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Yes thats it
I derived it and got f'(x) = 3x^(-2/3) + 6x^(1/3)
yes thats right
no you need to find the zeroes of this
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)
Am I on the right track?
the zero of f'(x) = -0.5
and its a minimum yep
Do I need the last step that I did or is that unnecessary?
no you need the last step - the second derivative is positive indicating a minimum
The function might have some inflection points though ( Non horizontal ones). I cant remember how to find those to be honest!
I'll graph it to see. Yes there is one close to x = 0
Okay so I solve the last step and then find the inflection points?
maybe cant remember I'd have to look that up
i think you equate f" to zero and solve
do you get x = 1 for that?
now you have to look for concavity each side of this point
Let me see
- no needn't do that They just want the x coordinate of the point of inflection
x = 0 will satisfy f" = 0 as well . That is the one i see on the graph.
So is it done as it?
i think so . there's a definite point of inflection at x =0 on the graph
Okay thank you!!! So x=0 and x=-.5 are my answers?
cant see one at x = 1 though
-0.5 is a minimum x = 0 is a point of inflection
Okay thank you!