x^3sqrt(x-9) find critical numbers
did you find the derivative?
I think is x(x-9)^1/3 I was wondering if we can go step by step
then I got it (1/3)x(x-9)^-2/3
hmm i am thinking it is \[f(x)=x^3\sqrt{x-9}\] but maybe it is something else
is it \[f(x)=x\sqrt[3]{x-9}\]?
yes
second one?
yes
ok so not x cubed but the cubed root got it you need the product rule for this because it is a product
so your derivative is wrong, looks like you just left the \(x\) there hanging about
\[\frac{ 1 }{ 3 }x(x-9)^{-2/3}\]
yeah that is part of it
\((fg)'=f'g+g'f\) 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 \[\sqrt[3]{x-9}\]
and I think then is \[(x-9)^{1/3}\]
right, but now you have to set it equal to zero and solve, so you need to get it out of that exponential notation
I have \[\frac{ 1 }{ 3 }x(x-9)^{-2/3}+(x-9)^{1/3} but then I get lost from there..\]
lol add
what you got is \[\frac{x}{3\sqrt[3]{x-9}}+\sqrt[3]{x-9}\] you got to add them up
Ahh ok..sorry this is my first time using openstudy a friend recommended it me
exponential notation is great for taking derivatives but not so good for solving equations
you are doing fine, it is going to be algebra from here on in
you got how to add them up, or no?
I think so 4x-27\[\frac{ 4x-27 }{ 3(x-9)^{2/3} }\]
wow ok then two steps and you are done
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
Wow thanks alot I got it now:)
yw but honestly you did all the work!
thank you:) I was wondering is this service free or do I have to pay after asking a certain amount of questions?
Free.
cool.
Ask away :) But close the question after it has been satisfactorily answered and post the next question separately.
really? i get $10 per answer
$20 for correct ones
LQTM
Laughing Quietly to Myself.
Join our real-time social learning platform and learn together with your friends!