Local Maximum-Minimum values

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Homework Statement


http://img4.imageshack.us/img4/8899/85779053.jpg [Broken]

Homework Equations





The Attempt at a Solution


could you please check my answers.. im pretty sure they are good but i just want to be sure.
 
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Answers and Replies

  • #2
lanedance
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i'm not so sure about number 4
 
  • #3
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i'm not so sure about number 4

i dunno i think it should be false. i tried putting true though and it didnt work. any other ideas?
 
  • #4
i dunno i think it should be false. i tried putting true though and it didnt work. any other ideas?

Statement 4 is correct, being "false."

If f(x) is on [a,b] and there is an absolute min at a or b, it does not mean f'(a)=0 or f'(b)=0.

Statement 5 is wrong. It should be "true."

Critical points are where f'(x)=0 and where f'(x) does not exist.

Let f(x)=x3 then f'(x)=3x2

f'(x)=0 at x=0. Therfore x=0 is a critical point. But is not a max or min.

A local extremum must occur at a critical point, but opposite is not true.

Image02.gif
 
  • #5
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it still shows me that its wrong.
i got everything 'true' but the 4th one which is 'false', and it doesnt work..
im sure 100% about the first 3 because they are straight from the textbook..
any ideas?
 
  • #6
Oh, I believe the last one should also be false.

I line with no extrema from x=a to x=b will have absolute min and max at the endpoints, being a and b. Just because it is continuous does not mean it has relative extrema at c and d between the endpoints a and b.
 
  • #7
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Oh, I believe the last one should also be false.

I line with no extrema from x=a to x=b will have absolute min and max at the endpoints, being a and b. Just because it is continuous does not mean it has relative extrema at c and d between the endpoints a and b.

nope..
i tried this:
true, true, true, false, true
true, true, true, false, false
true, true, true, true, true
true, true, true, true, false

all wrong. so it must be one of the first 3.. but they must be 100% right cuz they are from the text book..
 
  • #8
Dick
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Oh, I believe the last one should also be false.

I line with no extrema from x=a to x=b will have absolute min and max at the endpoints, being a and b. Just because it is continuous does not mean it has relative extrema at c and d between the endpoints a and b.

False? The question says c and d in [a,b]. Not in (a,b).
 
  • #9
False? The question says c and d in [a,b]. Not in (a,b).

I don't understand your argument. a and b are included in the interval. Therefore x=a and x=b are in the domain. Are they not?
 
  • #10
Dick
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I don't understand your argument. a and b are included in the interval. Therefore x=a and x=b are in the domain. Are they not?

Yes, a and b are in the domain, otherwise it wouldn't be true. They are also candidates to be c and d. It just says c and d are in [a,b], not (a,b). How could a continuous function on a closed interval not have a max and a min? Can you give me an example?
 
  • #11
Maybe I am reading the statement wrong.

I do not doubt that there will be a max or min. Just as you said, there has to be a max and min on [a,b]. But when the statement says that "f attains a max f(c) and min f(d) at some numbers c and d in [a,b]," I guess I am assuming that they have to have a max or min at c and d. I am also assuming c and d cannot equal a and b.

I say false because a straight line with say slope 3, would be continuous, but would have max and min at the endpoints, not between a and b, for c and d not equal to a and b.

Lots of letters being tossed around. I hope I'm not yapping in circles.
 
  • #12
Dick
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It says c and d are in [a,b]. I'm pretty sure that as it's stated they allow the case of c or d being endpoints.
 
  • #13
But aren't a and be the endpoints?

So if it said c and d were endpoints, then it would be [c,d].

I don't like that statement.
 
  • #14
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the thing is, false or true, looks like this isnt the problem anyways.. it must be the first 2.. which one could be wrong?
 
  • #15
Dick
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the thing is, false or true, looks like this isnt the problem anyways.. it must be the first 2.. which one could be wrong?

2 is wrong. Can you explain why? Suppose f is continuous?
 
  • #16
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wel because an abs.value is the highest point on the graph and it has to have a curve the it continues, otherwise it wouldnt be abs.value...

so right now i have everything set on 'true' and the 3rd and 4th are 'false' and its wrong.. :(
 
  • #17
Dick
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You seem to be just flipping back and forth without much discussion of why you think any particular line is true or false. There's 2^6=64 possible ways to frame an answer here. You probably aren't going hit on the correct combination by guessing. Or listening to opinions.
 
  • #18
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what do u mean.. i read every single comment and explained my point of view for some questions.
okay for the 3rd one,
" If f has a local maximum or minimum at c, then c is a critical number of f. "
it must be 'false' because critical number is the absolute max/min value isnt it?
 
  • #19
Dick
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I would define a critical number as a point where the derivative of f is zero or the derivative doesn't exist. What's your definition? That definitely doesn't mean it's automatically an absolute max/min.
 
  • #20
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I would define a critical number as a point where the derivative of f is zero or the derivative doesn't exist. What's your definition? That definitely doesn't mean it's automatically an absolute max/min.

well there is pretty much only one def and u gave it already..
i just didnt think about it for some reason.

that would mean that " If f has a local maximum or minimum at c, then c is a critical number of f. " is true because when there is a local maximum or minimum, its going to be 0 when taking the derivative.
am i right?

and that would pretty much be the same for "If c is not a critical point of f, then c is not a local minimum of f. " - true because if the derivative isnt 0 (or undefined) then there wouldnt be a local minimum (or anything else)
 
  • #21
Dick
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Sort of. But if c is NOT a critical point then the derivative IS defined and not equal to zero, so it's not a local max or min. You can have a critical point where the derivative is not defined, like f(x)=|x| when c=0.
 
  • #22
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Sort of. But if c is NOT a critical point then the derivative IS defined and not equal to zero, so it's not a local max or min. You can have a critical point where the derivative is not defined, like f(x)=|x| when c=0.

yeah thats what i meant.. sorry.. i was just rushing when i typed :b
i think i got it. thanks a lot for ur help buddy!
 

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