Discuss the continuity at the indicated value and sketch the graph of y=(x)

[Mackattack]

I already searched it on Wolfram Alpha, but I simply don't get why x is not equal to -1. Do I have to contradict it by saying x may be -1 or is there something I'm not getting? The table there is provided by the way and I have to fill it up (do I just use any value as long as it's not 1?).

 

[Dr.Peterson]

Discuss the continuity at the indicated value and sketch the graph of y=f(x)
View attachment 24937
I already searched it on Wolfram Alpha, but I simply don't get why x is not equal to -1. Do I have to contradict it by saying x may be -1 or is there something I'm not getting? The table there is provided by the way and I have to fill it up (do I just use any value as long as it's not 1?).

I don't think you've quoted the entire problem; but I'll suppose what you wrote in red is the definition of f, and they told you the "indicated value" is -1 (though what you show doesn't indicate any value!).

If I'm right, then they have just chosen not to define f for x=-1; they arbitrarily restricted the domain. So in graphing the function f, you will plot y=x^2-2x+1, and then make a "hole" at x=-1, to show it is not defined there. In your table, you'll fill in y for x=-1 with something like "undefined" or "does not exist". Then you can think about the definition of continuity at a point to decide whether f is continuous there.

It will be very helpful if you can show an image of the entire problem and instructions, so we can be sure what you were given and what you are to do. If you were just told to graph, then a table is not essential.

 

[Mackattack]

I don't think you've quoted the entire problem; but I'll suppose what you wrote in red is the definition of f, and they told you the "indicated value" is -1 (though what you show doesn't indicate any value!).

If I'm right, then they have just chosen not to define f for x=-1; they arbitrarily restricted the domain. So in graphing the function f, you will plot y=x^2-2x+1, and then make a "hole" at x=-1, to show it is not defined there. In your table, you'll fill in y for x=-1 with something like "undefined" or "does not exist". Then you can think about the definition of continuity at a point to decide whether f is continuous there.

It will be very helpful if you can show an image of the entire problem and instructions, so we can be sure what you were given and what you are to do. If you were just told to graph, then a table is not essential.

Sadly, I did quote the entire problem with instructions. The title is the full instructions and the attached file is everything given for this problem. I think your explanation is right though.

 

[Mackattack]

I don't think you've quoted the entire problem; but I'll suppose what you wrote in red is the definition of f, and they told you the "indicated value" is -1 (though what you show doesn't indicate any value!).

If I'm right, then they have just chosen not to define f for x=-1; they arbitrarily restricted the domain. So in graphing the function f, you will plot y=x^2-2x+1, and then make a "hole" at x=-1, to show it is not defined there. In your table, you'll fill in y for x=-1 with something like "undefined" or "does not exist". Then you can think about the definition of continuity at a point to decide whether f is continuous there.

It will be very helpful if you can show an image of the entire problem and instructions, so we can be sure what you were given and what you are to do. If you were just told to graph, then a table is not essential.

 

[pka]

View attachment 24942

\(\mathop {\lim }\limits_{x \to - 1} {x^2} - 2x + 1 = 4\)
Now can you discuss the continuity?

 

[Mackattack]

\(\mathop {\lim }\limits_{x \to - 1} {x^2} - 2x + 1 = 4\)
Now can you discuss the continuity?

Did I understand correctly? Also, what's the reason for equating the function to 4?

 

[Dr.Peterson]

Did I understand correctly? Also, what's the reason for equating the function to 4?

You weren't told to solve an equation.

Just ignore the restriction that [MATH]x \ne -1[/MATH], and find the limit of [MATH]x^2−2x+1[/MATH]. What do you get? (Hint: You'll discover why 4 was mentioned.)

Then consider that a limit at -1 ignores the value at -1, depending only on values near -1.

 

[pka]

Did I understand correctly? Also, what's the reason for equating the function to 4?

I simply do not understand why someone such as yourself is in a calculus course
and yet do not know that \(\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - 2x + 1} \right) = 4\)
given that \(f(x)=\begin{cases}x^2-2x+1 &: x\ne -1 \\ 6 &: x=-1\end{cases}\).
Note that \(\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - 2x + 1} \right) \ne f(-1)\)
Can you tell me why?

 

[JeffM]

I think the issue here is that the OP does not see any reason why a function should behave that way.

A function is a rule. In a math course, the teacher DEFINES the rule that a specific function must obey. There need not be any reason; it can be purely random. Here, however, the author‘s purpose is to get you to apply the definition of “continuous function“ to a specific case. You may think the specific case is silly, but it is highly relevant to the traditional foundations of calculus. The rationale for this function is to create an exercise that tests your understanding of the meaning of “continuous function.”

 

[Mackattack]

I simply do not understand why someone such as yourself is in a calculus course
and yet do not know that \(\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - 2x + 1} \right) = 4\)
given that \(f(x)=\begin{cases}x^2-2x+1 &: x\ne -1 \\ 6 &: x=-1\end{cases}\).
Note that \(\mathop {\lim }\limits_{x \to - 1} \left( {{x^2} - 2x + 1} \right) \ne f(-1)\)
Can you tell me why?

Hehehe. If I had a choice, I wouldn't be taking calculus.
Oh, dear. I just realized that 4 is the limit for that and that I just went in a circle hahahaha. That aside, I apologize for the incoming disappointment because I don't get how the limit of f(x) approaching -1 is not equal to f(-1). I get 4 for the limit and 4 for the evaluation of the function (1^2-2*-1+1=1+2+1=4). I actually don't understand the "given f(x) part that you put with the "1" and "6". I'm a still a bit new to this.

 

[JeffM]

This is an exercise in basics. (The fact that I do not personally like these basics is irrelevant).

A function is a rule that relates EACH element of set A (the "domain") to an UNAMBIGUOUS element of set B (the "range.") Sets A and B may be different sets or the same set.

The only restrictions on the rule are (a) it must apply to every element in set A, and (b) any one element in A is related to exactly one element in B.

There is no mathematical reason that a function must strike your intuition as plausible. As far as mathematics is concerned, a function is whatever it is defined to be.

It is a remarkably simple idea that for some reason is hard for students to grasp (mainly because they do not have the mathematical experience to see how it is useful). In large part, calculus is, as it is traditionally presented, the study of continuous and differentiable functions. (Differentiable functions are a subset of continuous functions). So they want you to understand what a continuous function is.

What I am about to say may not meet the approval of formal mathematicians. I think of continuity as being defined on an open interval.

[MATH]f(x) \text { is continuous on } (a,\ b) \iff[/MATH]
[MATH]a,\ b \text { are real numbers such that } a < b; \text { and}[/MATH]
[MATH]f(c) \text { is a unique real number if } a < c < b; \text { and}[/MATH]
[MATH]\lim_{x \downarrow c} f(x) = f(c); \text { and}[/MATH]
[MATH]\lim_{x \uparrow c} f(x) = f(c).[/MATH]
It is a very precise definition (for real functions of one variable).

The open-ended nature of the question permits you to respond to self-posed questions.

Where, if anywhere, is the f(x) given to you a continuous function?

Is the f(x) given to you continuous everywhere?

If not, why not?

(Basically, this whole response is just a rehash of Dr. Peterson's posts.)