{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
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.
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.
How can qualified in Mathematica the constraints to get the full answer? Thanks a lot
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
