem, we only want the values of X to be between 5 and 5. So in order to determine the behaviors, let's go ahead and use the number line rule or the number line trick. So, we will create a number line and we will go ahead and label our end points. The end points are going to be from -5 to 5. And now what we want to do is you want to label the restrictions of the domain of the derivative, and the critical points on this number line. That is going to be negative too. Positive 1 Positive 2 And 4. We specifically want to check the behaviors of the graph around 1 in 4, since those are critical points. So what we want to do is we want to take testing points around the neighborhoods of these values and determine the behavior of the graph by taking those testing points and plugging them into the derivative. A testing point between -2 and 1 is 0. A testing point between 1 and 2 will be 1.5. Between 2 and 4 will be the value of 3, and between 4 and 5 will be 4.5. We are going to take these testing points and plug them into the derivative. So F of 0, which is the 0 plugged into the derivative, is going to give us the value of negative 0.5. Since the output is negative, this tells us that the region between -2 and 1 will be decreasing. If we plug in 1.5 into the derivative, F of 1.5 gives us the output of 0.82. Since this is a positive value, the graph will be increasing within this region. By plugging in 3 into the derivative, we will get the output of 0.16. Since this is a positive value, the graph will be increasing within this region, and by plugging in 4.5 into the derivative, we get negative 0.05, and since this is a negative number, the graph is decreasing within this region. Now, why is this important? Notice that as the graph passes through the value of 1, the graph switches from decreasing to increasing. This means that we will have a local minimum value at the value X is equal to 1. And notice that as the graph passes through the value of 4, the graph switches from increasing to decreasing. This means that we have a local maximum at the value X is equal to 4. But now what we want to do is you want to determine the Y values of these local maximum and local minimum values. In order to do this, we will go ahead and plug in these values into the original function. By plugging in one into the original function. We will go ahead and get the value of one. And by plugging in 4 into the original function, we get the value of 14th. So what this means is that we have a local minimum. At the value X is equal to 1 with the Y value of 1. And we have a local maximum. At the value X is equal to 4, with the output of Y is equal to 1/4, and this is going to be the solution to this problem. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

g(x) = (x − 2) / (x²−1), 0 ≤ x < 1

Key Concepts

Local Extrema

Critical Points

Domain Restrictions

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