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).

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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).

Mathematics
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\[f(x) = 9x^\frac{ 1 }{ 3 } + \frac{ 9 }{ 2 }x^\frac{ 4 }{ 3 }\]
I have that \[f'(x) = 3x^\frac{ 1 }{ 3 }[x ^{-2} + 2]\]

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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
\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\] \[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\] \[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] \[\Large f \ '(x) = 3x^{1/3}\left(x^{-2} + 2\right)\] so I'm getting the same f ' (x) as you did
Then set it to 0, yes?
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
The portion \(\Large \pm\frac{1}{\sqrt{-2}}\) aren't real numbers, so we can ignore these two solutions
Now either do a first derivative test or a second derivative test to see if a local min or max is at x = 0
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
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
Why plug 0 back into f(x)?
\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\] \[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\] \[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] \[\Large f \ '(x) = 3x^{1/3}\left(x^{-\color{red}{1}} + 2\right)\]
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
it says `Find the x-coordinates of any relative extrema ` they don't want the y coordinates, so you can stop for this part
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 \[f''(x) = \frac{ 2(x-1) }{ x^\frac{ 5 }{ 3 } }\]
I'm getting \[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] \[\Large f \ ''(x) = 3(-2/3)x^{-2/3-1} + 6(1/3)x^{1/3-1}\] \[\Large f \ ''(x) = -2x^{-5/3} + 2x^{-2/3}\]
You can factor out -2x^(-5/3) \[\Large f \ ''(x) = -2x^{-5/3} + 2x^{-2/3}\] \[\Large f \ ''(x) = -2x^{-5/3}(1 - x)\]
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
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
don't use x = 0
I used -1 and 2... From decreasing to increasing I got
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
xD
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
Jeez, math is complicated
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Thank you!
no problem

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