Question

@jim_thompson5910

Answer 1

Is it where x = 6?

Answer 2

hmm, have you learned about integrals and areas under the curve yet?

Answer 3

if not, then I'll see if there's another way

Answer 4

no

Answer 5

well, I'm not sure. If so, I don't remember

Answer 6

one moment

Answer 7

k

Answer 8

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

Answer 9

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

Answer 10

what does area have to do with the absolute maximumt hough?

Answer 11

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?

Answer 12

the net change will help us figure out where the absolute max is

Answer 13

ah, I see where you're getting....The overall net change would be negative.

Answer 14

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

Answer 15

That makes sense. So the absolute maximum is at x = -2

Answer 16

sure we have an increasing interval on 5 < x < 6 but this increase is not enough to go higher than the starting point

Answer 17

yes at x = -2

Answer 18

as for how to do this without areas under the curve, I'm not sure

Answer 19

I think it is just logic in this situation, because there is just so much negative f '

Answer 20

I didn't even calculate the actual area things

Answer 21

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.

Answer 22

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

Answer 23

Yeah, that's actually sorta what I did @tkhunny

Answer 24

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.

Answer 25

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.