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.
33. y = (x² - x + 1) / (x - 1)

33. y = (x² - x + 1) / (x - 1)

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says graph the function F of X, which is equal to the quantity of X2 minus X plus 2 in quantity, divided by the quantity of X minus 2 in quantity. The 1st and 2nd derivatives are given, and we're given F of X, which is equal to the quantity of X multiplied by the quantity of X minus 4 in quantity, divided by the quantity of X minus 2 in quantity squared. And F of X is equal to 8 divided by the quantity of X minus 2 in quantity cubed. Below the problem we're given an empty graph on which to plot our function. Now, we need to determine a couple properties of our function F of X in order to graph it. And the first thing we need to do is determine the domain of our function. And if we look at our function F of X, we see that it will be defined everywhere except when the denominator is equal to 0. And so that's going to be when X minus 2 is equal to 0, so that's going to occur when X is equal to 2. So that means our domain is going to be the open interval from minus infinity to 2, union with the open interval from 2 to infinity. So function is defined everywhere except X equal to 2. Now, next we can look for symmetries. So we'll calculate F of minus X to check for even or odd symmetries. And so this is going to be equal to the quantity of minus X in quantity squared, minus minus X plus 2, and that whole quantity is divided by the quantity of minus X minus 2. So, simplifying this expression, this is going to be equal to. The quantity of X2 + X + 2 in quantity, divided by the quantity of minus X minus 2. Now, this is not equal to F of X. And it's not equal to minus F of X. So we have neither even nor odd symmetries, so symmetries are not going to help us to graph this problem. So next we need to look for asymptotes. And the first thing we'll look for are vertical aseptodes, 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 occurred at X equal to 2. So we'll have a vertical asymptote at X equal to 2. We will also have an oblique asotote, since if we look at a function F of X, the degree of the polynomial in the numerator is 1 degree higher than that of the denominator. So, in order to find the equation for that oblique asymptote, we're going to take our denominator of X minus 2, and divide it into our numerator of X2 minus X + 2, and we're going to do this by long division. So X will go into X2 x times. So we'll have X multiplied by the quantity of X minus 2, and that will give me X2 minus 2 X, and I'm going to subtract this quantity from X2 minus X + 2. In doing so, I will get X + 2. And X minus 2 will go into X + 21 time, so we'll have plus 1. And so we'll have 1 multiplied by the quantity of X minus 2, which will give me X minus 2, and I'm going to subtract this quantity from X + 2, and I will get a remainder of 4. Now, I'm not interested in the remainder since the quotient will give me the equation for the oblique asmatote. So that means that we'll have an oblique asptote that follows the equation Y is equal to x + 1. And we're going to go ahead and plot these on our graph. So, we'll have a vertical dash line. At X equal to 2, or our vertical asymptote, and we will have And another dash line that follows the equation. Of Y is equal to x + 1, so that means it's going to have a Y intercept of 1 and a slope of 1. So we'll draw that dash line on our graph as well. So, the next thing that we need to look for are 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 X. And so we need to find the critical points, and recall that the critical points occur whenever F is equal to 0 or undefined. Now, if we look at our function F of X, we'll see that it is undefined whenever the denominator is equal to 0, and this again occurs when X is equal to 2. However, X equal to 2 is not within the domain of our function, so it cannot be a critical point. So the only critical points are going to occur whenever our first derivative F of X is equal to 0, and that's going to occur only when the numerator is equal to 0. So we're gonna set the quantity of X multiplied by. The quantity of X minus 4 in quantity equal to 0, and solve for X. So to find the value of X, we need to set each factor equal to 0 and solve for X. So for the first factor, we'll just get X is equal to 0, and for the second factor, we'll have X minus 4 equal to 0. That means that X is equal to 4. So these here are critical points, and so we're going to use our 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 make a small little table. And so our table's going to have two rows. In the first row 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. Now, the sign of FXx is determined entirely by the numerator, since the denominator is always a positive quantity, since we have X minus 2 being squared. So, we only need to focus on the sign of the numerator, and that will give us the sign of F of X. So, the first interval that we're going to look at is going from minus infinity to X equal to 0. Now, on this interval, X is a negative value. So our factor of X is going to be negative, but also the quantity of X minus 4 is also going to be negative, since I'm subtracting 4 from a negative number, which will also give me a negative value. So multiplying two negative values together, which will give me an overall positive quantity. So F of X is going to be positive over this interval, 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 F of X is increasing over that interval. So that means we'll label the second row with an I for increasing. Now, the next interval that we want to look at is going from X equal to 0 to X equal to 2. Now, over this interval, our X value is positive, so our first factor in the numerator is going to be positive, but X is less than 2, so the quantity of X minus 4 is going to be negative. So I'll have a positive value multiplied by a negative value, which will give me an overall negative value. So we'll place a minus sign in the first row of our table, since F of X is going to be negative over the sentval. And recall that if F of X is negative over an interval, that means our function F of X is decreasing over the cent. 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 X equal to 2 to X equal to 4. And over this interval, our X value is still positive, so the first factor is going to be a positive quantity, but X is still less than 4, so the quantity of X minus 4 is going to be negative. So overall, we're going to have a negative value for F of X, and that means our function F of X is decreasing. And now for the last interval. We're gonna go from X equal to 4 to infinity. Again, this uh these X values are positive, so the first factor is going to be positive in the numerator, but X is going to be greater than 4, so the quantity of X minus 4 is going to be positive. So I'm multiplying two positive values together, which can be an overall positive quantity for F of X. And that means that F of X is going to be increasing over this interval. So we've now identified 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 equals 0 since F of X is changing from increasing to decreasing, and we have a local minimum at X equal to 4 since our function is changing from decreasing to increasing. Now, the next quantity that we want to look at are inflection points, and for this we'll need our second derivative F of X. And recall that inflection points occur whenever our second derivative is equal to zero. However, we see that FXx can never be equal to 0 in this case, since we have a positive constant value in the numerator. And so, since the um second derivative can never be equal to 0, we have no inflection points. But we still need to determine the intervals of concavity, since we have our domain being broken up into two different intervals. So, we'll just use our domain to determine the intervals of a concavity. And again, we'll use a table in order to do this. So, the first row of the table is going to be the sign of F of X, and the second row of the table is going to be the behavior of F of X. And so, the first interval that we're going to want to look at. Is going from minus infinity to 2. Now, if we look at our function F X, we see that the sign of the second derivative is determined entirely by the sign of the denominator, since the numerator is always a positive quantity. And so, in our denominator, we have the quantity of X minus 2 and quantity q. So, over this interval, X is going to be less than 2, so the quantity of X minus 2 is going to be negative, and when I cube that, I still get a negative quantity. So, in this case, F of X is going to be negative. So we'll plus a minus sign uh in the first row of our table. And recall that if the second root of it negative over an interval, that means our function F of X is concave down over that interval. So we'll label that with a D in the second row of our table. Now for the next interval we'll look at, it's going to be going from X equal to 2 to infinity. And on this interval, we see that X is going to be greater than 2. So, the quantity of X minus 2 is going to be a positive quantity, and I cube that I still get a positive quantity. So, FX is going to be positive, so we'll place a plus sign in the first row of our table, and recall that if the second derivative is positive over an infold, that means our function F of X is concave up. So we'll label that with a U. Now, the final thing that we want to do is determine the values of F of X at our two critical points at X equal to 0 and X equal to 4. So we'll calculate f of 0 1st, and this is going to be equal to the quantity of 0 squared minus 0 + 2 in quantity, divided by the quantity of 0 minus 2. And when we evaluate this this expression, this is going to be equal to -1. And so we'll have a point at 0-1. This also happens to be our Y intercept. We also need to calculate f of 4, and that's going to be equal to the quantity of 4 squad minus 4 + 2 in quantity, divided by the quantity of 4 minus 2. And when we evaluate this expression, this is equal to 7. So we'll have another point at 4.7. So we'll go ahead and plot those two points on our graph. So, we'll have 1 point at 0. minus 1. And we'll have another point at 4.7. And so now we're just going to use our concavity and our intervals of increasing decreasing to draw the rest of the function. So, on the interval, going from minus infinity to 2, our function is concave down, but our function is increasing on the interval going from minus infinity to X equal to 0. So we're gonna start near the oblique asymptote. Coming in from minus infinity, and our function is going to be increasing. Near that oblique a tote, and then it's going to approach our local maximum at X equal to 0, at which point our function is going to begin decreasing. And it's going to approach the vertical asymptote, approaching minus aphenidine. Now, when on the interval from X equal to 2 to infinity, our function is concave up, and our function is initially decreasing on the interval from 2 to 4. So, we're going to start by plotting our graph near the vertical isotope coming down from positive infinity. And then it's going to approach the local minimum at X equal to 4. And then at this point, our function is going to begin increasing and approaching the oblique asymptote. 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.
33. y = (x² - x + 1) / (x - 1)

Key Concepts

Graphing Rational Functions

Local Extreme Points

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.33. y = (x² - x + 1) / (x - 1)

Key Concepts

Graphing Rational Functions

Local Extreme Points

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.
33. y = (x² - x + 1) / (x - 1)