Acer's response addresses the main problem. Let me add a little more that should help down the road.

Maple gives you the choice to work with functions or expressions. It even allows you to mix the two. But, this is NOT advisable. In the original post, f is defined as a function, but the derivatives are defined as expressions.

This is very likely to lead to dificulties later. For example, to find the roots and stationary (critical) points of your function you would need to use something like

solve( f(x)=0, x );   # must include (x)
solve( df=0, x );     # would be wrong to include (x)

Instead, I suggest choosing to work only one type: expressions or functions. Here is how you would find the derivatives in both cases.

# Functions
f:= x->((x^3-x+1+(6/30))/(x^2-1));
df := D(f);
d2f := (D@@2)(f);
d3f := (D@@3)(f);

# Expressions
f:= ((x^3-x+1+(6/30))/(x^2-1));
df := diff(f,x);
d2f := diff(f,x$2);
d3f := diff(f,x$3);

I hope this will be helpful,

Doug

---------------------------------------------------------------------
Douglas B. Meade
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu/~meade/

As I was composing my previous response, you posted your new question. The answer to your question depends on exactly how you asked Maple to find a solution.

I am going to assume you are using functions.

solve( f=0, x );

This returns no solution. Nothing. Nada. Just a new prompt.

Include the (x) and you get the exact solutions.

solve( f(x)=0, x );
                         (1/3)                                 
  1  /             (1/2)\                    5              1  
- -- \2025 + 75 654     /      - -------------------------, -- 
  15                                                 (1/3)  30 
                                 /             (1/2)\          
                                 \2025 + 75 654     /          

                      (1/3)                              
  /             (1/2)\                     5             
  \2025 + 75 654     /      + ---------------------------
                                                    (1/3)
                                /             (1/2)\     
                              2 \2025 + 75 654     /     

                /                         (1/3)                            \  
     1    (1/2) |  1  /             (1/2)\                    5            |  
   + - I 3      |- -- \2025 + 75 654     /      + -------------------------|, 
     2          |  15                                                 (1/3)|  
                |                                 /             (1/2)\     |  
                \                                 \2025 + 75 654     /     /  

                         (1/3)                              
  1  /             (1/2)\                     5             
  -- \2025 + 75 654     /      + ---------------------------
  30                                                   (1/3)
                                   /             (1/2)\     
                                 2 \2025 + 75 654     /     

                /                         (1/3)                            \
     1    (1/2) |  1  /             (1/2)\                    5            |
   - - I 3      |- -- \2025 + 75 654     /      + -------------------------|
     2          |  15                                                 (1/3)|
                |                                 /             (1/2)\     |
                \                                 \2025 + 75 654     /     /

And, if you then make 0 into 0.0, you will see numerical solutions.

solve( f(x)=0.0, x );
 0.6848538719 + 0.6380238847 I, -1.369707744, 0.6848538719 - 0.6380238847 I

This was working with the original function. You would need to do something similar for the stationary (critical) points. I'll let you determine exactly what you should be doing.

Awaiting your next question,

Doug

---------------------------------------------------------------------
Douglas B. Meade
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu/~meade/

To determine if you believe Maple's answers, you can look at the graphs of these functions.

Also, re-check the definitions of critical points and inflection points. An inflection point is a point where the second derivative changes sign. One way this can happen is for the second derivative to be zero, but there is another way. I won't give you this answer, but I will tell you that I can see two points to check without doing any differentiation.

But, to get started, look at the graphs. You will need to specify the domain and the range, for example,

plot( df, x=-3..3, y=-10..10 );

You might also want to include discont=true as an optional argument in the plot command. That's enough hints.

Lastly, using RealDomain can have some unwanted consequences. This does not always work exactly as you would like. I prefer to let Maple give me the complex answers and then to ignore them if I don't want them. Something like this can be useful too:

remove( has,[ws1],I );

This will give you a list. To get just the elements of the list, append a [] to the end of the above command.

You should be getting close. Even though Maple can do a lot for you, you cannot stop thinking about what you trying to do and what Maple's responses mean.

Doug

---------------------------------------------------------------------
Douglas B. Meade
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu/~meade/