Question

Anyone good at finding critical points???

Answer 1

\[y=3x^2-96\sqrt{x}\]

Answer 2

Take it's derivative y'. Do you know how?

Answer 3

I know i have to take y' but that's the thing, would it be
\[6x-48(x)^{-1/2}\]
??

Answer 4

Correct.

Answer 5

then?

Answer 6

umm

Answer 7

I'm confused :(

Answer 8

Now, what does the derivative tells you about your original function?

Answer 9

Don't i have to solve for the independent variable?

Answer 10

You have to solve f'(x)=0.

Answer 11

yeah that's where i got stuck, because i'm used to questions where i favtor (x+y)(x+y)

Answer 12

Factor*

Answer 13

Well, you just need to solve f'(x)=0 really. Those points can be extrema points

Answer 14

So \[6x-48(x)^{-1/2} = 0 ?\]

Answer 15

Yes.

Answer 16

It has only one solution.

Answer 17

Square the whole thing?

Answer 18

Might do. Try it

Answer 19

Well, I have to leave soon, do you want me to give you the x coordinate?

Answer 20

sure

Answer 21

x=4. It shouldn't be hard to find it.

Answer 22

okay i got it, thank you so much

Answer 23

Now you just plug it in the original function and find its y coordinate.

Answer 24

i plug the 4 instead of x?

Answer 25

nvm that was a dumb question, i got it haha thanks

Answer 26

y=-30?

Answer 27

-144**