anonymous
  • anonymous
Anyone good at finding critical points???
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[y=3x^2-96\sqrt{x}\]
ChillOut
  • ChillOut
Take it's derivative y'. Do you know how?
anonymous
  • anonymous
I know i have to take y' but that's the thing, would it be \[6x-48(x)^{-1/2}\] ??

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

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