# Zeros of f''(x)

• Mar 23rd 2008, 07:38 PM
javax
Zeros of f''(x)
Hullo.
The function I was trying to draw was $\displaystyle f(x) = \frac{5-x}{9-x^2}$
The first derivative is $\displaystyle f'(x) = \frac{-x^2+10x-9}{(9-x^2)^2}$
When I come to determine Concave Up & Concave Down intervals...I calculate $\displaystyle f''(x)$ which is $\displaystyle f''(x) = \frac{-2(x^3-15x^2+27x-45)}{(9-x^2)^3}$
The problem now is that I can't find zeros of this equation, neither simplify anymore! I know there are rules to solve these but they take time I think! I had this function on test! There wasn't much time to calculate...!
So if you know any shorter way...tell me, I'd appreciate!
• Mar 23rd 2008, 07:45 PM
topsquark
Quote:

Originally Posted by javax
Hullo.
The function I was trying to draw was $\displaystyle f(x) = \frac{5-x}{9-x^2}$
The first derivative is $\displaystyle f'(x) = \frac{-x^2+10x-9}{(9-x^2)^2}$
When I come to determine Concave Up & Concave Down intervals...I calculate $\displaystyle f''(x)$ which is $\displaystyle f''(x) = \frac{-2(x^3-15x^2+27x-45)}{(9-x^2)^3}$
The problem now is that I can't find zeros of this equation, neither simplify anymore! I know there are rules to solve these but they take time I think! I had this function on test! There wasn't much time to calculate...!
So if you know any shorter way...tell me, I'd appreciate!

There is one real zero and it's (Swear) ugly. If you have a calculator you could have done a numerical approximation, but unless you know an advanced method the exact value would have been beyond you. (And also for some of us who do know the method as well. I've only done it correctly a handful of times. :) )

-Dan
• Mar 23rd 2008, 08:11 PM
javax
okay I see...!
Did I find $\displaystyle f''(x)$ right?
• Mar 23rd 2008, 08:26 PM
topsquark
Quote:

Originally Posted by javax
okay I see...!
Did I find $\displaystyle f''(x)$ right?

Yes. Good job. (Bow)

-Dan