f(x) graph based on derivative graph help

OnMyWayToBeAMathProffesor · #1 $Old$ December 31st, 2008, 10:32 PM

Hello MHF,

I am stuck on part c of this question and am not 100% sure on a and b either. There is a table involved. It is attached.

x.doc

a) What are the x-coordinates of all absolute maximum and absolute minimum points of $f$ on the interval $\left[-3, 3\right]$ . Justify your answer.

b)What are the x-coordinates of all points of inflection of $f$ on the interval $\left[-3, 3\right]$ ? Justify your answer.

c) Sketch a graph that satisfies the given properties of $f$ .

a-answer) i got $x=1$ for a maximum because the $f'(x)$ goes from negative y values to 0 back to negative y values. is this correct?

b-answer) point of inflections occurs when $f''(x)=0$ and thats and $x=1$ , I don't think there are any other points, correct?

c-answer) i could never figure out how to do that on MHF but i have seen other people do it. All my graphs don't seem to work, but from the chart i think there are 2 separate functions?

any help, especially with c would be greatly appreciated. Thanks.

Jhevon · #2 $Old$ December 31st, 2008, 11:40 PM

Quote:

Originally Posted by OnMyWayToBeAMathProffesor $View Post$

Hello MHF,

I am stuck on part c of this question and am not 100% sure on a and b either. There is a table involved. It is attached.

Attachment 9426

a) What are the x-coordinates of all absolute maximum and absolute minimum points of $f$ on the interval $\left[-3, 3\right]$ . Justify your answer.

b)What are the x-coordinates of all points of inflection of $f$ on the interval $\left[-3, 3\right]$ ? Justify your answer.

c) Sketch a graph that satisfies the given properties of $f$ .

a-answer) i got $x=1$ for a maximum because the $f'(x)$ goes from negative y values to 0 back to negative y values. is this correct?

b-answer) point of inflections occurs when $f''(x)=0$ and thats and $x=1$ , I don't think there are any other points, correct?

c-answer) i could never figure out how to do that on MHF but i have seen other people do it. All my graphs don't seem to work, but from the chart i think there are 2 separate functions?

any help, especially with c would be greatly appreciated. Thanks.

something confuses me about the table. you have two possibilities for x = 1. the derivatives can be zero or undefined. which is it?

for (c), does this help?
- where the first derivative is positive, the graph is increasing. (going up as you move from left to right)
- where the first derivative is negative, the graph is decreasing. (going down as you move from left to right).
- where the first derivative is zero or undefined, you have a critical point. these can be maximums, minimums, inflection points, etc.

- where the second derivative is positive, the graph is concave up (curved like a U or a smile)
- where the second derivative is negative, the graph is concave down (curved like a $\cap$ or a frown)
- if the second derivative is zero, you have a possible inflection point. you can test if this is the case. an inflection occurs if the second derivative changes sign on either side of the point you are concerned with

so drawing a graph based on the table means you have to follow the behavior described by these properties where appropriate

Grandad · #3 $Old$ December 31st, 2008, 11:41 PM

Hello OnMyWayToBeAMathProffesor

First, note that your answers to a and b are contradictory: you can't have a point of inflexion that's also a maximum. So, let's review what you need in order to determine which is which:

At a maximum or minimum turning point, $f'(x) =0$
At a maximum, $f'(x)$ goes from being positive on the left of the turning point to being negative on the right of it. So $f'(x)$ is decreasing, and hence $f''(x)$ is negative. And conversely...
At a minimum, $f''(x)$ is positive.
At a point of inflexion $f''(x)=0$

So, from the table, the only point at which $f'(x) =0$ is where $x=1$ , and at this point $f''(x)=0$ . So:

a) There are no maximum or minimum points in this interval

b) There is a point of inflexion where $x =1$

c) As far as the sketch is concerned:

For $-3<x<-1$ , the gradient of the graph is positive, and it's increasing (because both $f'$ and $f''$ are positive), and tends to infinity as $x$ tends to $-1$ . So there's going to be an asymptote at $x=-1$ .
For $-1<x<1$ , gradient is negative but the graph is levelling out since $f''$ is positive and hence the gradient is increasing (i.e. becoming less negative)
At $x=1$ , the graph is horizontal, but the gradient starts to go negative again after this point - hence a point of inflexion.
For $x>1$ , the gradient continues to decrease, and the graph therefore becomes steeper.

You should be able to sketch the graph now. (The way I would do it on MHF is to create a .jpg either using a drawing package or scanning a hand-drawn graph, and attach it using the 'Manage Attachments' button.)

Hope that helps.

Grandad

kalagota · #4 $Old$ December 31st, 2008, 11:48 PM

Quote:

Originally Posted by OnMyWayToBeAMathProffesor $View Post$

a-answer) i got $x=1$ for a maximum because the $f'(x)$ goes from negative y values to 0 back to negative y values. is this correct?

no. let $\varepsilon >0$ . suppose $f(a)$ exists. $(a,f(a))$ is a maximum point (minimum point) if
1. $f'(a)=0$ or $f'(a)$ does not exist
2. $f'(a-\varepsilon)> 0$ and $f'(a+\varepsilon)< 0$ ( $f'(a-\varepsilon)< 0$ and $f'(a+\varepsilon)> 0$ ).

Quote:

Originally Posted by OnMyWayToBeAMathProffesor $View Post$

b-answer) point of inflections occurs when $f''(x)=0$ ... correct?

no.. POI occurs at $a$ if $f(a)$ exists and $f''(a-\varepsilon)f''(a+\varepsilon)<0$

it doesn't really mean that if $f''(x)=0$ (or dne), then it is already a POI.

Quote:

Originally Posted by OnMyWayToBeAMathProffesor $View Post$

c-answer) i could never figure out how to do that on MHF but i have seen other people do it. All my graphs don't seem to work, but from the chart i think there are 2 separate functions?

any help, especially with c would be greatly appreciated. Thanks.

depending on the graph that you will make, either it will consists of one part or two part since there is no condition on $f(-1)$ .. however, if it exists, then the graph must be of one part only; otherwise, the graph is of 2 parts..

OnMyWayToBeAMathProffesor · #5 $Old$ December 31st, 2008, 11:49 PM

i am so sorry about the chart, i fixed the mistake now. Thank you Jhevon for pointing it out.

kalagota · #6 $Old$ January 1st, 2009, 12:04 AM

Quote:

Originally Posted by Grandad $View Post$

a) There are no maximum or minimum points in this interval

...

c) As far as the sketch is concerned:

For $-3<x<-1$ , the gradient of the graph is positive, and it's increasing (because both $f'$ and $f''$ are positive), and tends to infinity as $x$ tends to $-1$ . So there's going to be an asymptote at $x=-1$ .
For $-1<x<1$ , gradient is negative but the graph is levelling out since $f''$ is positive and hence the gradient is increasing (i.e. becoming less negative)

....

hmm, there is a possibility that at $x=-1$ , it is not asymptote. note that if $f(-1)$ exists, then the point $(-1,f(-1))$ is a pointed one (or is that a cusp?); thus if $f(-1)$ exists, then the point $(-1,f(-1))$ is a maximum point on this interval..

OnMyWayToBeAMathProffesor · #7 $Old$ January 1st, 2009, 12:05 AM

so this is the graph i got from the rules and tips you guys said. I hope you can see it.

$f-x-graph-based-derivative-graph-help-x1.bmp$

and thanks grandad for correcting my a).

kalagota · #8 $Old$ January 1st, 2009, 12:08 AM

yes, that one is a possible graph, and if you read my comments, there is another possible graph..

OnMyWayToBeAMathProffesor · #9 $Old$ January 1st, 2009, 12:18 AM

so i made my previous graph bigger to view here:

x2.doc

at this is the graph where f(-1) exists

x31.doc

are both graphs good? do they both work?

kalagota · #10 $Old$ January 1st, 2009, 12:21 AM

yes. these are the two possible graphs..

Grandad · #11 $Old$ January 1st, 2009, 06:25 AM

Hello everyone -

My feeling here is that for the most part, many of these posts are making things too complicated. Obviously, we're not told the level of the course on which this question appears, but as a simple set of tests for max, min and points of inflexion, I would re-iterate my original post. To introduce $\epsilon > 0$ into the discussion is probably not terribly helpful.

Also, note that in the original question, part (c) only asks for a graph that satisfies these conditions - not a full-blown discussion on all possible graphs. The most obvious example is that the curve is asymptotic to $x = -1$ .

As to the table showing two possible values at $x = 1$ , I thought that this was so obviously a typo that I didn't think it was worth passing comment.

Grandad

OnMyWayToBeAMathProffesor · #12 $Old$ January 1st, 2009, 01:23 PM

I am sorry, I should have said I am an AP Calc-AB student.

I have to admit, I wasn't entirely sure what $\epsilon$ was but made a graph with a cusp following what kalagota said. I will include both graphs to show both possibilities. Thank you everyone.

kalagota · #13 $Old$ January 1st, 2009, 10:23 PM

Quote:

Originally Posted by Grandad $View Post$

Hello everyone -

My feeling here is that for the most part, many of these posts are making things too complicated. Obviously, we're not told the level of the course on which this question appears, but as a simple set of tests for max, min and points of inflexion, I would re-iterate my original post. To introduce $\epsilon > 0$ into the discussion is probably not terribly helpful.

there is nothing special about the introduction of $\varepsilon > 0$ except for the fact that it is just a symbol for a positive number, in this case, nothing more, nothing less. so i don't think it is not helpful.

Quote:

Originally Posted by Grandad $View Post$

Also, note that in the original question, part (c) only asks for a graph that satisfies these conditions - not a full-blown discussion on all possible graphs. The most obvious example is that the curve is asymptotic to $x = -1$ .

of course, nothing is wrong about telling all the possible graphs. only that, if the teacher presented one of these possible graphs and the student drew the other one, the student might have confusions on what had gone wrong in his work. so before that happens, it would be better if the student knows has drawn both..

and talking about the "obvious", this is subjective, because for me, the obvious is the one with the cusp. but because i also know the your obvious is also a possible graph, i indicated on my post there are two possible graphs depending on another condition, (i.e. whether the function is continuous or not.) but since there is no condition about it, all i can say is "there are two possible graphs."

i hope, this wouldn't mind you.