- anonymous

???

- schrodinger

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- welshfella

|dw:1440801196586:dw|

- anonymous

Yes thats it

- anonymous

I derived it and got f'(x) = 3x^(-2/3) + 6x^(1/3)

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## More answers

- welshfella

yes thats right

- welshfella

no you need to find the zeroes of this

- anonymous

Well this is what I have so farf'(x) = 3x^(-2/3) + 6x^(1/3)
f"(x) = -2x^(-5/3) + 2x^(-2/3)
f'(x)= -(1/2)
f"(-1/2) = 6*2^(2/3)

- anonymous

Am I on the right track?

- welshfella

the zero of f'(x) = -0.5

- welshfella

and its a minimum
yep

- anonymous

Do I need the last step that I did or is that unnecessary?

- welshfella

no you need the last step - the second derivative is positive indicating a minimum

- welshfella

The function might have some inflection points though ( Non horizontal ones). I cant remember how to find those to be honest!

- welshfella

I'll graph it to see.
Yes there is one close to x = 0

- anonymous

Okay so I solve the last step and then find the inflection points?

- welshfella

maybe cant remember I'd have to look that up

- anonymous

Okay

- welshfella

i think you equate f" to zero and solve

- anonymous

Thats it?

- welshfella

do you get x = 1 for that?

- welshfella

now you have to look for concavity each side of this point

- anonymous

Let me see

- welshfella

- no needn't do that They just want the x coordinate of the point of inflection

- welshfella

x = 0 will satisfy f" = 0 as well . That is the one i see on the graph.

- anonymous

So is it done as it?

- welshfella

i think so . there's a definite point of inflection at x =0 on the graph

- anonymous

Okay thank you!!!
So x=0 and x=-.5 are my answers?

- welshfella

cant see one at x = 1 though

- welshfella

-0.5 is a minimum
x = 0 is a point of inflection

- anonymous

Okay thank you!

- welshfella

yw

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