{Use of Tech} Newton’s method and curve sketching Use Newton’s...

Question

{Use of Tech} Newton’s method and curve sketching Use Newton’s method to find approximate answers to the following questions.

Where are the inflection points of f(x) = (9/5)x⁵ - (15/2)x⁴ + (7/3)x³ + 30x² + 1 located?

Accepted Answer

Hello. In this video, we are going to be using Newton's method to locate the inflection points of the given function. So before we approach this function, let's just go ahead and review what Newton's method is stating. Newton's method is given to us as a recursive formula. XN + 1 is equal to XN minus. G of X G of X N divided by G of XN. So, in order to use Newton's method, first we want to identify what we are trying to approximate. Then we want to go ahead and create a function that gives us this approximation, and we will calculate the approximate value by using the recursive formula. So, first of all, we want to approximate the location of the inflection points. Now, normally, when we want to find inflection points, we want to find where the second derivative of the given function is equal to 0. So, in order to start this problem, let's go ahead and calculate the second derivative of the given function. Now, the function given to us is F of X equal to 3/5 multiplied by X 5th, minus 25 divided by 6 X 4th power, plus 7 X cubed plus 10 X 2 + 2. In order to find the first 2nd derivative, we first need to calculate the first derivative. So let's go ahead and take the first derivative of the function. So, by the power rule, the derivative of 3/5 X 5th power will be 3 x 4th power. By the power rule again, the derivative of negative 25 divided by 6 X4 will be negative 50 divided by 3 X3. By the power rule, the derivative of 7 x cubed will be 21 x 2. By the power rule, the derivative of 10 x 2 will be 20 X. And because 2 is a constant, the derivative of a constant will be 0. Now that we have the first derivative, we will take the derivative again in order to get the value or get the function of the second derivative. Now, again, we are going to apply the power rule to each of the terms in F of X. That is going to give us the derivative 12 X cubed minus 50 X2 + 42 X plus 20. This is going to be the 2nd derivative of the given function. Now, since we know that if we set this function equal to 0, it will give us the inflection points, and since we are using Newton's method to find the locations of the inflection points, we are going to call the double derivative or G of X for Newton's method. So G of X is equal to 12 x cubed minus 50 X 2 plus 42 X plus 20. Now for Newton's method, we also need the derivative of this function. So we will go ahead and take the derivative of G of X. That is going to be 24 X2 minus 100 X plus 42. So, we have the two pieces we need for Newton's method. We've identified the G of X and GX function that we are going to plug in, but now what we need to do is we need to identify a starting value of X. Well, in order to identify a starting value of X, we can go ahead and take a guess as to what X is. Now, the most convenient value of X to use in this case is going to be 1. So we're going to let X sub zero equal to 1 when N is equal to 0. Now, from the recursive formula, if we use this value of X, Then the recursive formula becomes XN + 1, which is going to be X1 equal to X 0, which is 1, minus 12 multiplied by 1 cubed, minus 50 multiplied by 1 squared, plus 42 multiplied by 1 plus 20, and this will be divided. By the derivative with XA been plugged in, and that is going to be 24 multiplied by 1 squared, minus 100 multiplied by 1 plus 42. With the help of a calculator, this is going to give us an approximate value of 2.0909090. This is the value to with respect to 7 decimal places. And now what we want to go ahead and do is we want to go ahead and repeat this process for different values of N, for the recursive formula, and pay attention to when we have two outputs that agree with each other. So, let's just go ahead and create a number line. We will have N as our inputs, and X of N as our outputs for the recursive formulas. Now let's go ahead and just plug in different values of N. Let's try all the values from 0 to 6. So, since we said that our starting value for the recursive formula X 0 is equal to 1, when N is equal to 0, XN will be 1. Now, when we plug this into the recursive formula, X1 gave us the output of 2.0909090 to 7 decimal places. Now, if we repeat this process with the recursive formula, when we plug in the value N is equal to 2, that output will be 1.9794021. When we plug in 3, we get to 1.9993872. When we plug in 4, we get to 1.999994. When we plug in 5, we get the value of 2, and when we plug in 6, we get the value of 2. Notice that 5 of 65 and 6 have the same output. By Newton's method, this tells us that the location of one of the inflection points. is going to be X is equal to 2. But now we need to ask ourselves, is there any other inflection points that we can identify from the original equation? Well, since X is equal to 2 was an inflection point by Newton's method, if we want to rewrite this as a factor of X, we can rewrite this as X minus 2. Now, what if we were to take the 2nd derivative. And divide by X minus 2. That will allow us to identify the other values of X that are going to be inflection points. So F of X, divided by the quantity X minus 2, which is one of the roots, is going to be 12 X cubed. Minus 50 X 2 plus 42 X plus 20 divided by X minus 2. And through polynomial long division. That is going to allow us to rewrite F of X as X minus 2 multiplied by the quantity, 12 X squared minus 30, 26 X. -10 Now, if we were to factor this further, we can rewrite F of X as X minus 2 multiplied by the quantity, X minus 5 divided by 2, multiplied by X plus 1/3. This tells us that there are two other roots that we can identify from the second derivative. That is going to give us X is equal to 5 divided by 2, and X is equal to -13. But now we need to verify that these are going to be inflection points. In order to verify this, we will go ahead and use the 2nd derivative test. We'll create a number line. And on this number line, we are going to plot the remaining 2 inflection points. We have 1/3. And we have 5 halves. And now what we want to do is you want to take a testing point on the intervals from negative infinity. To positive infinity, not including the inflection points, and depending on the outputs, that will tell us the behaviors of the concavity of the graph. A testing point that we can pick to the left of -13 is going to be -12. If we take -12 and plug it into the second derivative F, that is going to be F of -12, and that will give us the value of -15. Since this value was negative, this tells us that the concavity to the left of negative 1/3 is going to be concave down. Next, a testing point that we can pick between -1/3 and 5 halves is 0. If we plug in 0 to the double derivative F of X, that will give us F of 0, and that will give us the output of 20. Since the output is positive, the concavity in this region is going to be concave up. And finally, a testing point that we can pick to the right of five halves is 2.1. If we plug in 2.1 into the double derivative, F of 2.1 will give us the value of negative 1.168. Since this value of is negative, that tells us that the concavity is concave down within this region. Now, the behaviors around -1/3 switch from concave down to concave up. This verifies that X equal to -1/3 is going to be an inflection point. And the behavior around 5 halves is going to be a switch from concave up to concave down, and that verifies that 5 halves is going to be another inflection point. And we found a 3rd inflection point through Newton's method, which is X is equal to 2. So these three values of X are going to be the values of the inflection points through Newton's method. And with that being said, the solution to this problem is going to be a. 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.

{Use of Tech} Newton’s method and curve sketching Use Newton’s method to find approximate answers to the following questions.
Where are the inflection points of f(x) = (9/5)x⁵ - (15/2)x⁴ + (7/3)x³ + 30x² + 1 located?

Key Concepts

Newton's Method

Inflection Points

Second Derivative Test

Watch next