Need to find Critical Numbers/Concavity/Etc.

• Nov 11th 2011, 01:35 PM
APMAPM
Need to find Critical Numbers/Concavity/Etc.
Hey everyone, back again with another problem. So I have this:

(1/2)x^4 + x^3 - X^2 + 2

I need to find:
Critical Numbers
Critical Points
Relative Maxima and Minima
Concavity
Inflection Points

f'(x) = 2x^3 + 2x^2 - 2x. Taking out x I got (x)(2x^2 + 3x - 2)

The quadratic formula has given me the Critical Numbers: -2, 0, 1/2

Replaced them back in the original and after the 1DT got: (0, 2) = Rel. Max, (-2, 2) (1/2, 61/32(?)) = Rel. Min

Then f''(x) = 6x^2 + 6x - 2, and took out a two for 3x^2 + 3x - 1

Then for the quadratic formula I wound up with the concavity numbers of (-3 - square root 3)/6 and (-3 + square root 3)/6

Now I'm guess I did something wrong, since I don't think my professor would give me such messy numbers to work with. Otherwise I'd do the 2DT to get concavity and replace them in the original problem to get the inflection points... so I know what to do (I think) but I also think I'm messing up somewhere no matter how many times I retry it. Anyone know what's up?
• Nov 11th 2011, 02:06 PM
Siron
Re: Need to find Critical Numbers/Concavity/Etc.
The derivatives and the critical points are fine. With the concavity you probably have to determine an interval where the function is concave (and convex). The inflection points you can find by solving $\displaystyle f''(x)=0$
• Nov 11th 2011, 02:51 PM
APMAPM
Re: Need to find Critical Numbers/Concavity/Etc.
Thank you for the swift response. I understand how to find the inflection points and did set f''(x) to 0: 6x^2 + 6x - 2 = 0, and after the quadratic formula I wound up with those messy fractions. I believe I am going wrong there, as then I would use the inflection points to find concavity, but also after replacing them in the original problem I would wind up with a mess, and one which I don't believe my professor would have given me compared to the other problems on this sheet we are supposed to do. Or... is it correct?
• Nov 12th 2011, 12:47 AM
Siron
Re: Need to find Critical Numbers/Concavity/Etc.
Solving $\displaystyle 3x^2+3x-1=0$ gives 2 solutions with the quadratic formula:
$\displaystyle x_1=\frac{-3+\sqrt{21}}{6}$ and $\displaystyle x_2=\frac{-3-\sqrt{21}}{6}$

So they're indeed a little bit messy, but I don't think that's a problem.
Now make a sign table to determine where the function is convex or concave.
• Nov 13th 2011, 12:31 PM
APMAPM
Re: Need to find Critical Numbers/Concavity/Etc.
negative infinity to -3-squareroot(21)/6 and -3+squareroot(21)/6 to infinity = concave up
-3-squareroot(21)/6 to -3+squareroot(21)/6 = convex

I'm pretty sure that's right. Which leaves the inflection points.

Now I know what to do

1/2(x)^4 + x^3 - x^2 + 2

With x = (-3 + squareroot(21))/6 and (-3-squareroot(21))/6

So I'm a bit rusty with these square root binomial multiplication, so if I could just get some clarification that I'm doing this right:

(-3 + squareroot(21)/6 X (-3 + squareroot(21)/6

= (9 - 3 sr(21) - 3 sr(21) - sr(21)^2)/36

= (30 - 6 sr(21))/36

= (5 - sr(21))/6

And then I'd keep multiplying until I got to the right power. I was just hoping someone could confirm that if my math is correct or not right here. Thanks again for everything so far though, it's bee a huge amount of help.