StudyGurl14
  • StudyGurl14
@jim_thompson5910
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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StudyGurl14
  • StudyGurl14
Is it where x = 6?
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jim_thompson5910
  • jim_thompson5910
hmm, have you learned about integrals and areas under the curve yet?
jim_thompson5910
  • jim_thompson5910
if not, then I'll see if there's another way

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StudyGurl14
  • StudyGurl14
no
StudyGurl14
  • StudyGurl14
well, I'm not sure. If so, I don't remember
jim_thompson5910
  • jim_thompson5910
one moment
StudyGurl14
  • StudyGurl14
k
jim_thompson5910
  • jim_thompson5910
ok if you were to try to approximate the area under the curve, one way would be to draw in figures like this (trapezoids and triangles) see attached
jim_thompson5910
  • jim_thompson5910
you can use the area of a triangle formula A = b*h/2 and the area of a trapezoid formula A = h*(b1+b2)/2
StudyGurl14
  • StudyGurl14
what does area have to do with the absolute maximumt hough?
jim_thompson5910
  • jim_thompson5910
the area represents the net change notice how much of the areas are below the x axis. This means we have negative net change (ie the function is decreasing) the only area that is above the x axis is the pink triangle on the very right the question is: is the overall net change positive? or is it negative?
jim_thompson5910
  • jim_thompson5910
the net change will help us figure out where the absolute max is
StudyGurl14
  • StudyGurl14
ah, I see where you're getting....The overall net change would be negative.
jim_thompson5910
  • jim_thompson5910
yes, so this means that we start up somewhere high, we decrease, then come back up (like described in the last post) the overall net change is negative, so our starting point is higher than our ending point this means that the starting point is our abs max
StudyGurl14
  • StudyGurl14
That makes sense. So the absolute maximum is at x = -2
jim_thompson5910
  • jim_thompson5910
sure we have an increasing interval on 5 < x < 6 but this increase is not enough to go higher than the starting point
jim_thompson5910
  • jim_thompson5910
yes at x = -2
jim_thompson5910
  • jim_thompson5910
as for how to do this without areas under the curve, I'm not sure
StudyGurl14
  • StudyGurl14
I think it is just logic in this situation, because there is just so much negative f '
StudyGurl14
  • StudyGurl14
I didn't even calculate the actual area things
tkhunny
  • tkhunny
Might want to try a mean value. On [-2,0], average decrease is about -2.5, giving an approximate net decrease of -5 on [-2,0] On [0,2], average decrease is about -1.0, giving an approximate net decrease of -2 on [0,2]. Total decrease is now approximately -7 Continue on this way and see if it ever gets higher than where it started. Note: This is the same method proposed by Jim Thompson, just slightly restated.
jim_thompson5910
  • jim_thompson5910
I guess you could make the observation that much of f ' is under the x axis, so f spends much of its time decreasing. That's more of a qualitative answer than anything
StudyGurl14
  • StudyGurl14
Yeah, that's actually sorta what I did @tkhunny
tkhunny
  • tkhunny
Well, apparently you did not believe it. Try it with the "area things" and see if you get the same result. It will be a useful exploration.
tkhunny
  • tkhunny
Also, you are required to state the Absolute Maximum. What we have stated will help you find WHERE the absolute maximum is, but not WHAT it is.

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