Mackattack
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- Oct 14, 2020
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- 29
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!).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?).
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.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.
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.
\(\mathop {\lim }\limits_{x \to - 1} {x^2} - 2x + 1 = 4\)
Did I understand correctly? Also, what's the reason for equating the function to 4?\(\mathop {\lim }\limits_{x \to - 1} {x^2} - 2x + 1 = 4\)
Now can you discuss the continuity?
You weren't told to solve an equation.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 courseDid I understand correctly? Also, what's the reason for equating the function to 4?
Hehehe. If I had a choice, I wouldn't be taking calculus.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?