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anonymous
 5 years ago
determine all critical points
3x^296sqrt(x)
anonymous
 5 years ago
determine all critical points 3x^296sqrt(x)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey buddies am waiting

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i do . i know i have to find the max n min values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, so in general what makes a point a critical point? In other words, how are they defined?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they are thos points in a fuction that the behaviour changes. For instance change in inflection, or moving from max to min or viceversa

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i knw how to find the derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good, so what is the derivative? Or can you do the problem from here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well the derivative will be y=6x48

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if i say thats equal to zero then x=8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I SHUD B FINE FROM HERE. thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, glad I could be some help.
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