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Maximum/Minimum values of a graph on a restricted interval

  • Thread starter Julie H
  • Start date
  • #1
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Homework Statement


f(x) = x^3 + 12x^2 - 27x + 11
Absolute Maximum
Absolute Minimum
on the interval [-10,0]
(there are 3 different interval sets, but if I can do this one, I think I can figure out the rest.)

Homework Equations


Derivative, set equal to 0, then solve for the problem, but what I'm confused about is how the solving process differs as the interval changes.


The Attempt at a Solution


I have the derivative set as
3x^2 + 24x - 27

but what I'm unsure about is how finding the absolute maximum on the interval [-10,0] differs in process from finding the interval on, say, [-7, 2]

I think what I'm really trying to ask is how the restrictions on the intervals are reflected in the mathematical process to solve for an absolute maximum and minimum.
 

Answers and Replies

  • #2
LCKurtz
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The extremes of a continuous function on a closed interval must occur for values of x on the interval where one of the following is true:

1. f'(x) = 0
2. f'(x) fails to exist.
3. x is an end point of the closed interval.

You have to check all three cases, noting that values of x that aren't on the interval are irrelevant. Just list the possibilities and pick out the max and min.
 
  • #3
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edit: ah, someone just posted earlier delete some part of my post
 

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