Use the guidelines of this section to make a complete graph of...

Question

Use the guidelines of this section to make a complete graph of f.

f(x) = x + 2 cos x on [-2π,2π)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to graph the function F of X, which is equal to X minus 2 multiplied by sin of x, when they close the interval from -2 pi to 2 pi. Below the problem, we're given an empty graph on which to plot our function. Now we need to create a graph of F of X on the closed interval from -2 pi to 2 pi. So we'll need to determine some properties of our function F of X. The first thing we're going to need to know is there are the critical points for this function. So for that, we need to calculate the first derivative. So we're gonna take a derivative of F here with respect to X, and when I do so, I'll have F of X. And that's going to be equal to. 1 minus 2 multiplied by the cosine of X. So here we've taken a derivative with respect to x of the quantity of X minus 2 multiplied by the sin of x. The derivative of X with respect to X is 1, and the derivative with respect to X of sin of X is equal to cosine of x. So now to find the critical points, I need to determine whenever our first derivative is undefined. However, in this case, our first derivative is defined everywhere on the infold that we're given. But the other condition is when our first derivative is equal to zero. So I'll set 1 minus 2 multiplied by cosine of x. Equal to 0. And so I can solve this for X. So the first step would be to subtract one from both sides of the equation and then divide both sides by minus 2, and that would give me cosine of X is equal to 1/2. Now cosine of X is equal to 12, when X is equal to minus 5 pi divided by 3. Minus pi divided by 3. Pi divided by 3. And 5 pi divided by 3. And so here we've only considered values of X within our interval from minus 2 to 2 pi. So we're going to do now is use this information to determine whether we have a local maximum or a local minimum. So to do this, we're going to make a table, and in this table, one row is going to be the sign of our first derivative F of X, and then the second row is going to be the behavior of our function F of X, whether it's increasing or decreasing based upon the first derivative. So, we're going to look at the various intervals that are bounded by our critical points. So, the first interval that we're gonna need to look at is the values of X that are less than -5 pi divided by 3. And so for this, we'll actually choose the endpoint of our interval for my two to evaluate F of X. And so when I have -2 pi. In for X, and we calculate F of X. We get a value of -1. Which is negative. And with our first derivative is negative, that means our function is actually decreasing. So we'll label that as a D in our table, so D for decreasing. The next interval that we'll need to look at is the interval between -5 pi divided by 3 and minus pi divided by 3. Now, again, we'll choose a point in this info and we'll choose the point X equal to minus pi. And so when we substitute N X equals to minus pi into our first derivative, we get a value for F X of 3. Which is actually positive. And recall that if our first derivative is positive, that means that our function F of X is increasing, so we'll label this interval with an I to stand for increasing. So now the next interval that we need to look at is going to be the interval from minus pi divided by 3 to divided by 3. And we'll again choose an x value in this interval. So we'll choose X equal to 0. I choose X equal to 0, I get minus 14 F of X, and so that is a negative value, and so that means that our function F of X is decreasing on this interval. The next interval that we need to look at is going to be the interval from pi divided by 3 to 5 pi divided by 3. And I'll choose a point in this interval, so we'll choose X equal to pi. And we choose X equal to pi and calculate our firsttative, we get a value of 3, which is positive, and so that means our function is increasing on this interval. And finally, we need to look at values of X that are greater than 5 pi divided by 3. So we'll choose the endpoint of our interval of 2 pi to test for this interval. And so when we substitute in X equal to 2 pi into our first derivative, we get a value of -1, which is negative, so that means our function is decreasing on this interval. And so now we can use this information to determine whether we have a local minima or a local maxima at the various critical points. And so if our function is changing from decreasing to increasing, we actually have a local minimum. So we have a local minimum at X equal to -5 pi divided by 3, and pi divided by 3. And if our function is increasing and then decreasing, That means that we have a local maximum. So that means we have a local maximum at X equal to minus pi divided by 3 and 5 pi divided by 3. So now we have determined the intervals over which our function is increasing and decreasing, as well as our critical points, we now can look at concavity and inflection points, and for this we'll need our second derivative. So we need to calculate F of X, and to do this, we'll just take another derivative. We'll take a derivative with respect to X of FX. And when I do that, this is going to be equal to 2 multiplied by sine of X. So we call that the derivative with spec X of the quantity of 1 minus 2 multiplied by cosine of X, we take the derivative of 1 with respect to X, we get 0 since it's a constant, and we take the derivative with respect to x of cosine of X, we get minus sine of X. However, the minus sign is canceled out by the minus sign in front of the 2 factor of 2 out in front. And so that's the reason why we get just two side effects. And so now we need to do is calculate our possible inflection points. So to do that, we need to set our second derivative equal to 0. We'll set two sine of Xs. Equal to 0. And so that means that sin of X has to be equal to 0, and so that means that X is going to be equal to minus pi, 0, and pi. Since those are values of X, that sin of X is equal to 0, and here I'm ignoring the endpoints of our interval. So now we need to use this information along with the fact that these are possible inflection points to determine see if we actually have inflection points at this at these points, as well as to determine the concavity of our function F of X. So again, we'll make it a table where we'll have the sign of our second derivative F of X, and we'll have whether our function F of X is concave up or concave down as the next row. So we're gonna be looking at 4 intervals. So the first interval that we're going to look at is going to be values for X, that is less than minus pi. And so here we'll choose a value of minus. 3 pi divided by 2. And so when we calculate our second derivative at this point, we get F of X is equal to 2, which is a positive quantity. And we call that if our second derivative is positive, that means that our function F of X is concave up, so we'll label that with a U in our table. Now the next interval that I want to look at is going to be the interval from minus pi to 0. And so again, I'll choose a point in this interval, so I'll choose X equal to minus pi divided by 2. And so, when I calculate our second derivative at At that point, we get -2 as our value, so that means our second derivative is negative, and that means that our function F of X is concave down. So we'll label that as a D. So recall that if our second derivative is negative, then our function F of X is going to be concave down. Now the next interval that we need to look at. It's gonna be from 0 to high. And here we're going to choose a value of X being equal to pi divided by 2. And so when I calculate our second reus value, we see that is equal to 2. Which is a positive quantity, and that means our function F of X is concave up. And then finally, we're going to be looking at points greater than pi, so we'll choose a point of X equal to 3 pi divided by 2. And when we calculate our second derivative at this point, we get a value of -2, which is negative. So that means that our function is concave down at this point. So now we've determined the concavity over function F of X over these intervals. We see that we do have a change concavity at X equal to minus pi, X equal to 0, and X equal to pi. So all three of those points are inflection points since our concavity is changing. So now we just need to use this information to create our graph. And so we're gonna need to calculate our function F of X at all these points, as well as the end points. Of our interval. So, when we calculate F of X, when X is equal to -2 pi, we get a value of minus 6.28, and so we'll plot that point on our graph. When we calculate F of X, when X is equal to minus 5 pi divided by 3, we get a value of 6.97. And so we'll plot that or get a value of minus 6.97. We'll plot that point on our graph. The next value of X that we need to look at is X when when X is equal to minus pi. And so F of minus pi is going to be equal to -3.14. We'll plot that point on our graph as well. The next point we need to look at is when X is equal to minus pi divided by 3, and when we calculate the value of F of X at that point, we get 0.68, we'll plot that point on the graph. Next point we need to look at is when X is equal to 0, and F of 0 is equal to 0. The next point is when X is equal to pi divide by 3, and when we calculate F of I divide by 3, we get minus 0.68. We'll plot that point on the graph. The next point we need to look at is when X is equal to pi. And F of pi is equal to 3.14, so we'll Plot that point on the graph as well. The next point that we need to look at is when X is equal to 5 pi divided by 3, and its value of F of X, when X is equal to 5 divided by 3 is going to be equal to 6.97. We'll plot that point on the graph, and then finally, the last point that we need to plot is when X is equal to 2 pi, and the value of F of 2 pi is equal to 6.28. So we'll plot that value, that point on the graph. As well. So now all we need to do is to connect these points with smooth lines, making sure that we have minima where we're supposed to have minima, and inflection points where we're supposed to have inflection points and maxima where we're supposed to have maxima. And when we connect all these points together, We should get a graph that is similar to what is drawn on the screen. So this graph here is the final answer for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Use the guidelines of this section to make a complete graph of f.
f(x) = x + 2 cos x on [-2π,2π)

Key Concepts

Function Analysis

Graphing Techniques

Interval Notation

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