Use the guidelines of this section to sketch the curve. y =...

Key Concepts

Concavity and Inflection Points

Analyzing the second derivative provides insights into the concavity of the function, indicating whether the graph curves upwards or downwards. Points where the concavity changes, known as inflection points, are important for refining the understanding of the graph's shape and ensuring an accurate sketch.

Intercepts and Key Points

Identifying the intercepts, where the graph crosses the axes, is another critical component of curve sketching. These key points, along with symmetry properties if present, help in outlining the essential features of the graph and serve as anchors for constructing the overall curve.

Critical Points and Derivative Analysis

Finding the critical points by setting the first derivative equal to zero or undefined helps in identifying local extrema, such as relative minima or maxima. This information is used to determine where the function changes direction, which is fundamental in constructing an accurate sketch of the curve.

Asymptotic Behavior

Determining asymptotes is a key aspect of curve sketching. Vertical asymptotes occur where the function approaches infinity due to division by zero, while horizontal or oblique asymptotes describe the end behavior of the function as the independent variable tends to infinity or negative infinity. These asymptotes serve as guidelines that shape the overall structure of the graph.

Domain Analysis

In curve sketching, it is essential to first determine the domain of the function. This involves identifying all the values for which the function is well-defined, particularly by looking for restrictions such as divisions by zero or even roots of negative numbers. Knowing the domain helps in understanding where the function exists and where potential discontinuities might occur.

Limit Analysis and End Behavior

Evaluating limits allows us to understand the behavior of the function near points of discontinuity and at infinity. This step is crucial for identifying how the function behaves near its domain boundaries and for confirming the presence of any asymptotic lines, thereby complementing the asymptotic analysis.

Transcript

00:01 We want to sketch the graph of y is equal to 1 plus 1 x plus 1 over x squared.

00:07 Now, i went ahead and just added these up because this was really a rational function in disguise.

00:14 And we might need to realize this is a rational function, so we can start looking at horizontal and vertical asymptotes.

00:24 Now, normally the first thing i would do is try to factor the top and bottom, but we know that x squared plus x plus one does not factor so our numerator will never even to zero so we won't be able to factor and we won't be able to simplify this at all all right so first in this chapter it gives us a list of steps we should follow to sketch the graph of the function the first one is to find the domain now the only thing we will need to worry about is our denominator not being 0 .0 .0.

01:03 0.

01:06 So that will just be where x cannot be 0.

01:09 So it'll be negative infinity to 0 union 0 to infinity.

01:16 Next, we want to find our intercepts.

01:21 So interceptions.

01:27 Now we already said there's going to be no x intercepts just due to the fact our numerary cannot factor.

01:34 And to find our y intercept, we're going to go ahead and let x equal to 0.

01:40 Y intercept.

01:43 Is going to be 0 squared plus 0 plus 1 over 0, but that's undefined.

01:55 Undefined.

01:57 So there's no y intercept.

02:01 So we're going to have no intercepts for our function.

02:05 Next we want to look for whether there is any symmetry.

02:12 So rational functions aren't really known for being periodic.

02:15 So we will instead to check to see if this is an even or odd function.

02:19 We can determine if this is symmetric about the y -axis or the origin.

02:25 So just plug in negative x.

02:28 We get negative x squared minus x plus one over negative x squared.

02:36 Well, that's going to be x squared minus x plus one over x squared.

02:42 And this here is not equal to negative f of x nor f of x.

02:49 So that tells us we're going to have no symmetry, or at least no symmetry about the y -axis and origin.

02:58 The next thing they tell us to look for are asymptotes.

03:06 So we know when we have the same power, our horizontal asymptote will be just dividing the largest power's coefficients.

03:17 So the limit as x approaches both positive and negative infinity of f of x.

03:25 Is going to just be the lean coefficient at the top is one, and the bottom is one.

03:30 So it'll just be one.

03:33 Now, the limit as x approaches 0 from the right, and we want to look from the right because we'll have a vertical asymptote at x is equal to 0, since when we plug in 0, we would be dividing by 0.

03:50 Of f of x.

03:52 So let's just go ahead and plug this in.

03:54 So 0 from the right, squared, plus 0 from the right, plus 1 ,000, over 0 from the right squared.

04:03 Now, if i square something that is slightly larger than 0 is going to be positive, something slightly larger than 0 will be positive and 1 is positive.

04:15 So adding all those up in the numerator will have something that's positive in our numerator.

04:22 And when i square something slightly larger than 0, that will also be positive.

04:27 So this here will approach positive infinity.

04:34 Now we can look at the limit as x approaches zero from the left of f of x.

04:41 And so zero from the left squared plus zero from the left plus one over zero from the left.

04:56 So just like before, we can go ahead and look at what happens here.

05:04 So zero from the left will be a slightly small negative number.

05:08 Squaring that would make it positive.

05:11 Zero from the left is going to be negative and one is going to be negative.

05:14 So now we know adding two positive numbers will be positive, and then we have to decide if adding something slightly smaller than 0 to 1 will make it negative.

05:27 Well, if we make that term small enough, then it will get dwarfed by 1, so we can go ahead and say that our numerator is going to be positive.

05:37 And if we square something slightly to the left of 0, even though it's negative, it will become positive.

05:41 So this will also go to positive infinity.

05:46 So we have our acetopes done.

05:49 The next thing they suggest we find is our intervals of increasing and decreasing, and any local maxes or mince we may have.

06:00 So we're going to need to find y prime.

06:04 So let's go ahead and do that on the next page.

06:08 So y is equal to 1.

06:10 So for this, i'm going to use the original equation, since i don't want to have to use quotient rule to take this derivative.

06:17 You could use quotient rule and we'll get the same answer.

06:19 I just want to be lazy when i do this.

06:22 So taking the derivative of this, the derivative of 1 is going to be 0.

06:27 The derivative of 1 number x, so let's go ahead and rewrite this.

06:31 So this is x to negative 1st power and x to the negative 2nd power.

06:34 So to take the derivative of each of those, we use power rule.

06:37 So it would be negative 1, x to the negative second power, and then plus negative 2 x to the negative third power.

06:54 And then we can go ahead and rewrite this, using a little bit of algebra, as negative 1.

07:00 And actually, i want to factor that negative out.

07:02 So negative 1 over x plus 2 over x cubed.

07:08 And then we can add those to get x plus 2 over x cubed.

07:14 And there should have been x squared there.

07:18 All right, so now let's go ahead and figure out where this function is equal to zero at.

07:26 So the denominator doesn't matter.

07:27 So it's only matters when the numerator is equal to zero.

07:30 So we do x plus 2 is equal to 0 or x is equal to negative 2.

07:34 So this is one of our critical points.

07:36 And also we get a critical point when our denominator is equal to zero, since it would be undefined.

07:47 So x is equal to zero.

07:48 But we know our original function x cannot be equal to zero so we can just go ahead and ignore this but keep in mind normally when we have undefined points in our derivative those are still critical all right so just go ahead and use that so we get y prime is equal to negative x plus 2 over x cubed now we want to determine where this function is strictly larger than zero and strictly less than so y prime will be strictly larger than zero or this will be increasing when, so i went ahead and already solve this beforehand, just for the sake of brevity, and this would be negative 2 to 0 for when this function is decreasing...

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