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determine all critical points 3x^2-96sqrt(x)

Mathematics
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hey buddies am waiting
This early in the morning there may not always be someone who can answer a Caclulus question so it may take some time to get an answer. That said, I can help. So, because I won't work the problem for you, let's see if we can figure it out. Do you recall what critical points are?
yes i do . i know i have to find the max n min values

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Okay, so in general what makes a point a critical point? In other words, how are they defined?
yes
So, tell me. How are critical points defined? Understanding that is key to working this problem and we need the definition to proceed to the next step. I'm sorry, but I'm not going to just work the problem for you. I want you to understand how to work these problems in general so you can do them again during, say, an exam.
ok
they are thos points in a fuction that the behaviour changes. For instance change in inflection, or moving from max to min or viceversa
Well, yes and no. Places where a function changes concavity (infection points) are not always critical points. Minimum and maximum points are critical points, but there are also critical points that are not minimum and maximums. So, think of it instead in terms of the function itself. For a general function, say f(x), critical points are defined to be the values of x for which the derivative, f ' (x) is zero or does not exist. So, we need the derivative of your function to find the critical points. What is the derivative of your function?
ok. makes sense
i knw how to find the derivative
Good, so what is the derivative? Or can you do the problem from here?
well the derivative will be y=6x-48
if i say thats equal to zero then x=8
Yes, except your derivative of the second term isn't quite correct. Isn't the derivative of \( \sqrt{x} \), \[ \frac{1}{2}x^{-\frac{1}{2}} \] ? That will change your deriviatve and your critical points.
I SHUD B FINE FROM HERE. thanks
Okay, glad I could be some help.

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