Sketch the graphs of the rational functions in Exercises 53–60...

Question

Sketch the graphs of the rational functions in Exercises 53–60.

y = (x² + 1) / x

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says sketch the graph of the function. The 1st and 2nd derivatives are given, and we're given F of X, which is equal to the quantity of X2 + 4 in quantity divided by X, F of X, which is equal to 1 minus 4 divided by X2, and F of X, which is equal to 8 divided by x cubed. the problem we're given an empty graph on which to plot our function. So in order to sketch graph, we need to analyze our function F of X. The first thing we need to do is determine the domain of our function. And if we look at F of X, we see that we have a fraction. So that means our function is not going to be defined whenever our denominator is equal to 0. And since our denominator is just equal to X, that means our function will not be defined when X is equal to 0. Our function F of X is going to be defined everywhere except X equal to 0. So the domain for a function is going to be the open interval from minus infinity to 0, union with the open interval from 0 to infinity. Next, we can check for symmetries. And so we'll check for even our odd symmetries by calculating F of minus X, and this is going to be equal to the quantity of minus X in quantity squad + 4, and that whole quantity is divided by minus X. And if we simplify this expression, this is going to be equal to minus the quantity of X2 + 4 in quantity divided by X, which is equal to minus F of X. So here we see that F of minus X is equal to minus F of X, so that means that we have odd symmetry. So function is symmetric about the origin, they'll save us some time later. When we create our graph. Next, we need to determine the asymptotes for our function. And we already solved that our function is undefined when X is equal to 0, and that's because we will have 0 in the denominator of the fraction or a function F of X. So that means that we'll have a vertical asymptote at X equal to 0. We also noticed that the degree of the polynomial in the numerator is actually one larger than the denominator. So that means that we're going to have an oblique or slanted asymptote. So in order to find the equation for that line, we'll divide our denominator, which is X, into our numerator of X2 + 4. We'll do this by long division. So X will go into X2 x times. So X multiplied by X will be X2, and I'll subtract that from X2 + 4, and I'll get just a remainder of 4. Now, I'm not concerned about the remainder because the quotient here is going to give us the equation for our oblique asymptote. So that means the equation for our line is going to be Y is equal to X. And we'll go ahead and plot these asymptotes on our graph. So we'll have a vertical dash line. At X equal to 0 for a vertical asymptote. Then we will have our oblique aseptote. And that's gonna be a slanted line. Following the equation Y is equal to X. Oop, draw that with a dashed line. So now that we have our asymptotes drawn, now we can focus on drawing our function. And so the next step is to determine the intervals over which our function F of X is increasing or decreasing, and for this we'll need to focus on our first derivative. So, we first need to find our critical points. And recall that our critical points occur whenever our first derivative is equal to zero or undefined. Now, our first derivative is undefined when the denominator in the second term of our first derivative is equal to 0. And that occurs when X is equal to 0, but X equal to 0 is not in the domain of our function, so it cannot be a critical point. So the only critical points that we need to worry about are when F is equal to 0. So we're going to set 1 minus 4 divided by X2 equal to 0 and solved for X. So the first step is to add 4 divided by X2 to both sides. We'll have 1 is equal to 4 divided by X2. Then we'll multiply both sides by X2. We'll have X2 is equal to 4. And then taking the square root of both sides, we'll have X is equal to plus or minus 2. So, we actually have two critical points here. And we're going to take these critical points and divide up our domain using those critical points. And those are going to be the intervals over which our function is increasing or decreasing. Now, to determine whether a function is increasing or decreasing over a certain interval, we're going to create a table. And the table will have two rows. The first row will have the sign of FXX, and in the second row, we'll have the behavior of F of X, whether it's increasing or decreasing. And so we're gonna look at these two quantities over several intervals. So, the first interval is going to go from minus infinity to X equal to -2. Now, over this interval, the quantity of 4 divided by X squared is actually going to be less than 1. So when I subtract that from 1, I'm going to get an overall positive quantity. So that means our first derivative of X is going to be positive over this interval. So we'll level that with a plus sign in the first row of our table. Now recall that if the first derivative is positive, that means our function F of X is increasing over that interval. So we'll label that with I for increasing in our second row of our table. The next in that we want to look at is going to go from X equal to -2 to 0. And over this interval, The quantity of 4 divided by X2 is actually going to be greater than 1. So when I subtract that from 1, I'm going to get an overall negative quantity. So F of X is going to be negative, so we'll place a negative sign in the first row of our table, and recall that if the first derivative is negative, that means our function F of X is decreasing. So we'll live with that with a D in the second row of our table. The next interval that we want to look at is going from X equal to 0 to 2. And again, over this interval, the quantity of 4 divided by X2 is going to be greater than 1. So when I subtract that from 1, I get a negative quantity. So F of X is going to be negative, and that means that F of X is going to be decreasing over this interval. And then the final info that we want to look at. Is going from X equal to 2 to infinity. And over this interval, the quantity of 4 divided by X2d is going to be less than 1, so 1 minus this quantity is going to be a positive quantity. So, our first derivative is going to be positive over this interval, and since our first derivative is positive, that means that F of X is going to be increasing. So now we've determined all the intervals over which our function F of X is increasing or decreasing, and we see that we have a local maximum at X equal to -2, since our function is changing from increasing to decreasing at this point. We also have a local minimum at X equal to 2 since our function is changing from decreasing to increasing. Now, the next quantity that we want to look at are um inflection points, and for this we'll need our second derivative. So, if we look at our second derivative, FX, which is equal to 8 divided by X cubed, we need to find the points where our second derivative is equal to 0, because that's where our inflection points are located. However, FX is never equal to 0. And so we don't have any inflection points. And so, we still need to look at the intervals of concavity, but the interval of concavity can be broken up based upon our domain. So, again, we're going to create a table to simplify this process. So, the first row of our table is going to be F of X, and we'll be looking at the sign of that function, and the second row of our table is going to be the behavior of F of X, whether it's concave up or concave down. So, the first interval that we need to look at is going to be going from minus infinity to 0. And if we look at our function F Xx, which is equal to 8 divided by X cubed, the sign of FX is determined entirely by the sign of X cubed, since the numerator is always a positive constant. And here in this interval, we have X being less than 0, so X is negative. When X is negative, X cubed is also negative, and therefore F of X is also going to be negative. So we'll place a minus sign in the first row of our table, and recall that if the second derivative is negative, That means that our function F of X is going to be concave down, so we'll label that with a D in the second row of our table. The next interval that we want to look at is going to be going from 0 to infinity. And again we need to find the sign of FX, and here X is going to be a positive quantity. So therefore X cubed is also going to be positive, and that makes FXx also positive. So we'll place a plus sign in the first row of our table. And then recall that if the second derivative is positive, that means that our function F of X is going to be concave up, so we'll label that with a U in our table. So, now we have all the information that we need in order to create our graph. So, the first thing we're going to do is plot our critical points. And so, when X is equal to -2, we see that F of -2 is equal to -4. And by odd symmetry when X is equal to 2. F of 2 is equal to 4. And so now we're going to use the fact that For X less than 0, our function is going to be concave down, and on the interval going from minus infinity to -2, our function F of X should be increasing, so our functions could be coming in. From the oblique atote approaching. The point at X equal to -2, which is going to be a local maximum, then after that point, it's going to decrease, and it's going to approach the vertical asymptote. At X equal to 0. When X is greater than 0, that means our function is going to be concave up, and on the interval, going from 0 to 2, our function is decreasing. So, our function is going to be coming in from infinity. From the vertical acito approaching. The local minimum at X equal to 2, and at X equal to 2, our function is beginning to increase, and it's going to approach our oblique asymptote. And so the graph that we have drawn on the screen here is the answer to this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Sketch the graphs of the rational functions in Exercises 53–60.
y = (x² + 1) / x

Key Concepts

Rational Functions

Vertical Asymptotes

End Behavior

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