In Exercises 9–66, graph the function using appropriate method...

Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
56. y = x² + 2/x

56. y = x² + 2/x

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 X2 minus 2 divided by X. The 1st and 2nd derivatives are given, and they are F of X, which is equal to 2 X plus 2 divided by X2, and F of X, which is equal to 2 minus 4 divided by X cubed. Below the problem we're given an empty graph on which to plot our function. Now, in order to graph our function F of X, we need to determine a couple of its properties. First thing we need to do is determine the domain of our function. If we look at our function F of X, we see that the second term contains a fraction. And so that means that our function F of X is going to be undefined whenever the denominator of that fraction is equal to 0, and that occurs when X is equal to 0, since the denominator is just X. So that means that our function is going to be undefined when X is equal to 0. So the domain is going to be the open interval from minus infinity to 0. Union with the open interval from 0 to infinity. So next we need to check for symmetries, and for this we'll calculate F of minus X. To check for even or odd symmetries. This is going to be equal to the quantity of minus X in quantity squared, minus 2 divided by the quantity of minus X in quantity. And simplifying this expression, this is going to be equal to X2 + 2 divided by X, and that is not equal to F of X, and it's also not equal to minus F of X. So we have neither even nor odd symmetry. Now, the next thing we can look for are asymptotes. And here we already saw that our function F of X was undefined when X is equal to 0, and that's due to the denominator being equal to 0. So that means that we'll have a vertical asymptote at X equal to 0. Now we're not gonna have any oblique asymptotes, but we could check for horizontal asymptotes. So, if you look at the limit, As X approaches plus or minus infinity of F of X. We see that our function F of X is going to be dominated by the X2 term, and so X2d approaches infinity as X approaches plus or minus infinity. So that means that the limit as X approaches plus or minus infinity of F of X is going to be equal to infinity. So that means we have no horizontal asymptotes. So we just have the single vertical asymptote. So we'll go ahead and draw that on our graph. So we're going to draw a dashed vertical line. At X equal to 0 to denote our vertical aseptote. So now we need to determine the intervals over which our function F of X is increasing or decreasing, and for this we'll need to look at our first derivative of F of X. So the first thing we need to do is determine the critical points. Now the critical points are going to occur whenever F of X is undefined or equal to 0. Now again with FX, the second term here is a fraction, and so that means that F of X is going to be undefined whenever the denominator of that fraction is equal to 0, and that occurs when X is equal to 0. However, X equal to 0 is not within the domain of our function, so it cannot be a critical point. So the only time we're going to have critical points is when F of X is equal to 0. So we'll set 2 X plus 2 divided by X quad. Equal to 0 and solve for x. So the first thing I'll do is subtract 2 divided by x2 from both sides. We'll get 2 X is equal to minus 2 divided by x2. So multiplying both sides by x2d and dividing both sides by 2 we'll get X cubed is equal to -1, and taking the cubed root of both sides, we'll get X is equal to -1. So we have a possible extrema at X equal to -1. So what we're going to do is use our critical point here to break up our domain and use that to determine the intervals over which our function is increasing or decreasing. So we're going to do that using a table. In the first row of the table, we'll have the sign of F of X, and in the second row, we'll have the behavior of F of X, whether it's increasing or decreasing. And so the first interval that I want to look at is going to be going from minus infinity to -1. Now, the easiest way to see whether our function F of X is increasing or decreasing is to choose a point in this interval and calculate F of X and see if it's positive or negative. So we're going to choose X equal to minus 2. And when X is equal to -2, we get a value for F of -2, that is equal to -3.5. So that is a negative value. So we'll put a minus sign in the first row of our column, and recall that if the first derivative is negative, that means that our function F of X is going to be decreasing, so we'll label that with a D in the second row of our table. Now, the next interval that I am going to look at. Is going to be the interval going from X equal to -1 to X equal to 0. And again, we'll choose a point on this interval to calculate F of Xs, and we'll choose X equal to minus 0.5. And if we calculate F of minus 0.5, We get a value of 7, which is a positive quantity. So we'll place a plus sign in the first row of our table and recall that if the first derivative is positive over an interval, that means that our function F of X is going to be increasing over that interval. So we'll level that with an I in the second row of our table. And the last interval that we want to look at is going from X equal to 0 to infinity. Now in this case, X is a positive quantity, and so the quantity of 2 X is going to be positive, and the quantity of 2 divided by X quad is also going to be positive. So I'll be adding together two positive numbers, so I don't really need to calculate anything here. So our function FX is just going to be positive over this entire interval. So that means that F of X is going to be increasing over this interval as well. So now that we've evaluated and found the intervals over which our function is increasing and decreasing, we see that we do have a local minimum at X equal to -1, since our function is changing from decreasing to increasing at this point. Now the next thing we need to look for are inflection points, and for this we'll need our second derivative. So if we look at our second derivative F of X, this is going to be equal to 2 minus 4 divided by X cubed. Now recall that inflection points occur whenever our second derivative is equal to 0. So we're going to set the quantity of 2 minus 4 divided by X cubed. Equal to 0 and solve for X. So the first thing we'll do is add 4 divided by X cubed to both sides. We'll have 2 is equal to 4 divided by X cubed. Then we're going to multiply both sides of the equation by X cubed and divide both sides by 2, and so we'll get X cubed. is equal to 2, and then we'll take the cube root of both sides, and so we'll have X is equal to the cube root of 2. Which is approximately equal to 1.26. So this here is a possible inflection point. So now we just need to determine our intervals of concavity, and again, we'll use a table to do that. So we'll have two rows again in our table. We'll have F of X. We'll have it sign as the first row of our table, and the behavior of F of X as the second row of our table. And again, we'll be looking over various intervals. So the first interval that we're going to look at is going to be going from minus infinity to 0. So, in this case, X is always going to be a negative value, and so the quantity of 4 divided by X cubed is going to be a negative value. And so when I have to subtract that from 2, I'm going to get an overall positive quantity since I'm essentially going to be adding two positive numbers together. So that means that the sign of FX is going to be positive. So we'll place a plus sign in the first row of that table. And recall that if the second derivative is positive over an interval, that means our function F of X is going to be concave up on that interval. So we'll label that with the EU in the second row for our table. Now the next interval that we want to look at is going from X equal to 0 to the cube root of 2. Now, over this interval, the quantity. Of 4 divided by X cubed is actually going to be greater than 2. And so when I subtract it from 2, I'm going to get an overall negative value. So that means that F of X is going to be negative over this interval, so we place a minus sign in the first row of our table. And recall that if the second derivative is negative over an interval, that means that F of X is going to be concave down over that interval, so we'll label that with a D. And now the final interval that we want to look at is going from X equal to the cube root of 2 to infinity. Now, over this interval, the quantity of 4 divided by X cube is always going to be less than 2, and so when I subtract it from 2, I'm getting positive quantity. So we'll place a plus sign on our table for the sin of FX, and since we have a positive second derivative, that means that our function F of X is going to be concave up. So we see that we do have an inflection point at X equal to the key root of 2. So the last thing we need to do is actually find the value of F of X for our critical point as well as our inflection point. So, if we look at F of -1, this is going to be equal to. The quantity of minus 1 in quantity squared. -2 divided by -1, and when we evaluate this expression, this is going to be equal to 3. And we also need to calculate F of the cube root of 2. And this is going to be equal to the quantity of the cube root of 2 in quantity squared minus. 2 divided by the cube root of 2. And when we evaluate this expression, this is going to be equal to 0. So now we just need to plot our points for our our critical point, as well as our inflection point, and then use the fact that we have the Intervals over which our function is increasing and decreasing, as well as our interval of concavity to draw the function. So we'll start off by placing our critical point, and that occurs at X equal to -1 with a Y value of 3. And then we'll have an inflection point at the X equal to the Q root of 2, which is approximately equal to 1.26. And a y value of 0. So now we're going to use the fact that our function is decreasing on the interval, going from minus infinity to -1, as well as our function having to be concave up to begin to draw our function. So we're coming down from positive infinity to our local minimum, and then after this, our function is going to increase, still having a concave up function, but we're going to be approaching our vertical asymptote. And then for the next interval, we're also going to be increasing from 0 to infinity, but our function is going to be concave down on the interval from 0 to the cuber of 2. So that means we're going to begin from minus infinity from near our vertical asymptote. And we're going to increase until we reach the inflection point, in which case we're still going to continue to increase, but now our function is going to be concave up. And so our functions can be concave up and continue to increase, approaching positive infinity. So the graph that we have drawn on the screen here is going to be the answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
56. y = x² + 2/x

Key Concepts

Critical Points

Second Derivative Test

Absolute Extrema

Watch next

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.56. y = x² + 2/x

Key Concepts

Critical Points

Second Derivative Test

Absolute Extrema

Watch next

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
56. y = x² + 2/x