Anyone good at finding critical points???

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Anyone good at finding critical points???

Calculus1
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\[y=3x^2-96\sqrt{x}\]
Take it's derivative y'. Do you know how?
I know i have to take y' but that's the thing, would it be \[6x-48(x)^{-1/2}\] ??

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Other answers:

Correct.
then?
umm
I'm confused :(
Now, what does the derivative tells you about your original function?
Don't i have to solve for the independent variable?
You have to solve f'(x)=0.
yeah that's where i got stuck, because i'm used to questions where i favtor (x+y)(x+y)
Factor*
Well, you just need to solve f'(x)=0 really. Those points can be extrema points
So \[6x-48(x)^{-1/2} = 0 ?\]
Yes.
It has only one solution.
Square the whole thing?
Might do. Try it
Well, I have to leave soon, do you want me to give you the x coordinate?
sure
x=4. It shouldn't be hard to find it.
okay i got it, thank you so much
Now you just plug it in the original function and find its y coordinate.
i plug the 4 instead of x?
nvm that was a dumb question, i got it haha thanks
y=-30?
-144**

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