A community for students.
Here's the question you clicked on:
 0 viewing
burhan101
 2 years ago
Determine the extreme values of this function on the given interval.
f(x)= 1+ (x+3)² ; 2<x<6
burhan101
 2 years ago
Determine the extreme values of this function on the given interval. f(x)= 1+ (x+3)² ; 2<x<6

This Question is Closed

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0Calc or precalc?

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0precal @SithsAndGiggles

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, that method's a bit less work. So you're given a parabola in vertex form: \[f(x)=1+(x+3)^2\] \(f(x)\) has vertex (an extreme value point) at \((3,1)\). \(3\) is not in the interval \(2<x<6\). So, you have something that looks like this: dw:1371696732887:dw This means that only the endpoints of this interval will contain the extreme values.

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.0Which means you have only to check \(f(2)\) and \(f(6)\).

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0where did you get (3,1) from ? :S
Ask your own question
Sign UpFind more explanations on OpenStudy
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.