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.
36. y = (x³ + x - 2) / (x - x²)

36. y = (x³ + x - 2) / (x - x²)

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 the quantity of X cubed minus X2 minus 4 in quantity, divided by the quantity of 2 X minus X squared in quantity. The 1st and 2nd derivatives are given, and we're given FX, which is equal to -1 + 2 divided by X2, and FX, which is equal to -4 divided by X cubed. Below the problem, we get an empty graph on which to plot our function. Now, in order to graph this function, the first thing we want to do is rewrite our function F of X, and so we can factor our numerator and denominator. So doing so, we'll have F of X. is going to be equal to in the numerator we can factor it to be X, the quantity of X minus 2 in quantity multiplied by the quantity of X2 plus X plus 2 in quantity. That's going to be divided by the quantity, and here in the denominator we'll factor out an X minus X. So we'll have minus x multiplied by the quantity of X minus 2 in quantity. And we see that we can, we can cancel out the factor of X minus 2, in doing so, this is going to be equal to minus the quantity of X + 1 + 2 divided by X in quantity. So the first thing we need to do is determine the domain of our function. So, our function F of X is going to be defined whenever our denominator is not equal to 0. So our denominator is going to be equal to 0 when X is equal to 0, and also when X is equal to 2 if we look at the original form of our function. So that means that our domain is going to be the open interval from minus infinity to 0, union with the open interval from 0 to 2, union with the open interval from 2 to infinity. Next, we're going to look at asymptotes. Now, we first check for vertical asymptotes, and that's going to occur whenever the denominator of our function F of X is equal to 0, and we already found that that occurs when X is equal to 0 and X equal to 2. However, since I can cancel out the factor of X minus 2 out of our original function of original form for F of X, that means that that is a removable discontinuity. So I'm not going to have a vertical asymptote at X equal to 2. I'm only going to have a vertical asymptote at X equal to 0. But also, if we look at our function F of X, we see that the degree of our polynomial in the numerator is actually one larger than that of our denominator. So, that means that We're going to have an oblique asymptote. Now, to find the equation for that line, I'm wanting to go ahead and cancel out the factor of X minus 2, and what I'm going to do then is divide our remaining denominator, which is minus X, into our remaining numerator, which is going to be the quantity of X2 + X + 2. So doing this by long division, we'll have minus X will go into X2 minus x times. And so minus x multiplied by minus X will give me X2, and I'll subtract that from X2 + x + 2, and that will give me x + 2. Now minus X will go into X + 2 minus 1 times. So I'll have -1 multiplied by minus X, which will give me X, and I'll subtract that from X + 2, which will give me a remainder of 2, which I'm not concerned about since the quotient is going to give me the equation for my oblique asmatote. So we'll have an oblique asptote with Y is equal to minus X minus 1. So we'll go ahead and plot these asymptotes on our graph. So, we'll have a vertical asptote which we'll draw with a vertical dash line. At X equal to 0. And then we'll have Another dash line that follows the equation. of Y equal to minus X minus 1. And That's going to be a line that has a slope of -1, with a Y intercept of -1. So we'll go ahead and draw that on our graph as well. Now, we will also have a removable discontinuity at X equal to 2. So what we need to do is determine the value of F of 2 after we have removed the discontinuity. So we'll calculate F of 2, and that's going to be equal to minus the quantity of 2 + 1 + 2 divided by 2, and evaluating this expression, this is going to be equal to -4. So we're going to have a removable discontinuity at 2.4. And so what we're going to do is plot an open circle at that point. So the open circle means that our function is going to approach that point but never actually take on that value. So that's going to be an open circle at 2.4. Next, we need to determine the intervals over which our function F of X is increasing or decreasing. And for this, we'll need to use our first derivative F of X. And we'll first need to find the critical points, and recall that the critical points occur whenever the first derivative is either undefined or equal to zero. Now our first derivative is undefined whenever the denominator in the second term of minus 1 + 2 divided by X2 is equal to 0, and that occurs when X is equal to 0, but X equal to 0 is not within the domain of our function. So therefore, it cannot be a critical point. So the only critical point that we're going to have is when our first derivative is equal to 0. So we're going to set -1 + 2 divided by x2 equal to 0 and solved for X. So the first thing we want to do is add 1 to both sides of this equation. So we'll have 2 divided by X2 is equal to 1. Multiplying both sides by X2, we'll get 2 is equal to X2, and then taking the square root of both sides, we'll get X is equal to plus or minus the square root of 2, which is approximately equal to plus or minus 1.41. So now we want to do is use these critical points to divide up our domain, and determine the intervals over which our function is increasing or decreasing. And for this, we're going to actually use a table, and in the first row of the table, we're going to have the sign of F X. And in the second row, we're going to have the behavior of F of X. So, the first interval that we want to look at, is going to be the interval going from minus infinity to minus the square root of 2. So we just need to determine the sign of FX over this interval. Now, over this interval, the quantity of 2 divided by X2 is actually going to be less than 1, since X2 is always going to be greater than 2. So when I subtract one from this quantity, I'm actually going to get a negative value. So F of X is going to be negative over this interval, so we'll put a minus sign in the first row of the table, and recall that if the first derivative is negative over an interval, that means our function F of X is going to be decreasing over this interval. So we'll label that with a D in the second row of our table. Now, the next interval that we want to look at is going from minus the square root of 2 to 0. And over this interval, the quantity of 2 divided by X word is actually going to be greater than 1. So when I subtract 1 from it, I'm going to get a positive quantity. So we'll place a plus sign in the first row of our table, and recall that if FXx is positive over an interval, that means F of X is going to be increasing over that interval. So we'll label that with an I. So, the next in that we want to look at is going from 0 to the square root of 2. And over this interval, again, the quantity of 2 divided by X2d is going to be greater than 1, so subtracting one from it, we'll still get a positive quantity. And so that means our function F of X is increasing over this interval. The next interval that we want to check is going to be Going from X equal to square root of 2, to X equal to 2. And over this interval, the quantity of 2 divided by X2 is going to be less than 1. So subtracting one from it will give me a negative value. So we'll have a minus sign in the first row of the table, and that means that our function is decreasing over the interval. And then finally, if we look at the interval going from X equal to 2 to infinity. Again, the quantity of 2 divided by X2 is going to be less than 1, and so subtracting 1 from it will give me a negative value for FX, and that means that our function is decreasing over the standardval. So now we've identified all the intervals over which our function is increasing or decreasing, we see that we have a local minimum, when X is equal to minus the square 2, and a local maximum when X is equal to the square root of 2. Now, the next thing that we want to look for are inflection points, and for this we'll need our second derivative FX, which is equal to -4 divided by X cub. Now recall that we have inflection points whenever our second derivative is equal to 0. However, F X here cannot be equal to 0 in this case for any value of X. So that means we have no inflection points. So, the intervals of concavity are going to be determined by the intervals of our domain. So again, we're going to use a table to determine the intervals of concavity, and it, again, it's going to have two rows. The first row is going to be the sign of F of Xs. And the second row is going to be the behavior of F of X, whether our function is concave up or concave down. So, the first interval that we want to look at is going to be going from minus infinity to X equal to 0. Now, to determine the sign of FXx, we just need to determine the sign of minus X cubed since the numerator is going to be a positive constant. So here, in our case, X is going to be negative. So that means that X cube is also going to be negative, and so minus X cubed is going to be a positive quantity. So F of X is positive, so we'll place a plus sign in the first row of our table. And recall that if the second derivative is positive, that means that our function F of X is concave up. So we'll label that with a U in the second row of our table. Now, the next level that we want to look at is going from X equal to 0 to X equal to 2. Now in this interval, X is positive, so that means that X cubed is positive, but minus X cubed is going to be negative. So F of X is going to be negative. So we'll place a minus sign in the first row of the table, and since 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. And then the last interval that we want to look at is going from X equal to 2 to X to infinity. And over this interval, FXx is still going to be negative, since X is positive and minus X cube is going to be negative. And so therefore, our function is concave down on this interval as well. So, the last thing we need to do is just to determine a couple points of the plot, so we'll determine the values of F of X for our critical points. And so we'll have F of minus the square root of 2. And this is going to be equal to minus the quantity of minus square root of 2 + 1 + 2 divided by minus the square root of 2 in quantity. And evaluating this expression, this is approximately equal to. 1.83. And then if we look at F of square root of 2. This is going to be equal to minus the quantity of the square root of 2, plus 1 + 2 divided by the square root of 2 in quantity, and this is approximately equal to minus 3.83. So we'll go ahead and plot those two points on our graph. And so we'll have the point of -1.14. And 1.83. And then the other point that we're going to have is 1.14. And minus 3.83. We'll plot that point as well. So now we just need to use our intervals of concavity, as well as our intervals of increasing and decreasing to draw the rest of our function. So, when X is negative, our function is going to be concave up, and our function is initially decreasing, coming in from minus infinity as it approaches X equal to 22. So we'll start near the oblique asymptote with our function decreasing. And our function is going to continue to decrease and approach the local minimum at X equal to. Minus the square of 2. At this point, our function is going to be increasing. And it's going to approach our vertical asymptote. Now when X is positive, our function is concave down and On the interval from 0 to the square root of 2, our function is increasing. So our function is going to be coming in near the vertical isotope from minus infinity. It's going to be increasing, approaching our local maximum at X equal to the square root of 2. Our function is then going to decrease, approaching the open point at X equal to 2, and then continuing to decrease. After that, approaching the 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.

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.
36. y = (x³ + x - 2) / (x - x²)

Key Concepts

Graphing Rational Functions

Local and Absolute Extrema

Inflection Points

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.36. y = (x³ + x - 2) / (x - x²)

Key Concepts

Graphing Rational Functions

Local and Absolute Extrema

Inflection Points

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.
36. y = (x³ + x - 2) / (x - x²)