Each of Exercises 67–88 gives the first derivative of a contin...

Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
74. y' = (x² - 2x)(x - 5)²

74. y' = (x² - 2x)(x - 5)²

Accepted Answer

Hello. In this video, we are going to be using the given information to sketch the approximate shape of the F of X function. We are told that the first derivative of a continuous function F of X is F X equal to X2 minus 2 X multiplied by X minus 1 quantity squared. We want to find the second derivative of F of X and sketch the general shape of its graph. Let's go ahead and approach this problem by first finding the second derivative of F of X. Now, we are given the first derivative, which is F X equal to this product of functions of X. If we want to take the derivative of F of X, we are going to need to use the product rule. Furthermore, we are going to need to use the chain rule, since we have functions inside of functions. So we have F of X equal to the first function X2 minus 2 X multiplied by the chain rule, the derivative of X minus 1 quantity squared. That will be 2 multiplied by X minus 1 multiplied by 1. Then we will add the second function, X minus 1 quantity squared, and multiply this by the derivative of X2 minus 2 X, which will be 2 X minus 2. Now what we are going to do is we are going to simplify the function. So, the first thing we are going to do is we are going to factor out an X minus 1 from the entire function. That is going to give us X minus 1 multiplied by the quantity, 2 multiplied by X2 minus 2 X plus X minus 1 multiplied by 2 X minus 2. The next thing we are going to do is we are going to combine all the like terms together in the brackets by first expanding what we have inside the brackets. So we are going to distribute 2 into X2 minus 2 X, and we are going to foil X minus 1 with 2 X minus 2. That is going to leave us with X minus 1 multiplied by the quantity, 2 X2 minus 4 X. Plus 2 X2 minus 2 X. Minus 2 X. Plus 2 Now, what we are going to do is we are going to combine like terms. That is going to leave us with X minus 1 quantity squared, or just X minus 1, multiplied by 4 X2 minus 8 X plus 2. Notice that we can factor out a 2 from this polynomial. Once we factor out the 2, the second derivative F of X will equal to 2 multiplied by x minus 1 multiplied by 2 X2 minus 4 X plus 1. This is going to be the simplest form of the second derivative of FX, and this is the first solution to this problem. OK. Now that we have the first derivative and second derivative, we are going to use these derivatives to approximate the shape of F of X. Let's go ahead and start by trying to get as much information as we can from the first derivative F of X. Now, from the first derivative, we can use this derivative to find the the critical points and the intervals of decreasing and increasing. In order to do this, we first need to calculate the critical points, and we can get the critical points by solving for X by setting the derivative equal to 0. This will allow us to set both of the factors of X equal to 0. We will have X2 minus 2 X equal to 0, and the quantity X minus 12 equal to 0. On the left-hand equation, we can factor out an X, leaving us with X multiplied by X minus 2 equals 0. We can set both factors of X equal to 0 and solve, and that is going to give us X's equal to 0 and X is equal to 2. Furthermore, if we take the square root of X minus 12, that is going to give us X minus 1 equal to 0, and solving for this value of X gives us X's equal to 1. So, there are 3 critical points in total, and what we want to do now is we want to go ahead and determine the intervals of increasing and decreasing by using the first derivative test. We are going to plot our critical points on a number line. And what we are going to do is we are going to take testing points from the 4 intervals on the number line, plug those testing points into the derivative, and depending on the output, we will know whether the function is increasing or decreasing in that region. The testing points we can pick are -1, 0.5, 1.5, and 3. When we plug in -1 into the derivative, F of -1 is going to give us the output of -2. This output is negative, oh sorry, not 2, it is going to be 12. Now, since this output is positive, that means that the graph will be increasing on the region from negative infinity to, or to 0. Next, we will plug in 0.5 into the derivative, and F of 0.5 is going to give us the output of negative 0.1875. Since this value is negative, the graph will be decreasing within this region. When we plug in 1.5 into the derivative, we get the output of negative. 0.1875. Again, this is negative, so the graph is decreasing in this region, and when we plug in the value of 3 into the first derivative, F3 gives us 12, which tells us that the graph will be increasing from 2 to infinity. OK. We have the intervals of increasing and decreasing, but one thing to note too is that as we pass through the value of 0, the graph switches from increasing to decreasing, meaning that we have a local minimum at the value X is equal to 0. The graph stays decreasing as we pass through one, so it is neither a local minimum or local maximum. But the graph switches from decreasing to increasing as we are passing through the value of 2, and that means that we have a local. Minimum at X is equal to 2, and I made a mistake here. There is not a local minimum at X is equal to 0, it is a local maximum, since we are switching from positive to negative. OK, so we have the local maximum and local minimum values, and we have the intervals of decreasing increasing from the first derivative. Now, what we are going to do is we are going to check the concavity of the graph by setting the second derivative equal to 0 and solving for X. That is going to give us 2 multiplied by X minus 1, multiplied by 2 X2 minus 4 X + 1 equal to 0. Now, setting the first factor of X X minus 1 equal to 0, will give us X is equal to 1, which is a possible inflection point. By using the quadratic equation on 2 X2 minus 4 X + 1, we get two values of X. X is equal to 1 plus or minus the square root of 2 divided by 2. This gives us two approximate decimal values for X. X is approximately equal to 1.71, and X is approximately equal to 0.29, which are going to be two possible critical points for the graph. Now, just like the number line that we made for the first derivative, we are going to do the same thing for the 2nd derivative to check the concavity of the graph. Plotting these points, we will have 0.29. We will have 1, and we will have 1.71. And again, we are going to pick testing points to plug into the second derivative. These points are going to be 0, 0.5, 1.5, and 2. Now, when we plug in 0 into the second derivative, F 0 will have the output of -2. This tells us that the graph will be concave down from 0 to 0.24. Now, when we plug in 0.5. Into the second derivative, we get 0.5, which tells us the graph is concave up in this region. Plugging in 1.5 into the second derivative gives us an output of -0.5, which tells us this region is concave down, and plugging in 2 into the second derivative gives us 2, which tells us that the graph is concave up within this region. Furthermore, 0.24 0.29. 1 and 1.71 are all going to be inflection points since the concavity switches as we pass through these points. So, all 3 points are going to be inflection points. Now that we've determined this information, let's go ahead and use this information to create a rough sketch of FF X. Now to help us, let's go ahead and create an axis. Now, we know that we have a local maximum at the value X is equal to 0, and a local minimum at X is equal to 2. We will go ahead and approximate X equal to 0 to be here, and let's just go ahead and place X equal to 2 on the bottom here. Now, we want to go ahead and plot our inflection points. 01 more, yeah, our inflection points. Now, we know that we have inflection points at 1 point or 0.24, which is the equivalent. To 1 minus. Square root, 2 divided by 2. We have an inflection point at X is equal to 1. And we have an inflection point. At 1 plus the square root 2 divided by 2. Now, from the first derivative, we know that the graph is going to be increasing from negative infinity to zero, and it is going to be concave down. So we can represent that behavior as the following. And if we just connect the dots. This is going to be the rough sketch of the graph of F of X. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
74. y' = (x² - 2x)(x - 5)²

Key Concepts

First Derivative

Second Derivative

Graphing Procedure Steps

Watch next

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.74. y' = (x² - 2x)(x - 5)²

Key Concepts

First Derivative

Second Derivative

Graphing Procedure Steps

Watch next

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.
74. y' = (x² - 2x)(x - 5)²