Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
if \[f(x)=\frac{ 1 }{ 3 }x ^{3}4x ^{2}+12x5\] and the domain is the set of all x such that \[0\le x \le9\], then the absolut emaximum value of the function f occurs when x is.....
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
my teacher gave me the answer, and the answer is 9.
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
and haha yes i figured out how to do the last one :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
So we're closed within a small interval. To find max and min, we'll do a couple things. We'll take the derivative of \(f\) to find critical points. Then we simply plug those points into the original function to see which gives us the biggest answer. But we ALSO have to plug in the END POINTS x=0 and x=9, and compare those values as well. The biggest output will be our answer.
 one year ago

mlddmlnog Group TitleBest ResponseYou've already chosen the best response.0
ah , ok. because I found the derivative, and 9 was not one of the values i got when i set the derivative equal to zero and solved for x. but thank you! I didnt know that you have to include the endpoints :p
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
ya c; good times.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.