Find the x-coordinates of any relative extrema and inflection point(s) for the function f of x equals 9 times x raised to the one third power plus 9 halves times x raised to the four thirds power . You must justify your answer using an analysis of f ′(x) and f ′′(x).

- zeesbrat3

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- zeesbrat3

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

- zeesbrat3

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

- zeesbrat3

@ganeshie8

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## More answers

- zeesbrat3

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

- ganeshie8

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

- zeesbrat3

I'm confused...

- zeesbrat3

@jim_thompson5910 @amistre64

- zeesbrat3

Hi, Jim_thomspon5910

- jim_thompson5910

\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\]
\[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\]
\[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\]
\[\Large f \ '(x) = 3x^{1/3}\left(x^{-2} + 2\right)\]
so I'm getting the same f ' (x) as you did

- zeesbrat3

Then set it to 0, yes?

- jim_thompson5910

if you set f ' (x) equal to zero, and solve for x, you do get
\[\Large x = 0 \text{ or } x = \pm \frac{1}{\sqrt{-2}}\]
so I agree there as well

- jim_thompson5910

The portion \(\Large \pm\frac{1}{\sqrt{-2}}\) aren't real numbers, so we can ignore these two solutions

- jim_thompson5910

Now either do a first derivative test or a second derivative test to see if a local min or max is at x = 0

- zeesbrat3

The number line check?

- jim_thompson5910

yeah you can do that for a first derivative test

- jim_thompson5910

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

- zeesbrat3

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

- jim_thompson5910

correct on both

- zeesbrat3

So 0 is a min

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

I'm not thinking
if x = 0, then f ' (x) is undefined because of the x^(-2) portion. That leads to a division by zero error. So x = 0 cannot be a critical value.
The graph of f(x) also shows a vertical tangent at x = 0

- zeesbrat3

Why plug 0 back into f(x)?

- ganeshie8

\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\]
\[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\]
\[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\]
\[\Large f \ '(x) = 3x^{1/3}\left(x^{-\color{red}{1}} + 2\right)\]

- jim_thompson5910

oh my bad, I factored incorrectly

- zeesbrat3

So would the second zero be a critical number?

- jim_thompson5910

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

- zeesbrat3

So x = -1/2

- jim_thompson5910

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

- zeesbrat3

Using 0 as well, or not?

- jim_thompson5910

x = -1/2 is the critical value now

- jim_thompson5910

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

- zeesbrat3

Would it be a min again?

- jim_thompson5910

yes

- zeesbrat3

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

- jim_thompson5910

it says `Find the x-coordinates of any relative extrema `
they don't want the y coordinates, so you can stop for this part

- jim_thompson5910

yeah

- zeesbrat3

Can you help me find the points of inflection please?

- jim_thompson5910

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

- jim_thompson5910

so differentiate f ' (x)

- jim_thompson5910

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

- zeesbrat3

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

- jim_thompson5910

I'm getting
\[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\]
\[\Large f \ ''(x) = 3(-2/3)x^{-2/3-1} + 6(1/3)x^{1/3-1}\]
\[\Large f \ ''(x) = -2x^{-5/3} + 2x^{-2/3}\]

- jim_thompson5910

You can factor out -2x^(-5/3)
\[\Large f \ ''(x) = -2x^{-5/3} + 2x^{-2/3}\]
\[\Large f \ ''(x) = -2x^{-5/3}(1 - x)\]

- jim_thompson5910

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

- zeesbrat3

Yay!

- jim_thompson5910

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

- jim_thompson5910

that will give potential x coordinates of the point of inflection

- jim_thompson5910

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

- zeesbrat3

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

- zeesbrat3

@jim_thompson5910

- zeesbrat3

So x = 1 is the point of inflection?

- jim_thompson5910

it's at the possible point of inflection

- jim_thompson5910

set up a number line
then see if f '' changes sign for a value to the left of x = 1 and to the right of x = 1

- jim_thompson5910

don't use x = 0

- zeesbrat3

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

- jim_thompson5910

when x = 0.5, f '' (x) = -3.1748
when x = 2, f '' (x) = 0.62996
f '' (x) changes from negative to positive as we move through x = 1
so we definitely have an inflection point at x = 1

- zeesbrat3

xD

- jim_thompson5910

it turns out x = 0 is also at an inflection point. I'm reading now that if f '' (c) doesn't exist, then x = c could be a possible inflection point (assuming f(c) is defined and f '' changes sign)
http://www.sosmath.com/calculus/diff/der15/der15.html
since f '' (x) changes in sign going from say x = -0.5 to x = 0.5, this means that another inflection point is at x = 0

- zeesbrat3

Jeez, math is complicated

- jim_thompson5910

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- jim_thompson5910

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- zeesbrat3

Thank you!

- jim_thompson5910

no problem

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