{f[x] == (a x^3 + 6 x^2 + b)/(c x^2) with a?0 and b?0, Maximum relative (-2,0), oblique asymptotes ( y=2/3x + 2 )} Thanks Mike
I see. Continuing from above
In[3]:= y[x_] := (2 x^3 + 6 x^2 + b)/(3 x^2) In[4]:= y'[-2] Out[4]= (8 + b)/12 (* y'[2]=0 gives b=-8 *) In[5]:= Plot[(2 x^3 + 6 x^2 - 8)/(3 x^2), {x, -4, 4}]
This gives your function
In[4]:= y[x_] := (a x^3 + 6 x^2 + b)/(c x^2) In[5]:= y[x] Out[5]= (b + 6 x^2 + a x^3)/(c x^2) In[6]:= Expand[%] Out[6]= 6/c + b/(c x^2) + (a x)/c
As x->Infinity, asymptotic form implies c=3, a=2. Don't understand " Maximum relative (-2,0)"
local maximum (relative maximum): a value of a function that is greater than those values of the function at the surrounding points, but is not the greatest of all values.
In Mathematica execute
Maximize[ {(a x^3 + 6 x^2 + b)/(c x^2), -2 < x < 0} , x]
to get the full answer qualified by constraints on the parameters.
How can qualified in Mathematica the constraints to get the full answer? Thanks a lot
