Derivative Graphing HELP!!!!!!

[rjw292]

How do you go about graphing a graph that satisfying conditions with

f(x)
f'(x)
f''(x)


There are various points given but not sure how I go about doing it.

All three have DNE at one point and f'(x) and f"(x) have DNE at another point

 

[Bob Brown MSEE]

Use a Taylor Polynomial. When everything is given at one point.
Use a Vandermonde Matrix otherwise. With coefficients consistant with f, f' and f''

 

[rjw292]

Not sure how to use that or if thats what I need as we have not done cos, sin, etc

here is what I am given

Draw graph satisfying the following conditions.

x -2 -1 0 1 2
f(x) DNE 0 3 5 6
f'(x) DNE 2 0 1 DNE
f"(x) DNE 2 -4 0 DNE

 

[Bob Brown MSEE]

Hi rjw

What does this mean to you? Does DNE = "do not erase"?
Was this written on a blackboard?

 

[Bob Brown MSEE]

DNE = Does not exist, oh I get it -- sorry

x -2 -1 0 1 2
f(x) DNE 0 3 5 6
f'(x) DNE 2 0 1 DNE
f"(x) DNE 2 -4 0 DNE

x	-2	-1	0	1	2
f(x)	DNE	0	3	5	6
f'(x)	DNE	2	0	1	DNE
f"(x)	DNE	2	-4	0	DNE

Is this it? What have you tried? Forget what I said about the Vandermonde Matrix
It doesn't apply

Do you have graph paper?

 

[JeffM]

Not sure how to use that or if thats what I need as we have not done cos, sin, etc

here is what I am given

Draw graph satisfying the following conditions.

x -2 -1 0 1 2
f(x) DNE 0 3 5 6
f'(x) DNE 2 0 1 DNE
f"(x) DNE 2 -4 0 DNE

I am reasonably confident that DNE stands for "does not exist."

 

[rjw292]

x -2 -1 0 1 2
f(x) DNE 0 3 5 6
f'(x) DNE 2 0 1 DNE
f"(x) DNE 2 -4 0 DNE

x	-2	-1	0	1	2
f(x)	DNE	0	3	5	6
f'(x)	DNE	2	0	1	DNE
f"(x)	DNE	2	-4	0	DNE

Is this it? What have you tried? Forget what I said about the Vandermonde Matrix
It doesn't apply

It is Does Not Exist

and I have tried nothing as we are just into derivatives and concavity so not sure what do here or how to go about drawing this correctly

 

[Bob Brown MSEE]

Hint 1

I would start by plotting {x, y} points as dots.
{-1,0}, {0,3}, {1,5} and {2,6}

 

[rjw292]

I would start by plotting {x, y} points as dots.
{-1,0}, {0,3}, {1,5} and {2,6}

Did that and definitely know the straight line is not the graph he wants for f(x)

 

[Bob Brown MSEE]

Hint 2

Good! You see they are not in a straight line.
Put a very short line segment through each point with the slope indicated by the row of f'(x) values.
Do you see why?
What do you notice?
What do you think you should do next?

 

[rjw292]

Good! You see they are not in a straight line.
Put a very short line segment through each point with the slope indicated by the row of f' values.
Do you see why?
What do you notice?
What do you think you should do next?

Put a very short line segment through each point with the slope indicated by the row of f' values.

Use the f' points or the f points?

 

[rjw292]

Good! You see they are not in a straight line.
Put a very short line segment through each point with the slope indicated by the row of f'(x) values.
Do you see why?
What do you notice?
What do you think you should do next?

Do I use the f'(x) or the F(x) values for the slope?

 

[rjw292]

Not seeing it sorry.

Any help is greatly appreciated.

 

[Bob Brown MSEE]

Do I use the f'(x) or the F(x) values for the slope?

f'(x) is slope

 

[Bob Brown MSEE]

x	-2	-1	0	1	2
f(x)	DNE	0	3	5	6
f'(x)	DNE	2	0	1	DNE
f"(x)	DNE	2	-4	0	DNE

Should look something like this (but better)
Does yours look like this?
I could be wrong -- what questions do you have?
Do you see any misstakes?

 

[rjw292]

View attachment 2365

x	-2	-1	0	1	2
f(x)	DNE	0	3	5	6
f'(x)	DNE	2	0	1	DNE
f"(x)	DNE	2	-4	0	DNE

Should look something like this (but better)
Does yours look like this?
I could be wrong -- what questions do you have?
Do you see any misstakes?

I didnt even have a graph drawn as I have no idea how to do this to be honest

 

[Bob Brown MSEE]

Honesty is the only way to learn.
Ask questions about the graph -- use what you heard in class about "derivatives and concavity"

 

[mmm4444bot]

Did you plot the four points, as Bob instructed? You may be in over your head, if you're unable to plot four given (x,y) points.

Do you remember something from precalculus called a removable discontinuity? Open circles vs solid dots, when plotting such discontinuous graphs? Do you remember any other types of discontinuities? When you're told that f(x) DNE at some x-value, then that value of x is not in the domain. Many different ways to conjure up such a situation, there are.

Have you learned the meaning of first derivative with respect to slope of tangent line to graph of function f(x)? If you're told that f`(x) at some point is 2, what does that mean in terms of the behavior of function f's curve at that point? What if the first derivative is zero -- then what's happening to curve at that point? You need to draw a curve consistent with this info.

Have you learned about situations where function exists at a point but first derivative does not? Different ways this happens, too. Have you heard of cusps or vertical inflection points?

Have you learned the phrases "concave up" or "concave down"? These are related to your second derivative data.

This exercise requires you to bring a lot of knowledge together, in ways consistent with all of the given information. You need both the new calculus knowledge and the precalculus graph-experience, first.

 

[lookagain]

I would start by plotting {x, y} points as dots.
{-1,0}, {0,3}, {1,5} and {2,6}

I would start by plotting (x, y) points as dots.
(-1,0), (0,3), (1,5) and (2,6)

Do you see any misstakes?

Yes, "mistakes" is spelled incorrectly.


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View attachment 2365

x	-2	-1	0	1	2
f(x)	DNE	0	3	5	6
f'(x)	DNE	2	0	1	DNE
f"(x)	DNE	2	-4	0	DNE

I looked at that graph, and I looked at what I plotted.

There should be points at (-1, 0), (0, 3), (1, 5), and (2, 6).

There should be a horizontal tangent at (0, 3).

There should be a vertical asymptote line x = -2.

There should be an undefined slope at x = 2.

There should be a point of inflection at (1, 5).