Expert: Paul Klarreich - 6/22/2006

Question
I am really confused about how to do the following problem:

PLEASE HELP!
Analyze the function f(x) = x / (x^2 ? 4) as if you
were going to graph it.  
*What are all of the x and y intercepts?
*Compute and simplify the 1st and 2nd derivative.  
*Describe the vertical and horizontal asymptotes.  
*Where is the function increasing and decreasing?
*Where is it concave up and concave down?
*Find all critical points (what kind are they)
*What are the inflection points?


Answer
Hi, Todd,

Subject:  Calculus - function analyzation...
Question:  I am really confused about how to do the following problem:

PLEASE HELP!
Analyze the function f(x) = x / (x^2 ? 4) as if you
were going to graph it.  
*What are all of the x and y intercepts?
*Compute and simplify the 1st and 2nd derivative.  
*Describe the vertical and horizontal asymptotes.  
*Where is the function increasing and decreasing?
*Where is it concave up and concave down?
*Find all critical points (what kind are they)
*What are the inflection points?
-----------------------------------
I am not sure about two things:
A) Do you just want an answer, or do you need to know what all those things are?
B) What's in between the x^2 and the 4?  Looks like a question mark.

I will assume the answer to A is: You need to know what this is all about.  And the answer to B is 'minus'

This problem really says: Sketch the graph of ....

The term 'sketch' means to locate and mark the important features of the graph and then sketch in a figure that is consistent with all those features.  What are these important features?  Your list tells you what they are.  How do you compute them?  Here goes:

*What are all of the x and y intercepts?

Get your x-intercepts by setting f(x) = 0 and solving.  Remember that with any fraction, the answer is zero when the top is zero.  In this case, x = 0 is the intercept, so you will make sure the graph passes through the origin.

Get your y-int's by setting x = 0 and computing y.  In this case, it's y=0, or the origin again.

*Compute and simplify the 1st and 2nd derivative.  

This is just hackwork, which I will leave to you. (Could be quite a bit.)
       - 4 - x^2
f'(x) = ----------
       (x^2 - 4)^2

        2x^3 + 24x
f''(x) = ----------
           ()^3


*Describe the vertical and horizontal asymptotes.  
That should say 'determine any...'

Vertical asymptotes: When you have a fraction, set the bottom = 0 and solve.  In this case  x^2 - 4 = 0  gives x = +- 2, so those are your vertical asymptotes.  

Draw those vertical lines and make real sure your graph approaches them on the left and right.

Howizontal asymptotes: Does the lim(x -> +- infinity) exist?  In this case, you have lim(x->inf) f(x) = 0, which is the x-axis.  So make real sure your graph approaches the x-axis as it goes way out towards the right and left.


*Where is the function increasing and decreasing?

Increasing: Take your first derivative and see where it is a positive number.

       - 4 - x^2
f'(x) = ----------
       (x^2 - 4)^2
Since the denominator is a square, it is never negative, so take the top and write:

-4 - x^2 > 0  Never going to happen.  The graph never rises.  (Cute title for a math example, right?)

-4 - x^2 < 0  Always going to happen.  The graph is always falling. (Function is decreasing.)


*Where is it concave up and concave down?

Use your second derivative.  By the way, I prefer to use the phrases
'turning to the left' instead of 'concave up.'  and
'turning to the right' instead of 'concave down.'

        2x^3 + 24x
f''(x) = ----------
           ()^3

Now this is harder to analyze, because the bottom changes sign at +-2, and the top is:

2x(x^2 + 8), which changes sign only at x=0.  I think the rest of the analysis will give you these answers.

*Find all critical points (what kind are they)

A critical point is determined by the first derivative.  It works like this:

A. If f'(x) = 0, you have a stationary point.  This could be:

 A1: A max if  f'' < 0
 A2: A min if  f'' > 0
 A3: A 'who knows what?' point, (likely an inflection point) if f'' = 0

B. If f'(x) is undefined, you have a singular point.  The graph behaves badly, perhaps being discontinous.  A vertical asymptote is a good example of this.

In this example, you will get  x = 0 to be a stationary point, and x=+-2 as singular points.  (Not actually points on the graph.)



*What are the inflection points?

Determined by the second derivative.  Set f''(x) = 0 and solve. In this case, you get x = 0 as your inflection point.

Now you are ready.  You will plot a figure that:

A. Passes through (0,0).
B. Always falls.
C. Is discontinuous at x = +- 2, approaching those lines.
D. Gets closer and closer to the x-axis the farther away from the origin we are.
E. Does not 'wiggle' anywhere, because there are no other special points.

Now you can go to work.  If you like you can send me your email address and I'll send you a picture of the graph.  (I can't do it through this site.)