
function question
This is the function:
f(x)=(x^2Bx+4)/(xA)
given:
x=2 critical point
y=x+5 is an oblique asymptote
I need to find A and B and decide if x=2 is min or max.
i got to f'(x)= (x^22Ax4+BA)/{(xA)^2}
but I'm not even sure that I need a derivative here.
can someone explain what would be the way to solve this?
Please, if you can, show your work so that I can understand it.
THANKS!

Indeed, the derivative is $\displaystyle \displaystyle f'(x)=\frac{x^22Ax+AB4}{(xA)^2}$.
If x is a critical point then $\displaystyle f'(2)=0\Rightarrow A(B4)=0$. (1)
The oblique asymptote is $\displaystyle y=mx+n$ where $\displaystyle \displaystyle m=\lim_{x\to\infty}\frac{f(x)}{x}$ and $\displaystyle \displaystyle n=\lim_{x\to\infty}[f(x)mx]$.
Now, $\displaystyle \displaystyle m=\lim_{x\to\infty}\frac{x^2BX+4}{x^2Ax}=1$.
$\displaystyle \displaystyle n=\lim_{x\to\infty}\left(\frac{x^2Bx+4}{xA}x\right)=\lim_{x\to\infty}\frac{(AB)x+4}{xA}=AB=5$ (2)
From (1) and (2) we get:
I) $\displaystyle A=0,B=5\Rightarrow x=2$ point of minimum
II) $\displaystyle A=9,B=4\Rightarrow x=2$ point of maximum.

Thanks for your reply!
I still don't understand the second part.
How did you decide what A goes with what B
and how do you know what's max and whats min?
If you could clear this up for me that would be great
thanks!

For $\displaystyle A=0,B=5\Rightarrow \displaystyle f'(x)=\frac{x^24}{x^2}$
The roots of f'(x) are $\displaystyle x_1=2,x_2=2$.
Then $\displaystyle f'(x)<0,\forall x\in(0,2)$, so f is decreasing and
$\displaystyle f'(x)>0,\forall x\in(2,\infty)$, so f is increasing.
Then x=2 is a point of minimum.
For $\displaystyle A=9,B=4\Rightarrow \displaystyle f'(x)=\frac{x^218x+32}{(x9)^2}$.
The roots of f'(x) are $\displaystyle x_1=2,x_2=16$.
$\displaystyle f'(x)>0,\forall x\in(\infty,2)$, so f is increasing and
$\displaystyle f'(x)<0,\forall x\in(2,9)$, so f is decreasing.
Then x=2 is a point of maximum.

thanks for all your help :D