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zeesbrat3
 one year ago
Find the xcoordinates 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
 one year ago
Find the xcoordinates 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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zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x) = 9x^\frac{ 1 }{ 3 } + \frac{ 9 }{ 2 }x^\frac{ 4 }{ 3 }\]

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1I have that \[f'(x) = 3x^\frac{ 1 }{ 3 }[x ^{2} + 2]\]

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So, I set f'(x) = 0 and I got \[x = 0, \pm \frac{ 1 }{ \sqrt{2} }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=%289x%5E%281%2F3%29+%2B+9%2F2*x%5E%284%2F3%29%29%27

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1@jim_thompson5910 @amistre64

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Hi, Jim_thomspon5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\] \[\Large f \ '(x) = 9(1/3)x^{1/31} + \frac{9}{2}*\frac{4}{3}x^{4/31}\] \[\Large f \ '(x) = 3x^{2/3} + 6x^{1/3}\] \[\Large f \ '(x) = 3x^{1/3}\left(x^{2} + 2\right)\] so I'm getting the same f ' (x) as you did

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Then set it to 0, yes?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.2The portion \(\Large \pm\frac{1}{\sqrt{2}}\) aren't real numbers, so we can ignore these two solutions

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Now either do a first derivative test or a second derivative test to see if a local min or max is at x = 0

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1The number line check?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yeah you can do that for a first derivative test

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2see if f ' (x) changes sign for a value less than 0, and for a value greater than 0

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1when it is less than 0, it is decreasing, when it is greater than 0, it is increasing

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2correct on both

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2when x < 0, f ' (x) < 0 when x > 0, f ' (x) > 0 we have a local min at x = 0 dw:1437349622635:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2the min is at x = 0 the actual min value itself is unknown at this point

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2plug x = 0 back into f(x) to find the min value

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I'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
 one year ago
Best ResponseYou've already chosen the best response.1Why plug 0 back into f(x)?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\] \[\Large f \ '(x) = 9(1/3)x^{1/31} + \frac{9}{2}*\frac{4}{3}x^{4/31}\] \[\Large f \ '(x) = 3x^{2/3} + 6x^{1/3}\] \[\Large f \ '(x) = 3x^{1/3}\left(x^{\color{red}{1}} + 2\right)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2oh my bad, I factored incorrectly

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So would the second zero be a critical number?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2x = 0 won't work solve x^(1) + 2 = 0 for x to get the actual only critical value that will work

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yep, do another first derivative test to see if that is a local min, local max, or neither

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Using 0 as well, or not?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2x = 1/2 is the critical value now

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you cannot use x = 0 because f ' (x) is undefined when x = 0

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Would it be a min again?

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So, when x = 1/2 there is a local minimum

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2it says `Find the xcoordinates of any relative extrema ` they don't want the y coordinates, so you can stop for this part

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1Can you help me find the points of inflection please?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you'll need to find f '' (x)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so differentiate f ' (x)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large f \ '(x) = 3x^{2/3} + 6x^{1/3}\] will be easier to work with

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1I got \[f''(x) = \frac{ 2(x1) }{ x^\frac{ 5 }{ 3 } }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I'm getting \[\Large f \ '(x) = 3x^{2/3} + 6x^{1/3}\] \[\Large f \ ''(x) = 3(2/3)x^{2/31} + 6(1/3)x^{1/31}\] \[\Large f \ ''(x) = 2x^{5/3} + 2x^{2/3}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2You 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
 one year ago
Best ResponseYou've already chosen the best response.2so it looks like we got the same things, just in slightly different forms

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2now solve f '' (x) = 0 for x

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2that will give potential x coordinates of the point of inflection

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2to see if they actually are at the inflection point, you need to see if f '' changes sign

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1So x = 1 is the point of inflection?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2it's at the possible point of inflection

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2set 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
 one year ago
Best ResponseYou've already chosen the best response.2don't use x = 0

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.1I used 1 and 2... From decreasing to increasing I got

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2when 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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.1Jeez, math is complicated
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