# Local Maxima, Minima and Points of inflections

• Mar 24th 2013, 10:11 PM
iFuuZe
Local Maxima, Minima and Points of inflections
So far this is what i got:
Original eq:
f(x)=(x+5)/(x^2-16)

By using quotient rule i got:

f'(x)=0=x^2-16-2x^2-10x
0=(x+8)(x+2)
x=-8, x=-2

Now subbing them back into f(x) i got max and min as (correct me if im wrong):
(-8,3/80) and (-2, -3/20)

Point of inflection:

f''(x)=2(x^3+15x^2+48x+80)/(x^2-16)^3

the second derivative is right im pretty sure i just took out 2 as common factor:

Now this is where im confused do i make the whole thing equal 0, or the top line? and how can i simplify it afterwards?
• Mar 25th 2013, 12:34 AM
Prove It
Re: Local Maxima, Minima and Points of inflections
By the quotient rule, your derivative should be $\displaystyle \displaystyle \frac{x^2 - 16 - 2x^2 - 10x}{\left( x^2 - 16 \right) ^2}$. The evaluation of the x values of critical points is correct though. Note that f(-8) = -3/48 = -1/16 and f(-2) = -3/12 = -1/4. How did you determine which is a local max and which is a local min?
• Mar 25th 2013, 03:46 AM
iFuuZe
Re: Local Maxima, Minima and Points of inflections
i factorised the top line and subbed -2 and -8 into f(x)
• Mar 25th 2013, 03:49 AM
Prove It
Re: Local Maxima, Minima and Points of inflections
Well you made some mistakes that I already pointed out. And what you did is not enough to show if it's a local maximum or minimum. I suggest you look up the second derivative test.