@acer 

I thought i could do the task step by step (learning step by step)

The calculations made on reaction on question are very useful for me to see basis calculus with plots to learn more

Here is the task where i am working on , could be interesting to see if there are more answer are possible opposite  the direction of the book is aiming for. 

But that is a lot work and that's why i tried to do it step by step and come up with a own solution 

Download betounes_ex_set2_task_3_vraagforum.mw

CD2-5.aangepast.mwsCD2-5.aangepast.mws

@acer 

Thanks

Impressive..

The task is draw in y the tangentlines in all 6 inflection points of y on  interval  0.1..16  ( so the y '' is needed) 
So inspection by sight shows then 6 infllection points for ( your calculation use 0..10 ) 

Yes, change sign must then be shown in  y '' to conclude were y  has inflectionpoints with a particular direction.

https://www.hhofstede.nl/modules/buigpunten.htm

@tomleslie 
Thanks 

I could chance in the procedure the functions : it is y and y ' , so the y plot with the inflection point depicted then

@tomleslie 

Thanks

@tomleslie 

Thanks 

The start function is f(x) = x^sinx  and the prime : x^sin(x)*(cos(x)*ln(x)+sin(x)/x  ,  and has the form  x^n
The task ask me to use fsolve command  

I think for polynomial the exponent must be a integer number ..i must read more about this 

Thanks

I noticed already dat y '' = 0 was calculated ( that's good too for learning) : what gives all special points for y ( maxs,mins, inflection points), but i need only the inflection points of y for now. 

RootFinding:-Analytic(diff(fprime_expr,x,x)=0, re=0..16, im=-1..1);

If a replace fprime_expr (= y ' ) for y then i get the maxs and mins, right?  

y

y ' =0 :  max or min y

y '' = 0  : max or min +inflection points -> gives all points of direction chance of graph y

note: i must try to keep this simple, because the task was draw in  y the tangentlines in all 6 inflection points of y ( so the y '' is needed) 

Thanks

Amazimg with Maple 

Now i must this relate with the fprime graph in my worksheet.

@acer 

When the 2-dimensional array is evaluated you can see the content. 
Now with a one-dimensional array it seems to be not possible to see the content inside as a whole  ?..why is that ?

@acer 

Thanks

Some extra examples makes it more clear.

@Carl Love 

Thanks 

@Kitonum 

Thanks 

It was asked the do this by two pair of nested loops
Those nested loops do have more uses in other tasks i see and it is to get used of them

Maybe i see also your array specify in this book examples?

What makes it easier  ..to come up with this?   

@acer 

Thanks

Could not come up on this array

The elements are:

A[1,1]=1^1 , A[1,2]=1^2,   A[1,3]=1^3  A[4,1]= 1^4 

A[2,1]=2^1, A[2,2]=2^2,   A[2,3]=2^3  A[2,3]=2^4  

........................

De outer loop is for from top to bottom en inner loop from left to right

@Carl Love 

Download betounes_ex_set2_1.mw

@Carl Love 

Thanks

I made a overhaul of task 1c and get stuck for drawing cross-hatching one symmetry part of the graph. 

Also i used a piecewise function A__right , but don't understand it  

All integrals ( symbolic, numeric (Int and int ) give the same answer

Download betounes_ex_set_1_andereaanpak_1_opg_5.mw