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Chipset3600
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Hello MHB, the f(0) of this function doesn't exist, so I am i wrong or this question don't hv solution?
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View attachment 406
Chipset3600 said:Hello MHB, the f(0) of this function doesn't exist, so I am i wrong or this question don't hv solution?
View attachment 406
Sudharaka said:Hi Chipset3600, :)
Your function seem to have some strange symbols which I don't understand. Is it,
\[f(x)=\begin{cases}\frac{2}{\pi}\mbox{arctan}\left(\frac{x+1}{x^2}\right),&\,x=0\\
(x+1)^{1/\sec(x)}-\frac{\cos 2x}{x+1},&\,x<0\\
\frac{\sec^{2}x}{x.3^x-9x^2},&x>0\\
\end{cases}\]
In that case the definition seem to be problematic since \(f(0)\) is not defined properly. \(f(0)\) should be a constant value whereas in the definition it's not.
Kind Regards,
Sudharaka.
Chipset3600 said:Well, my language is portuguese, i forgot the translate of symbols: sen^2(x) = sin^(x), arctg(x)=arctan(x)..
and in the exercise is sin^2(x) and not sec^2(x).
"f(0)" refers to the value of a function at the input of 0. For a function to exist at a specific input, it means that the function is defined and has a corresponding output at that input value.
The continuity of "f(0)" is determined by evaluating the left-hand and right-hand limits of the function at the input of 0. If both limits are equal, then the function is continuous at 0 and "f(0)" exists.
No, in order for "f(0)" to exist, the function must be continuous at the input of 0. If the function is not continuous, then "f(0)" does not exist.
The continuity of "f(0)" can be affected by the presence of a removable or non-removable discontinuity at the input of 0, as well as the behavior of the function near 0. Other factors such as the type of function (e.g. polynomial, rational, exponential) and the existence of a limit at 0 can also impact the continuity of "f(0)".
No, it is possible for a function to be defined at 0 but not have continuity at that point. For example, a function with a removable discontinuity at 0 is still defined at 0 but is not continuous. However, if "f(0)" does not exist, then the function is not defined at 0.