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\[f(x) = 9x^\frac{ 1 }{ 3 } + \frac{ 9 }{ 2 }x^\frac{ 4 }{ 3 }\]

I have that \[f'(x) = 3x^\frac{ 1 }{ 3 }[x ^{-2} + 2]\]

So, I set f'(x) = 0 and I got \[x = 0, \pm \frac{ 1 }{ \sqrt{-2} }\]

http://www.wolframalpha.com/input/?i=%289x%5E%281%2F3%29+%2B+9%2F2*x%5E%284%2F3%29%29%27

I'm confused...

Hi, Jim_thomspon5910

Then set it to 0, yes?

The number line check?

yeah you can do that for a first derivative test

see if f ' (x) changes sign for a value less than 0, and for a value greater than 0

when it is less than 0, it is decreasing, when it is greater than 0, it is increasing

correct on both

So 0 is a min

when x < 0, f ' (x) < 0
when x > 0, f ' (x) > 0
we have a local min at x = 0
|dw:1437349622635:dw|

the min is at x = 0
the actual min value itself is unknown at this point

plug x = 0 back into f(x) to find the min value

Why plug 0 back into f(x)?

oh my bad, I factored incorrectly

So would the second zero be a critical number?

x = 0 won't work
solve x^(-1) + 2 = 0 for x to get the actual only critical value that will work

So x = -1/2

yep, do another first derivative test to see if that is a local min, local max, or neither

Using 0 as well, or not?

x = -1/2 is the critical value now

you cannot use x = 0 because f ' (x) is undefined when x = 0

Would it be a min again?

yes

So, when x = -1/2 there is a local minimum

yeah

Can you help me find the points of inflection please?

you'll need to find f '' (x)

so differentiate f ' (x)

\[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] will be easier to work with

I got \[f''(x) = \frac{ 2(x-1) }{ x^\frac{ 5 }{ 3 } }\]

so it looks like we got the same things, just in slightly different forms

Yay!

now solve f '' (x) = 0 for x

that will give potential x coordinates of the point of inflection

to see if they actually are at the inflection point, you need to see if f '' changes sign

So \[0 = 2(x-1)\]

So x = 1 is the point of inflection?

it's at the possible point of inflection

don't use x = 0

I used -1 and 2... From decreasing to increasing I got

xD

Jeez, math is complicated

Thank you!

no problem