Question

Hi, can someone please explain how they would derive f(x)=x^(2/3)*(x^2-4)

Answer 1

you want to differentiate?
if so, use product rule first.

Answer 2

derive or derivate? your question is kind of vague!

Answer 3

Omg, everything I typed disappeared because I pressed the back button -.-

Answer 4

hate when that happens

Answer 5

\[f(x)=x^{2/3}(x^2-4)\]your \(f = x^{2/3}\) and your \(g=(x^2-4)\) so yes, you would use product rule as @freckles stated :)

Answer 6

\[\frac{ d }{ dx }(x^\frac{ 2 }{ 3 })(x^2-4) = \frac{ 2 }{ 3 }x^{(-\frac{ 1 }{ 3 })}(x^2-4)+x ^{2/3}(2x)\]

Answer 7

It just got so complicated, and I got lost at this part. :(

Answer 8

\[(fg)'=f'g+fg' \\ [(x^\frac{2}{3})(x^2-4)]'=(x^\frac{2}{3})'(x^2-4)+x^\frac{2}{3}(x^2-4)'\]

Answer 9

so first what are
\[(x^\frac{2}{3})' =? \\ (x^2-4)'=?\]

Answer 10

2x?

Answer 11

oh it loks like you already have the derivatives above

Answer 12

looks*

Answer 13

Yes I just dont know how to make it simpler

Answer 14

Im looking for the minimum and maximum values

Answer 15

\[\frac{ d }{ dx }(x^\frac{ 2 }{ 3 })(x^2-4) = \frac{ 2 }{ 3 }x^{(-\frac{ 1 }{ 3 })}(x^2-4)+x ^{2/3}(2x) \\ x^\frac{-1}{3} [ \frac{2}{3}(x^2-4)+x^\frac{3}{3}2x ]\]

Answer 16

that should help

Answer 17

Oh, then you want to change your function the have all positive values. So then, you can solve for your critical points and test them

Answer 18

Your local mins and max's can be found by the first derivative test, but first you need to find your crit points.

Answer 19

^ I was hoping to get that answer.

Answer 20

mmhmm. letme resolve it so you can see how it works.

Answer 21

\[ f(x)=x^{2/3}(x^2-4)\]\[f'(x) = \frac{2}{3}x^{-1/3}(x^2-4) +2x(x^{2/3})\]\[f'(x) = \frac{2(x^2-4)}{3x^{1/3}} +2x^{5/3}\]\[f'(x) = \frac{2(x^2-4)}{3x^{1/3}}+\frac{2x^{5/3}\cdot 3x^{1/3}}{3x^{1/3}}\]\[f'(x) = \frac{2x^2-8+6x^2}{3x^{1/3}}\]\[f'(x) = \frac{8x^2-8}{3x^{1/3}}\]\[f'(x)=\frac{8(x^2-1)}{3x^{1/3}}\]

Answer 22

You just kind of have to keep going with it.

Answer 23

Oh and \(x^{1/3} = \sqrt[3]{x}\)

Answer 24

Aaaaamzing!!!! Thank you!!!

Answer 25

Woooot!

Answer 26

Can you find your critical points and then use them to find your ma and mins? :)

Answer 27

Im just looking between [-1,2] so if x=1 then there the tangent line slop at that point is 0.

At 0.5 the slop is -2.5 and at 2 the slop is 6.3, so x=1 must be the minimum, and x=2 must be the maximum. When I plug these x values back into the original equation I get f(1)=-3 and f(2)=0

Answer 28

But what about f(0)? That would also equal to a minimum but f'(0) D.N.E....

Answer 29

you should have 3 critical numbers

Answer 30

what values did you get for critical numbers?

Answer 31

you find when f'=0 and when f' dne
the values that satisfy those equations where the x values actually exist for the original function are your critical numbers

Answer 32

\[\frac{ d }{ dx }(x^\frac{ 2 }{ 3 })(x^2-4) = \frac{ 2 }{ 3 }x^{(-\frac{ 1 }{ 3 })}(x^2-4)+x ^{2/3}(2x) \\ x^\frac{-1}{3} [ \frac{2}{3}(x^2-4)+x^\frac{3}{3}2x ] \\ \frac{1}{x^\frac{1}{3}} (\frac{2}{3}x^2-\frac{8}{3}+2x^2) \\ \frac{2}{x^\frac{1}{3}} (\frac{1}{3}x^2-\frac{4}{3}+x^2) \\ \frac{2}{x^\frac{1}{3}}(\frac{4}{3}x^2-\frac{4}{3}) \\ \frac{8}{3 x^\frac{1}{3}}(x^2-1)\]

so x^2-1=0 when x=? or x=?
x^(1/3)=0 when x=?

Answer 33

and yes I know @Jhannybean already put it into that form
I was just finishing the way I started earlier :)

Answer 34

:P Of course. Both methods work!

Answer 35

\[f'(x)=\frac{8(x^2-1)}{3x^{1/3}}\]\[8(x^2-1)=0\]\[3x^{1/3}=0\]These should give you two values, 2 crit points from the first function, and the same for the bottom.

Answer 36

Yes so x=1 and x=-1 are the minima? and x=2 is the maximum?

Answer 37

Can you tell me where the x=2 came from?

Answer 38

ok but x^(1/3)=0 when x=?

Answer 39

Im solving for the domain section of [-1,2]

Answer 40

what number raised to any power (not including 0) will give you 0?

Answer 41

To my knowledge only zero....

Answer 42

so critical numbers are -1,1,0
endpoints are -1,2
---
now only local extrema can happen at non-endpoint numbers

Answer 43

Good, and if you are evaluating at those end points then you would not only test your critical points but those end points as well to mind your maximum/minimum values, which I believe you did?

Answer 44

Yes, i tested for -1 and 2.

2 was a maximum endpoint

Answer 45

doesn't f(2)=f(0)?

Answer 46

This is interesting.

Answer 47

Is it right to say \[x^\frac{ 2 }{ 3 } = x^\frac{ 1 }{ 3 }\times x^ \frac{ 1 }{ 3 }\]

Answer 48

\[\text{ yes you can say that } \text{ or } (x^\frac{1}{3})^2 \]

Answer 49

Mmhmm. Because when functions multiply, powers add.

Answer 50

I just want to say you have an absolute max value occurring at two numbers not just one

Answer 51

So at x= 0, 2?

Answer 52

yes since f(0)=f(2)=0

Answer 53

The interesting thin is the screenshot i posted. Its from wolfram alpha
-1^0.6666 = -1

Answer 54

But \[-1^{\frac{ 2 }{ 3 }}= 1\]

Answer 55

Because you're squaring (-1) and then taking it's cube root

Answer 56

Which is simply \(1\)

Answer 57

\[(-1)^\frac{2}{3}=(-1)^2=1 \\ -1^\frac{2}{3}=-1\]

Answer 58

Yes, I understand.

Answer 59

and wolfram's graph I have seen doesn't always agree with the graphs we make

Answer 60

if that makes sense

Answer 61

Math is fun like that haha. I would like to think we are right and the computer at wolfram is wrong :D

Answer 62

But I'm guessing there is more too it