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#1 2016-01-28 15:25:14

Mathegocart
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Registered: 2012-04-29
Posts: 2,226

Help... Concave Downwards?

http://prntscr.com/9w849a
http://prntscr.com/9w846x Would I be correct on this? Also how do you tell wether a (-infinity,-3) is concave up or down

Last edited by Mathegocart (2016-01-28 15:29:26)


The integral of hope is reality.
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He will be sorely missed.

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#2 2016-01-28 22:22:23

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Help... Concave Downwards?

hi mathgogocart

I'm in the process of working out the functions for f' and f.  I've got the graph for f' ..... still de-bugging the graph for f.

I'll post later when I'm more confident of what I'm saying.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2016-01-29 02:08:27

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: Help... Concave Downwards?

Ok.  I'll post what I've got.  (Bear in mind I've never heard of this concavity test.)

Step 1.  What is the function for f'' ?  (Assuming it is a polynomial)

It has four roots so a good trial is

There is an objection to this;  it doesn't go through (0,11)

This is easily corrected by introducing a multiplier of 11/18.  As we are only concerned with the shape of the graph a re-scaling such as this is irrelevant so I'll not bother.

I have tried this on the MIF function grapher and it looks ok.

Step 2.  What does the graph f' look like?  We have enough information from the given graph to make a good 'stab' at this.

It should have four turning points at x = -3, -2, 1 and 3.  Looking closely at the behaviour close to x = -3 we can see that f' will have a positive gradient just to the left of x = -3, and a negative gradient just to the right.  So f' has a local maximum at x = -3.  Similarly the next three are minimum, maximum and minimum respectively.

That's enough to sketch the graph.

But I thought I would integrate f'' to get an algebraic function:

6uembH5.gif

I have re-scaled to make a better image of the shape of the graph.

There could also be a constant of integration but all this does is shift the graph up or down parallel to the y axis, so it doesn't affect the shape.

Step 3.  What does the graph f look like?

Depending on that constant of integration it might more than one turning point.  Since the test should work for any graph that double differentiates to f'', I'll just try integrating the f' function.

d03BPqo.gif

f'' is negative in the interval (1,3) .  To establish whether the graph is downwardly concave in this interval it is necessary to determine the exact position of the points of inflexion. 

For f', the gradient is negative from 1 to 3 so that seems to confirm the concavity.

In the interval (-3,-2) the gradient is similarly negative.  I think that is it although my head is aching so much I'll have to come back to this later to check what I'm saying.  dizzy

Bob

Last edited by Bob (2016-01-29 02:44:06)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2016-01-29 03:39:50

Mathegocart
Member
Registered: 2012-04-29
Posts: 2,226

Re: Help... Concave Downwards?

bob bundy wrote:

Ok.  I'll post what I've got.  (Bear in mind I've never heard of this concavity test.)

Step 1.  What is the function for f'' ?  (Assuming it is a polynomial)

It has four roots so a good trial is

There is an objection to this;  it doesn't go through (0,11)

This is easily corrected by introducing a multiplier of 11/18.  As we are only concerned with the shape of the graph a re-scaling such as this is irrelevant so I'll not bother.

I have tried this on the MIF function grapher and it looks ok.

Step 2.  What does the graph f' look like?  We have enough information from the given graph to make a good 'stab' at this.

It should have four turning points at x = -3, -2, 1 and 3.  Looking closely at the behaviour close to x = -3 we can see that f' will have a positive gradient just to the left of x = -3, and a negative gradient just to the right.  So f' has a local maximum at x = -3.  Similarly the next three are minimum, maximum and minimum respectively.

That's enough to sketch the graph.

But I thought I would integrate f'' to get an algebraic function:

http://i.imgur.com/6uembH5.gif

I have re-scaled to make a better image of the shape of the graph.

There could also be a constant of integration but all this does is shift the graph up or down parallel to the y axis, so it doesn't affect the shape.

Step 3.  What does the graph f look like?

Depending on that constant of integration it might more than one turning point.  Since the test should work for any graph that double differentiates to f'', I'll just try integrating the f' function.

http://i.imgur.com/d03BPqo.gif

f'' is negative in the interval (1,3) .  To establish whether the graph is downwardly concave in this interval it is necessary to determine the exact position of the points of inflexion. 

For f', the gradient is negative from 1 to 3 so that seems to confirm the concavity.

In the interval (-3,-2) the gradient is similarly negative.  I think that is it although my head is aching so much I'll have to come back to this later to check what I'm saying.  dizzy

Bob

Yeah but how do you do it on the go? Khan says that it can be solved by looking at the graph.


The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.

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#5 2016-01-29 05:38:12

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Help... Concave Downwards?

OK, I'll have a go.

here I have put three graphs in one.

17CWvqN.gif

I'll start at the bottom with the blue graph.

Let's say that is part of a double differentiated graph f''

At A and B the function is zero and at C it has its lowest local value.

So the single differentiated graph must have turning points at the equivalent points to A and B and a largest negative gradient in line with C.  That's the red graph.

So the function f has a certain gradient in line with A, lower values as we track right and lower still in line with B.  That's the black graph.

The black graph is downwards concave.

The test is to look for where the f'' graph is fully negative in an interval.  In this case, from A to B.

You can explore upwards concavity similarly by looking for positive bits of f''

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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