Graph of the function

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rachelmaddie

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Hi. I need help with this problem. I don’t have attempted work for this one because I’m having trouble with understanding how to find the critical points.
545C7218-74EE-48BD-BA5F-6CA86F36D089.jpeg
 
Subhotosh Khan said:
Please tell us the definition of a critical point (in the domain of the function).
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Critical point of a single variable function. A critical point of a function of a single real variable, f(x), is a value x0 in the domain of f where it is not differentiable or its derivative is 0 (f ′(x0) = 0).
 
rachelmaddie said:
Critical point of a single variable function. A critical point of a function of a single real variable, f(x), is a value x0 in the domain of f where it is not differentiable or its derivative is 0 (f ′(x0) = 0).
Click to expand...
Correct.

Commensurate with the definition above (response #4),

where would you estimate the critical point/s to be for the given graph?​
 
pka said:
I would argue that the only critical point shown on the given graph is \((-1,2)\). WHY?
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How is this?

(1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc.

I’d say they want you to estimate the function as:

f(x)=(x+1)^3+2
 
rachelmaddie said:
The slope of every tangent line that passes through a critical point is always 0.
Click to expand...


Look at the graph of \(\displaystyle \ y \ = \ \sqrt[3]{x}. \ \) It has a critical point at (0, 0)
that is a point of inflection. However, there is a vertical tangent line there.
The slope of the tangent line is not zero. The slope is undefined.
 
rachelmaddie said:
How is this?

(1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc.

I’d say they want you to estimate the function as:

f(x)=(x+1)^3+2
Click to expand...
lookagain said:
Look at the graph of \(\displaystyle \ y \ = \ \sqrt[3]{x}. \ \) It has a critical point at (0, 0)
that is a point of inflection. However, there is a vertical tangent line there.
The slope of the tangent line is not zero. The slope is undefined.
Click to expand...
How does this look?
 
Hi. I need this checked please.
C20A7461-2F5A-4EF8-A84E-F9AA503663DE.jpeg
1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc.

I’d say they want you to estimate the function as:

f(x)=(x+1)^3+2
 
rachelmaddie said:
Hi. I need this checked please.
View attachment 20958
1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc.

I’d say they want you to estimate the function as:

f(x)=(x+1)^3+2
Click to expand...
I’d say they want you to estimate the function as:.................f(x)=(x+1)^3+2

But the problem did not ask for that. The problem asked for specifically:

What are the critical point/s? What are the class/es of those critical point/s?

You did not respond to those "explicitly"

What you wrote is good and correct but unnecessary verbiage (actually incorrect - considering the "find" of the problem).
 
Subhotosh Khan said:
I’d say they want you to estimate the function as:.................f(x)=(x+1)^3+2

But the problem did not ask for that. The problem asked for specifically:

What are the critical point/s? What are the class/es of those critical point/s?

You did not respond to those "explicitly"

What you wrote is good and correct but unnecessary verbiage (actually incorrect - considering the "find" of the problem).
Click to expand...
Can you help me write it correctly?
 
rachelmaddie said:
Hi. I need this checked please.
View attachment 20958
1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc.

I’d say they want you to estimate the function as:

f(x)=(x+1)^3+2
Click to expand...
It looks to me like you totally ignored what they asked for, and just answered the question you felt like answering. None of what you wrote is relevant.

So tell us: What critical point(s) do you see? How would you describe each of them?
 
Dr.Peterson said:
It looks to me like you totally ignored what they asked for, and just answered the question you felt like answering. None of what you wrote is relevant.

So tell us: What critical point(s) do you see? How would you describe each of them?
Click to expand...

It has a critical point at (0, 0)
that is a point of inflection. However, there is a vertical tangent line there.
The slope of the tangent line is not zero. The slope is undefined.
 
rachelmaddie said:
It has a critical point at (0, 0)
that is a point of inflection. However, there is a vertical tangent line there.
The slope of the tangent line is not zero. The slope is undefined.
Click to expand...

Well, no. The point (0, 0) is not on the graph shown. And there are no vertical tangents or undefined slopes, either. On the other hand, the graph is not quite clear, so I would assume, as you did initially, that it is meant to be a translation of y = x^3, which tells you some important things.

I suspect that you meant the right things, but didn't write carefully. Try again. And if you still make the same statements, tell us the observations on which you base them (e.g. where you see a vertical line), so we can see whether it is your thinking or your vision that disagrees with us. Or maybe mark up the picture.
 
rachelmaddie said:
I’d say they want you to estimate the function …
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Hi Rachel. The exercise doesn't ask you to estimate the function. It asks you to estimate the point on the graph where the derivative is zero.

The instructions ask you to classify that critical point, also. Classifying a critical point generally means to state whether its location is a function maximum, minimum, or inflection.

Therefore, the answer is the coordinates of the critical point, followed by the classification.

?
 
rachelmaddie said:
How is this?

(1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc.

I’d say they want you to estimate the function as:

f(x)=(x+1)^3+2
Click to expand...
I see nothing in this problem to suggest that the function is a cubic and the problem doesn't ask about that. Looking at the graph the curve "levels" and the derivative is 0 at (-1, 2). Since the graph rises to (-1, 2) then continues going up (-1,, 2) is neither a "maximum" nor a "minimum". It is an "inflection point".
 
Now that the answer has been given [grumble], I might as well state explicitly why I suggested using the assumption that it is meant to look like a cubic:

The graph has a quirk at the inflection point, where it steps up a pixel so that it might be taken not to be exactly horizontal. We all look at it and see it as looking like the cubic, which we know has a horizontal inflection point, so we can overlook the imperfect graph and still see what is intended. I rather dislike such problems that depend on seeing a graph as we see it, and assuming things from it that might not in fact be true. This problem does say "estimate", admitting that we can't be sure of the exact location; if I wrote it, I would make some comment permitting the student to expect that there is, in fact, a critical point.

Another deficiency of the graph is that there are no arrows (or dots) at the ends, so we aren't really sure whether they are endpoints of the domain.
 
Dr.Peterson said:
Now that the answer has been given [grumble], I might as well state explicitly why I suggested using the assumption that it is meant to look like a cubic:

The graph has a quirk at the inflection point, where it steps up a pixel so that it might be taken not to be exactly horizontal. We all look at it and see it as looking like the cubic, which we know has a horizontal inflection point, so we can overlook the imperfect graph and still see what is intended. I rather dislike such problems that depend on seeing a graph as we see it, and assuming things from it that might not in fact be true. This problem does say "estimate", admitting that we can't be sure of the exact location; if I wrote it, I would make some comment permitting the student to expect that there is, in fact, a critical point.

Another deficiency of the graph is that there are no arrows (or dots) at the ends, so we aren't really sure whether they are endpoints of the domain.
Click to expand...
Should I include this whole statement in my answer?

4)
(1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc. Looking at the graph the curve "levels" and the derivative is 0 at (-1, 2). Since the graph rises to (-1, 2) then continues going up (-1,, 2) is neither a "maximum" nor a "minimum". It is an "inflection point".
 
rachelmaddie said:
Should I include this whole statement in my answer?

4)
(1,1) and (-1,-1) are on the graph of x^3. The fact that the points (0,3) and (-2,1) are on this graph (and since they are the points right one and up one, and left one and down one relative to the vertex position (-1,2), that suggests that the x^3 is only shifted in position and that there’s no stretch/compression/reflection etc. Looking at the graph the curve "levels" and the derivative is 0 at (-1, 2). Since the graph rises to (-1, 2) then continues going up (-1,, 2) is neither a "maximum" nor a "minimum". It is an "inflection point".
Click to expand...
Most of that is utterly irrelevant to the problem. Which part do you think is appropriate?

Also, I wish you wouldn't copy what we say! You need to learn to write your own answers, or you will never succeed on your own. Put the ideas in your own words, at the least.
 
Dr.Peterson said:
Most of that is utterly irrelevant to the problem. Which part do you think is appropriate?

Also, I wish you wouldn't copy what we say! You need to learn to write your own answers, or you will never succeed on your own. Put the ideas in your own words, at the least.
Click to expand...
This part?
Looking at the graph the curve "levels" and the derivative is 0 at (-1, 2). Since the graph rises to (-1, 2) then continues going up (-1,, 2) is neither a "maximum" nor a "minimum". It is an "inflection point".
 
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