24–34. Curve sketching Use the guidelines given in Section 4.4...

Question

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.

ƒ(x) = 4cos (π (x-1)) on [0, 2]

ƒ(x) = 4cos (π (x-1)) on [0, 2]

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to draw the graph of the given function on the specified in volt. And we're given the function F of X is equal to 4 multiplied by sine of the quantity of pi multiplied by the quantity of X minus 1 in quantity, and this is only the closed interval from 0 to 2. Now, in order to graph a function, we need to determine a couple properties of our function F of X, and the first thing we're going to look at is the domain of our function. And so here, if we look at our function F of X, we see that it involves the sine function and recall that the sine function is defined for all real numbers. That means our function F of X is also defined on all real numbers. So the domain here is just going to be equal to our specified intervals. So that's going to be the close interval from 0 to 2. Now, the next quantity that we want to look for are critical points, and for this we'll need to calculate our first derivative. So we'll calculate F of X, which is equal to the derivative with respect to X. Of the quantity of 4 multiplied by sine of the quantity of pi multiplied by the quantity of X minus 1 in quantity. And so taking this derivative, this is going to be equal to 4. Multiply by and here taking the derivative of sine function and recall that when you take the derivative of a assign function, you get a cosine function. We'll have 4 multiplied by cosine of the quantity of pi multiplied by the quantity of X minus 1 in quantity, and then this is going to get multiplied by the derivative of the argument of the sine functions we need to apply our chain rule. So this is going to get multiplied by the derivative with respect to x of the quantity of i multiplied by the quantity of X minus 1 in quantity. I want to take that derivative, I get pi. So simplifying this expression, this is just going to be equal to 4 pi, multiplied by cosine of quantity of pi, multiplied by the quantity of X minus 1 in quantity. And so in order to find critical points, recall that those are the X values which are first derivatives either undefined or when it's equal to zero. Now the cosine function here is defined for all real numbers, so that means our first derivative here is defined for all real numbers. So the only critical points are going to be due to our first derivative being equal to 0. So we set or pi multiplied by cosine. Of the quantity of pi multiplied by the quantity of X minus 1 in quantity, equal to 0. So this is going to be true whenever our cosine function is equal to 0, and we call that our cosine function is equal to 0 when it's argument, which in this case is pi, multiplied by the quantity of X minus 1 in quantity, is equal to pi divided by 2 plus an integer multiple of pi. So we'll label that integer as N and then it will get multiplied by pi. So we need to do is solve this for X, so we'll divide both sides by pi, and we'll have X minus 1 is equal to 1 divided by 2 plus N. And then we'll just add one to both sides, and so we'll get X is equal to 3 divided by 2. Plus. So we need to find the values for X, which fall within our domain, which is between 0 and 2. So, in here has to be an integer, so we'll start off by setting N equal to 0. In that case, we'll get X equal to. 3 divided by 2, which is equal to 1.5, which is between 0 and 2. So this is a critical point in our domain. If you look at N equal to 1, we'll get X is equal to 5 divided by 2, which is 2.5, which is outside of our domain, and any values of n that are greater than that are also give us values for X that are outside of our domain. N can also be negative, so if we set N equal to -1, we get X equal to 1 divided by 2, which is 0.5, which falls within our domain. And if we set N equal to -2, we actually get a negative value for X. And since negative values are less than 0, that falls outside of our domain. So these two values are our critical points. So we're going to do now is use the critical points to determine the animals over which our function F of X is increasing or decreasing. So for this, we're going to create a table. And here we're gonna have in one row, the sign of our first derivative F of X, and in the second row, we're going to have the behavior of F of X, whether it's increasing or decreasing. So, the first interval that we want to look at is going to be going from 0 to X equal to 1 divided by 2. And if we look at our first derivative, the sign of the first derivative here, F of X is determined by the sign of the cosine function. And if we look at the argument of our cosine function on the interval from 0 to 12, we see that the argument of the cosine function is going from minus pi to minus pi divided by 2. That means that our cosine function is going to be in the 3rd quadrant, and our cosine function is negative in the 3rd quadrant. That means that FX over this interval is going to be negative, so we'll label a minus sign in that first row. And recall that if our first derivative is negative over an interval, that means our function F of X is decreasing over that info. So we'll label that with a D in our table. Now the next interval that we want to look at is going to be going from X equal to 1/2 to X equal to 3 divided by 2. And again, we'll look at the sign of our cosine function to help us determine the sign of our first derivative. So over this interval, going from X equal to 1 divided by 2 to X equal to 3 divided by 2, the argument of the cosine function is going from minus pi divided by 2 to pi divided by 2. That means our cosine function is in the 4th and 1st quadrant, and in both of those quadrants, Our cosine function is positive. That means that our first derivative F of X is also positive over this interval. And since our first derivative is positive, that means our function F of X is increasing, so we'll label that with an I in our table. And then the last interval that we want to look at is going from X equal to 3 divided by 2 to X equal to 2. And if we look at the sign of the cosine function, which will allow us to determine the sign of our first derivative here, we see that the argument of the cosine function over this info is going from pi divided by 2 to pi, and so that means our cosine function is in the second quadrant. And our cosine function is negative in the second quadrant, so that means that our first derivative here is also negative over this interval. And since our first ro is negative, that means our function here is decreasing, FX is decreasing over this interval. So now we've identified all the intervals over which our function is decreasing and increasing. We also see that we have a local minimum at X equal to 1 divided by 2, and a local maximum at X equal to 3 divided by 2. Now, the next quantity that we want to look at are the intervals of concavity and inflection points. So for this, we're going to need our second derivative. We'll calculate F double prime effects. Which is equal to the derivative with respect to X. Of the quantity Of 4 pi multiplied by cosine. Of the quantity of pi, multiplied by the quantity of X minus 1. In quantity And so taking this derivative, this is going to be equal to or pie. Multiplied by, and here I'm taking the derivative of a cosine function. So recall that when I take the derivative of a cosine function, I get a minus sine function. So I'll have 4 i multiplied by minus sign. Of the quantity of pi, multiplied by the quantity of X minus 1 in quantity. But here I still have to apply our chain rule, so we need to multiply this by the derivative of our argument. So we'll have the derivative with respect to x of the quantity of i multiplied by the quantity of X minus 1 in quantity, and when I take that derivative, I get I. And so simplifying this expression, this is going to be equal to minus or pi squared multiplied by sine of the quantity of pi, multiplied by the quantity of X minus 1 in quantity. So now to look for possible inflection points, I want to set our second derivative here equal to zero. So that means we'll have minus 4 pi squared. Multiplied by sine of quantity of pi, multiplied by the quantity of X minus 1 in quantity is equal to 0. So, this is going to be equal to 0 when our assign function is equal to 0. And so that means that the argument of our assign function, which is pi, multiplied by the quantity of X minus 1 in quantity, has to be equal to an integer multiple pi, which will label as N multiplied by pi. So again, we're gonna solve for X, so define both sides by pi, we get X minus 1 is equal to N. And then solving for X will get X is equal to N + 1. So again, we need to find values for X that fall within our domain between 0 and 2. So when N is equal to 0, we see that we get X is equal to 1. So that's gonna be a possible inflection point. Now we also need to look for other values of n. So when N is equal to 1, we get X equal to 2, and when N is equal to minus 1, we get X is equal to 0, but those correspond to the endpoints of our domain. So we really can't look for inflection points at the end points. So this is the only inflection point that we need to be worried about. So now what we're going to do is determine the intervals of concavity for a function F of X using this possible inflection point. So we're going to again create a quick little table where the first row is the sign of our second derivative F of X, and then the second row is going to be the behavior of F of X, whether it's concave up or concave down. So the first interval that we're going to look for is going to be the interval going from 0 to 1. And so to turn the sign of our second derivative, we noticed that it's gonna be the opposite sign of our sign function. Now, the sine function over this in, if we look at the argument of the sine function, we see that the argument of the sign function is going to be going from minus pi to 0. And so our sign function is going to be in the Third and 4th quadrant. In both of those cases, our sine function is negative, but our second derivative is going to be the opposite as the sign on our sine function. So that means that our second derivative is going to be positive over this interval, so we'll label that with a plus sign in our table, and recall that if our second derivative is positive, that means our function F of X is going to be concave up, so we'll label that with a U. In our table. Now, the next interval that we want to look at is going to be going from X equal to 1 to X equal to 2. And again, we're going to look at the Sign of our second derivative and again looking at the sign of the sine function and taking the opposite of it. So if we look at the argument of the sine function over this interval from X equals to 1 to 2, we see that the argument of the sine function goes between 0 and pi. That means that the sine function is going to be in the 1st and 2nd quadrants, and the sine function is positive in the 1st and 2nd quadrants. But since we have a minus sign out in front of the sine function, that means that our second derivative here, FX is going to be negative. And since our second derivative here is negative, that means our function F of X is going to be concave down, so we'll label that with a D. So now we have enough information now to actually create our graph. So, on an empty graph, we're gonna have to plot a couple of points. So the first thing we're going to do is start with our first endpoint, which is X equal to 0, and we calculate F of 0, we get 0. The next point that we want to look for is X equal to 1 divided by 2, which is 0.5. So when we calculate F of 0.5, We get -4, so we'll plot that point on the graph. The next point is going to be our inflection point at X equal to 1. And if we look at F of 1. We get 0 as our value, so we'll plot that point on the graph as well. The next point is our next critical point. Which is X equal to 3 divided by 2, which is 1.5. So we calculate F of 1.5, we get a value of 4. We'll plot that on the graph as well. And then finally, our other endpoint at X equal to 2, so F of 2 turns out to be equal to 0. So now we want to do is connect these points, using the fact that our function F of X is going to be decreasing on the interval. Going from 0 to 0.5. But it's also going to be concave up. Then the function is going to increase from 0.5 to X equal to 1. And it still will be concave up on that interval, but then it's going to switch concavity but still increasing, so it's gonna be concave down on the interval from X equal to 1. To X equal to 1.5, but our function is going to be increasing. Then finally, our function is still going to be concave down, but decreasing on the interval from X equal to 1 to X equal to 2. So now we have drawn the graph, the graph here that we have drawn is the final answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.ƒ(x) = 4cos (π (x-1)) on [0, 2]

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24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.

ƒ(x) = 4cos (π (x-1)) on [0, 2]