# Concavity & Inflection point(s)

• Mar 18th 2010, 05:42 PM
myngo9191
Concavity & Inflection point(s)
This is the problem: https://webwork.math.lsu.edu/webwork...6cd14f7771.png

I have to find:
A) critical values = 0
B) where f(x) is increasing = (0,inf)
C) where f(x) is decreasing = (-inf,0)
D) local maxima at x = none
E) local minima at x = 0
F) concave up =
G) concave down =
H) inflection point(s) at x =
I) horizontal asymptote(s) y = 3
J) vertical asymptote(s) x = none

I was able to figure out everything excluding F, G, and H -as you may have noticed. I know that I have to find the second derivative of the original function however when I tried this was what I got:
$\displaystyle \frac{-150(x^4-2x^3+50x^2-50x+625)}{(x^2+25)^2}$
• Mar 18th 2010, 05:49 PM
DarkestEvil
Usually, to find concavity, you must set the second derivative equal to zero. Use those x values to find where it is positive, where it is negative. Positive means concave up, negative means concave down.
• Mar 18th 2010, 05:56 PM
myngo9191
i set everything equal to zero and i end up with $\displaystyle +/-\sqrt{-25}$ and $\displaystyle \frac{2+/-\sqrt{-96}}{2}$ and i really dont think they're looking for complex roots
• Mar 18th 2010, 08:45 PM
drumist
Your second derivative is not correct. Try reworking it.

But first, did you get the first derivative correct?

$\displaystyle f'(x) = \frac{150x}{(x^2+25)^2}$
• Mar 18th 2010, 08:48 PM
drumist
$\displaystyle f''(x) = \frac{(x^2+25)^2(150) - 2(x^2+25)(2x)(150x)}{(x^2+25)^4}$

Factoring out a $\displaystyle (x^2+25)$ first will make the algebra easier:

$\displaystyle f''(x) = \frac{(x^2+25)(150) - 2(2x)(150x)}{(x^2+25)^3}$
• Mar 19th 2010, 06:27 AM
myngo9191
[drumist] yup that was what i got for the first derivative.where i messed up was that i didn't take the derivative of $\displaystyle x^2$ for g prime.i fixed my error and found the inflection point to be $\displaystyle +/-\frac{5\sqrt3}{3}$ and it is correct.i plugged in points near the interval and found that it is concave up at $\displaystyle (-\inf,\frac{5\sqrt3}{3})$ and concave down at $\displaystyle (\frac{5\sqrt3}{3},\inf)$ however it said i was wrong.i dont understand how its wrong
• Mar 19th 2010, 07:21 AM
drumist
Since you have two inflection points, you have three intervals to consider!

$\displaystyle \left(-\infty,-\tfrac{5\sqrt{3}}{3}\right)$

$\displaystyle \left(-\tfrac{5\sqrt{3}}{3},\tfrac{5\sqrt{3}}{3}\right)$

$\displaystyle \left(\tfrac{5\sqrt{3}}{3},\infty\right)$
• Mar 19th 2010, 08:15 AM
myngo9191
0o0k.i got it now.thank you!
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