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.
82. y' = sin t, for 0 ≤ t ≤ 2π

82. y' = sin t, for 0 ≤ t ≤ 2π

Accepted Answer

Hello. In this video, we are going to be sketching the approximate shape of F of X using the given information. We are told that the first derivative of a continuous function F of X is F of X equal to two cosine of X on the interval from 0 to 2 pi. We want to find the second derivative of F of X and sketch the general shape of its graph. So, let's go ahead and approach this problem by first finding the second derivative to the function F of X. Now, we are given the first derivative, which is F of X is equal to 2 cosine of X. In order to take the derivative, by definition, the derivative of cosine is negative sine. So F of X. is equal to -2 multiplied by sin of X. This is going to be the second derivative of this function, and this is going to be one of the solutions we need for the problem. OK. Now what we want to do is we want to use the 1st and 2nd derivative in order to create a rough sketch of the FXX function. By using the first derivative F of X, we can find the regions of increasing and decreasing. For the function f of X. We will start by finding our critical points, and in order to do this, we will go ahead and solve for when the derivative is equal to 0. So, by dividing both sides of this equation by 2, we are left with cosine of X equal to 0. Now, the question here is, when is cosine of X 0 on the first rotation of the unit circle from 0 to 2 pi? Well, if we draw a unit circle, recall that cosine of X is the X values of all the points on the unit circle. There are two main angles where X is equal to 0. Those angles are at pi divided by 2 and 3 pi divided by 2. So there are two solutions to this problem. X is equal to pi divided by 2 and 3 pi divided by 2. These are going to be our potential critical points. Now that we have our critical points, we are going to use the first derivative test by plotting these points on the number line and determining the behaviors of the intervals they form. So what are some testing points that we can pick? On the interval between 0 and pi divided by 2, pi divided by 2 and 3 pi divided by 2, and 3 pi divided by 2 to 2 pi. While these testing points can be pi divided by 4. Pie And 37 pi divided by 4. Now what we are going to do is we are going to plug in these values into the first derivative and determine the behavior of the graph based off the output. So, when we plug in pi divided by 4 into the first derivative, our output is going to be positive square root of 2. Since this value is positive, the graph will be increasing within this region. If we plug in pi into the first derivative, F of i will give us the value of -2. Since this value is negative, the graph will be decreasing on this interval. And if we plug in 7 pi divided by 4 into the first derivative, the output of the derivative is going to be positive square root of 2. And again, because this is a positive output, the graph will be increasing within this region. One more thing to note too. Not that as we pass through pi divided by 2, the graph switches from increasing to decreasing. This means that we have a local maximum at the value X is equal to pi divided by 2. And the graph switches from decreasing to increasing as we are passing through the 03 pi divided by 2, and that tells us that we have a local minimum at X is equal to 3 pi divided by 2. Now that we have the locations of the local maximum and local minimum values, and we have the regions of increasing and decreasing, we need to check the concavity of this function. Now, again, to check the concavity, we are going to use the second derivative we calculated, that is F of X equal to -2 sin of X. And what we want to do is we want to find the locations of the possible inflection points. The inflection points occur when the second derivative is equal to 0. Dividing both sides of this equation by -2 leaves us with sin of X equal to 0. Now, the question is, where is sin of X equal to 0 on the interval from 0 to 2 pi? On the unit circle, sine is the Y value of any coordinate point along the unit circle, and there are two locations where sin of X is equal to 0. That is going to be at 0 and it is going to be at pi. It'll also be at 2 pi, but since the end points of our boundary cannot be the inflection points, that means that there is only one solution to this function, X is equal to pi. Now we are going to check the concavity around pi. So, we will draw a number line from 0 to 2 pi. And we will place our possible inflection point pi here. Now, just like before, we are going to take testing points to plug into the second derivative. Those testing points can be pi divided by 2 and 3 pi divided by 2. Now, when we plug in pi divided by 2 in the second derivative, the output of the derivative is going to be -2, which tells us that the graph is going to be concave down from 0 to pi. Also, if we plug in 3 pi divided by 2 into the function, the second derivative, that is going to give us an output of 2, which tells us that the graph will be concave up from the region of pi to 2 pi. Furthermore, because the concavity switches from concave down to concave up as you pass through i, this tells us that Pi is an inflection point. OK, now what we want to do is we want to use the information we got in order to uh craft the first. The original function So, let's just go ahead and create an axis. And we know that this axis is going to be from 0 to 2 pi. Now, we have a local minimum at pi divided by 2. And 3 pi divided by 2, and we also have an inflection point at pi. We have a local maximum at pi divided by 2, which we can estimate to be about here, and we have a local minimum at 3 pi divided by 2, which we can estimate to be here. Now, from 0, we know that the graph is going to increase to pi divided by 2, and the graph is going to be concave down as it passes through pi divided by 2. So, the shape of that graph in this region is gonna look like this. And again we're going to decrease down to pi. From i, we are going to decrease to 3 pi divided by 2, but we are going to be concave up for the rest of the graph. So we can estimate the shape of the graph to be roughly like this. And notice that this is the shape of the sine wave, which means that F of X is most likely some form of sine of X, and this is going to be the graph to the function. 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.

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.
82. y' = sin t, for 0 ≤ t ≤ 2π

Key Concepts

First Derivative

Second Derivative

Graphing Procedure

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.82. y' = sin t, for 0 ≤ t ≤ 2π

Key Concepts

First Derivative

Second Derivative

Graphing Procedure

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.
82. y' = sin t, for 0 ≤ t ≤ 2π