5
$\begingroup$

First of all thank you for visiting this question! I believe it's a pretty simple problem but get's kinda hairy and time consuming on each step as I have done it, so my question (the one you are here for) is the following :

What is the fastest and most reliable way to sketch a given equation determining zeroes, asymptotes, extreme values, and points of inflection?

(While doing this problem here , I realized of other doubts I had about the procedure I applied, I've listed those doubts at the bottom)

I was given by the university I'm looking forward to join an example of my math admission exam and on the last section and the one that it's worth the most I encountered the following :

Let $f(x) = \frac{x}{x^2+1}$

Determine the zeroes, asymptotes, extreme values, and points of inflection of the graph. Use this information to sketch the graph of $f(x)$ supported by complete arguments.

My first idea was to create a table that I would be filling while doing this problem, the table was the following :

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes :                        |
|Vertical Asymptote :            |
|Horizontal Asymptote :          |
|Extreme Values :                |
|Points of Inflection :          |
+--------------------------------+

Now I will separate each procedure by sections on the post , in the same order I did the problem.

1.- Determining the zeros


I proceeded to find the zeroes by doing the following

$$\frac{x}{x^2+1}=0$$ $$·(x^2+1) ·(x^2+1)$$ $$x=0$$

So when $x=0$ the equation is $0$ as well (Yes, I could have determined this by just looking at the pure fraction, but I'm trying to be as general as possible , my test might contain some other than just x)

We proceed to fill the table

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes : 0                      |
|Vertical Asymptote :            |
|Horizontal Asymptote :          |
|Extreme Values :                |
|Points of Inflection :          |
+--------------------------------+

2.- Determining Vertical Asymptote


I proceeded to determine the vertical asymptote by setting the denominator to zero as I illustrate next

$$x^2+1=0$$ $$-1 -1$$ $$x^2=-1$$

By having this I determined that there was no vertical asymptote.

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes : 0                      |
|Vertical Asymptote : None       |
|Horizontal Asymptote :          |
|Extreme Values :                |
|Points of Inflection :          |
+--------------------------------+

3.- Determining the horizontal asymptote.


To do this, I knew that

Given $$\frac{t^a}{d^b}$$ if $b>a$ then we have an horizontal asymptote at $y=0$

Consequently

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes : 0                      |
|Vertical Asymptote : None       |
|Horizontal Asymptote : y=1      |
|Extreme Values :                |
|Points of Inflection :          |
+--------------------------------+

4.- Determining the extreme values.


This is where it gets time consuming and complicated , to proceed I differentiated my $f(x)$ as illustrated next

$$y = \frac{x}{x^2+1}$$ $$\frac{dy}{dx}=\frac{dy}{dx}[\frac{x}{x^2+1}]$$ $$\frac{dy}{dx}[\frac{x}{x^2+1}] = \frac{dy}{dx}[x·((x^2+1)^-1)]$$

Note : On the last illustration the last term on the right is raised to the negative one power, which is equal to $\frac{1}{x^2+1}$ (I do not know why I can't express the negative one power here..)

And I also knew that

Given $$\frac{dy}{dx}[f(x)g(x)]=f'(x)g(x)+g'(x)f(x)$$

So I proceeded first to find the derivatives of each term being multiplied , the first one resulting as below : $$\frac{dy}{dx}[x] = 1$$

The second one resulting as below as well (Using chain rule) $$\frac{dy}{dx}[(x^2+1)^-1] = -1((x^2+1)^-2)·2x$$ Or equivalently $$\frac{dy}{dx}[(x^2+1)^-1] = -\frac{2x}{(x^2+1)^2}$$

So ended up having $$\frac{dy}{dx}[x] = 1$$ and $$\frac{dy}{dx}[(x^2+1)^-1] = -\frac{2x}{(x^2+1)^2}$$

Proceeded to use the product rule as illustrated below

$$\frac{dy}{dx}[f(x)g(x)]=f'(x)g(x)+g'(x)f(x)$$ $$\frac{dy}{dx}[x·\frac{1}{(x^2+1)}]=(1·(\frac{1}{(x^2+1)}))+(-\frac{2x}{(x^2+1)^2} · x)$$ $$\frac{dy}{dx}[x·\frac{1}{(x^2+1)}]=(\frac{1}{(x^2+1)})-(\frac{2x^2}{(x^2+1)^2})$$

I proceeded to find a common denominator, and accomplished this by multiplying $\frac{1}{(x^2+1)}$ by $(x^2+1)$ as illustrated next

$$\frac{1}{(x^2+1)}·\frac{x^2+1}{x^2+1} = \frac{x^2+1}{(x^2+1)^2}$$

Substituting it and resulting in

$$\frac{dy}{dx}[x·\frac{1}{(x^2+1)}]=(\frac{x^2+1}{(x^2+1)^2})-(\frac{2x^2}{(x^2+1)^2})$$ $$\frac{dy}{dx}[x·\frac{1}{(x^2+1)}]=\frac{(x^2+1)-(2x^2)}{(x^2+1)^2}$$ So our final derivative is $$\frac{dy}{dx}[x·\frac{1}{(x^2+1)}]=\frac{-x^2+1}{(x^2+1)^2}$$

To find it's extreme values we set the derivative equation equal to zero

$$0=\frac{-x^2+1}{(x^2+1)^2}$$

And solve for x doing the following

$$0=\frac{-x^2+1}{(x^2+1)^2}$$ $$·(x^2+1)^2·(x^2+1)^2$$ $$0=-x^2+1$$ $$-1 -1$$ $$-1=-x^2$$ $$·-1 ·-1$$ $$1=x^2$$ $$\sqrt(1)\sqrt(x^2)$$ $$\pm1=x$$

So finally, we have determined the extreme values which are $x=1$ and $x=-1$

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes : 0                      |
|Vertical Asymptote : None       |
|Horizontal Asymptote : y=1      |
|Extreme Values : x=1 & x=-1     |
|Points of Inflection :          |
+--------------------------------+

5.- Determining nature of extreme values.


This is were I was completely frustrated on how long it was taking for me to determine the nature of the extreme values I had obtained (Maxima or minima), I firstly remembered that to determine this I needed to determine the second derivative of the first equation and evaluate it at the obtained extreme values , if when doing so the value resulted positive that meant that It would be a minimum value, if negative it would be a maximum value.

So I would need to do as illustrated next

$$y=\frac{-x^2+1}{(x^2+1)^2}$$ $$\frac{dy}{dx}=\frac{dy}{dx}[\frac{-x^2+1}{(x^2+1)^2}]$$

To do this I decided to use the quotient rule as It was easier for me to derive with no negative terms on this more complicated functions, so the quotient rule states

$$\frac{dy}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$$

So I proceeded to derive each term of my equation, the top term resulting in

$$\frac{dy}{dx}[-x^2+1] = -2x$$

And the bottom term resulting in

$$\frac{dy}{dx}[(x^2+1)^2] = 2·(x^2+1)·2x$$

or $$\frac{dy}{dx}[(x^2+1)^2] = 4x(x^2+1)$$

So ended up having

$$\frac{dy}{dx}[-x^2+1] = -2x$$

and $$\frac{dy}{dx}[(x^2+1)^2] = 4x(x^2+1)$$

Now proceeded to use the quotient rule

$$\frac{dy}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$$

$$\frac{dy}{dx}[\frac{-x^2+1}{(x^2+1)^2}] = \frac{(x^2+1)^2·(-2x)-(-x^2+1)·(4x(x^2+1))}{((x^2+1)^2)^2}$$

Evaluate at our extreme values $x=1$ and $x=-1$

$$\frac{((1)^2+1)^2·(-2(1))-(-(1)^2+1)·(4(1)((1)^2+1))}{(((1)^2+1)^2)^2} = -\frac{1}{2}$$

So we can say that at point $x=1$ theres a maximum value.

$$\frac{((-1)^2+-1)^2·(-2(-1))-(-(-1)^2+-1)·(4(-1)((-1)^2+-1))}{(((-1)^2+-1)^2)^2} = \frac{1}{2}$$

So we can say that at point $x=-1$ theres a minimum value.

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes : 0                      |
|Vertical Asymptote : None       |
|Horizontal Asymptote : y=1      |
|Extreme Values:x=1 Max x=-1 Min |
|Points of Inflection :          |
+--------------------------------+

6.- Determining Inflection Points


It's taking me longer than expected to write this post so I'll leave for anyone interested this part as a practice, what I basically did was simplify the second derivative , in this case the following

$$0 = \frac{(x^2+1)^2·(-2x)-(-x^2+1)·(4x(x^2+1))}{((x^2+1)^2)^2}$$

After taking a bunch of time simplifying it I ended up with

$$2x^4+6=0$$

and guessed that for the function to be equal to zero, then 2x^4 should be equal to exactly -6

$$2x^4=-6$$

I ended up with an imaginary number so guessed that there was no points of inflection ?

NOTE: This part of determining the inflection points , I'm not pretty sure that I could have come with such a value (Imaginary) , please let me know of my mistakes.

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes : 0                      |
|Vertical Asymptote : None       |
|Horizontal Asymptote : y=1      |
|Extreme Values:x=1 Max x=-1 Min |
|Points of Inflection : None?    |
+--------------------------------+

6.- Graphing


With my final information table

+--------------------------------+
|Known Information               |
+--------------------------------+
|Zeroes : 0                      |
|Vertical Asymptote : None       |
|Horizontal Asymptote : y=1      |
|Extreme Values:x=1 Max x=-1 Min |
|Points of Inflection : None?    |
+--------------------------------+

I graphed the function by using the extreme values (evaluating it's y value on the function) and the zeroes (zero) , and ended up having the following

drawn_graph

that was actually a good result.

My doubts left after this entire procedure in conjunction with the main question (which I'm excluding here on the bottom) of this topic are the following

  • Is it quite possible that I could have gotten an imaginary number as my inflection point?
  • Is trying close numbers to the found extreme values with the purpose of determining if the point is a maxima or minima a valid argument ?

And thank you very much for reading the question! (:

$\endgroup$
  • 4
    $\begingroup$ Speed when drawing functions is a rather useless ability to acquire, really. $\endgroup$ – Mariano Suárez-Álvarez May 24 '14 at 20:38
  • $\begingroup$ Drawing functions means determining zeroes, asymptotes, extreme values, and points of inflection of the graph , being fast at it would mean doing things the most efficient and correct way, which is truly a handy ability to acquire. @MarianoSuárez-Alvarez $\endgroup$ – Joel Hernandez May 24 '14 at 20:40
  • $\begingroup$ Juggling or back somersaults or even card tricks are a better objective! 《Quick! Sketch this function FAST or Gotham City is doomed, Batman!》is something we'll never hear in a movie. $\endgroup$ – Mariano Suárez-Álvarez May 24 '14 at 20:41
  • $\begingroup$ Guessing you're not a student anymore , when it comes to tests time is a most precious thing.@MarianoSuárez-Alvarez $\endgroup$ – Joel Hernandez May 24 '14 at 20:43
  • $\begingroup$ Your final sketch has at least one inflection point at $x=0$. In fact, between two extrema there is always an inflction point (assuming no gaps in the domain) $\endgroup$ – Hagen von Eitzen May 24 '14 at 20:49
5
$\begingroup$

The fastest, most reliable way to graph a function may well be to (intelligently) use a computer. If your circumstances prohibit that, then more experience with algebra and taking the derivative will make graphing faster. Here are more comments:

1.The zeros

A fraction is zero iff the numerator is zero (assuming the denominator isn't simultaneously zero). So in fact, you can simply say that $$ \frac{x}{x^2+1} = 0 \implies x=0.$$

2. Vertical asymptotes

In 1, I implicitly used the fact that $x^2+1$ is never zero.

3. Horizontal asymptotes

This one didn't take you long; good. You should probably have the numerator and denominator both as functions of the same variable. Hopefully you also know what happens if the numerator and denominator grow at the same approximate rate, such as knowing $$\lim_{x \to \infty}\frac{13x^{3}-100x+1}{-2x^{3}+9} = -\frac{13}{2}.$$

4. Extreme values

I didn't read all of this section because I think it's too long. To find the derivative faster, Hagen von Eitzen is correct in his comment above; use the quotient rule. What you're doing is basically rederiving it.


Let me insert a comment that applies to more than just bullet 4. First, thank you for putting significant effort into the question and for showing your work. However,

You really should be showing less work.

Being concise is good. For example, your potential readers probably already know calculus. Therefore, you don't need to remind us what the product and quotient rules are. Also, you don't need to show every step. For instance, to find $$\frac{d}{dx} \left[\frac{-x^2+1}{(x^2+1)} \right],$$ you can just apply the quotient rule in one step and then simplify in one or two more steps.

5. Nature of extreme values

Although finding the second derivative works, it is not required. Yes, trying numbers close to the extreme values will show what kind they are, but then you'd have to know how close is close enough. Another way would be to see if the first derivative goes from positive to negative or vice versa, as you plug in numbers smaller then larger than the extreme values.

6. Points of inflection

As Hagen von Eitzen mentions, there is at least one. To find where the numerator of your fraction is zero, the first thing you should do is factor out a $2x(x^2+1)$. Without much trouble at all, this gives $2x(x^2-3)$ (after cancelling the $x^2 + 1$), and this has three zeros. We therefore have three points of inflection.

7. Graphing the function

Overall, you did do a decent job following the algorithm given in calculus texbooks on how to graph a differentiable function.


Final Remarks

First, about imaginary points of inflection, I'm not sure if the concept is useful for functions whose argument is a complex number. Maybe, but I'm not familiar with it. In most introductory, single variable calculus classes (at least in America), all functions have domains and ranges that are subsets of $\mathbb{R}$. So, anytime all proposed answers are complex, then you know there aren't real solutions. Also, my impression is that the concept of concavity doesn't directly apply to complex valued functions.

Also, let me comment on the use of $\frac{dy}{dx}$. That expressions is itself the derivative of the function $y$. But $\frac{dy}{dx} y$ is in fact $y'y$ (i.e. $y'$ times $y$). So if $y = f(x)g(x)$, then $\frac{dy}{dx}[f(x)g(x)] = y'y$.

The symbols $$\frac{d \text{blah}}{dx} \text{ and } \frac{d}{dx}[\text{blah}]$$ both mean the derivative of "blah" with respect to x. When "blah" is short (such as "$y$") we usually choose the notation on the left. When it's long, the one on the right is usually preferred.

At the risk of offering too many suggestions, let me add that you may find the MathJax basic tutorial helpful. See for example bullet point 5.

$\endgroup$
  • $\begingroup$ Thank you very much Andrew for your precise and honest answer! I will surely take in mind all points stated! $\endgroup$ – Joel Hernandez May 25 '14 at 0:40
  • $\begingroup$ You're welcome Joel. And thank you again for putting significant effort into your question. $\endgroup$ – Andrew Kelley May 25 '14 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.