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OpenStudy (anonymous):

x^3sqrt(x-9) find critical numbers

OpenStudy (anonymous):

did you find the derivative?

OpenStudy (anonymous):

I think is x(x-9)^1/3 I was wondering if we can go step by step

OpenStudy (anonymous):

then I got it (1/3)x(x-9)^-2/3

OpenStudy (anonymous):

hmm i am thinking it is \[f(x)=x^3\sqrt{x-9}\] but maybe it is something else

OpenStudy (anonymous):

is it \[f(x)=x\sqrt[3]{x-9}\]?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

second one?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok so not x cubed but the cubed root got it you need the product rule for this because it is a product

OpenStudy (anonymous):

so your derivative is wrong, looks like you just left the \(x\) there hanging about

OpenStudy (anonymous):

\[\frac{ 1 }{ 3 }x(x-9)^{-2/3}\]

OpenStudy (anonymous):

yeah that is part of it

OpenStudy (anonymous):

\((fg)'=f'g+g'f\) is the product rule you have \(x\sqrt[3]{x-9}\) so one part of the derivative is \[\frac{1}{3}x(x-9)^{-2/3}\] but also you have at add \[\sqrt[3]{x-9}\]

OpenStudy (anonymous):

and I think then is \[(x-9)^{1/3}\]

OpenStudy (anonymous):

right, but now you have to set it equal to zero and solve, so you need to get it out of that exponential notation

OpenStudy (anonymous):

I have \[\frac{ 1 }{ 3 }x(x-9)^{-2/3}+(x-9)^{1/3} but then I get lost from there..\]

OpenStudy (anonymous):

lol add

OpenStudy (anonymous):

what you got is \[\frac{x}{3\sqrt[3]{x-9}}+\sqrt[3]{x-9}\] you got to add them up

OpenStudy (anonymous):

Ahh ok..sorry this is my first time using openstudy a friend recommended it me

OpenStudy (anonymous):

exponential notation is great for taking derivatives but not so good for solving equations

OpenStudy (anonymous):

you are doing fine, it is going to be algebra from here on in

OpenStudy (anonymous):

you got how to add them up, or no?

OpenStudy (anonymous):

I think so 4x-27\[\frac{ 4x-27 }{ 3(x-9)^{2/3} }\]

OpenStudy (anonymous):

wow ok then two steps and you are done

OpenStudy (anonymous):

critical points are where the derivative is 0, set \(4x-27=0\) and solve, and also where the derivative is undefined, set \(x-9=0\) and solve

OpenStudy (anonymous):

Wow thanks alot I got it now:)

OpenStudy (anonymous):

yw but honestly you did all the work!

OpenStudy (anonymous):

thank you:) I was wondering is this service free or do I have to pay after asking a certain amount of questions?

OpenStudy (aum):

Free.

OpenStudy (anonymous):

cool.

OpenStudy (aum):

Ask away :) But close the question after it has been satisfactorily answered and post the next question separately.

OpenStudy (anonymous):

really? i get $10 per answer

OpenStudy (anonymous):

$20 for correct ones

OpenStudy (aum):

LQTM

OpenStudy (aum):

Laughing Quietly to Myself.

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