Function that is non-zero only at one point. I am searching for, if there exists, a continuous function $f(x)$ such that $f(x) = 0$ for all values of $x$, with the exception of one point (say $\tilde x$) where $f(\tilde x)\neq0$.
 A: There is no continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x)=0$ for all $x\not=x_0$ and $f(x_0)\not=0$, since by definition of continuity 
$$f(x_0)=\lim_{x\rightarrow x_0,x\not=x_0} f(x)=0$$
A: This is possible if and only if $\tilde x$ is an isolated point of the domain.
Example 1: For example, the domain could be $$(-\infty,\tilde x - \epsilon)\cup \{\tilde x\}\cup (\tilde x +\epsilon,\infty)$$ if it is a subset of $\mathbb R$ (which, incidentally, is not specified) and the function rule could be
$$f(x) = \begin{cases}
0 & \textrm{ if }x <\tilde x - \epsilon\\
c & \textrm{ if }x = \tilde x \\
0 & \textrm{ if }x > \tilde x +\epsilon
\end{cases}
$$
where $\epsilon$ is a positive constant and $c$ is a nonzero constant.
For example, you could have $$f:(-\infty,-1)\cup \{0\}\cup (1,\infty) \rightarrow \mathbb R$$
with
$$f(x) = \begin{cases}
0 & \textrm{ if }x <-1\\
1 & \textrm{ if }x = 0 \\
0 & \textrm{ if }x >1
\end{cases}
$$
Example 2: An even simpler example of such a function would be
$$f:\{0,1\} \rightarrow \mathbb R$$
with
$$f(x) = \begin{cases}
1 & \textrm{ if }x = 0 \\
0 & \textrm{ if }x =1
\end{cases}
$$
A: This function cannot exist, if it must be continuous.
Suppose such a function $f:R \to R$ exists, and for some $x \in R$, $f(x) = c \ne 0$. Let $y \ne x$. By the Intermediate Value Theorem, then $\exists z \in [x, y]$ (or $[y, x]$) such that $0 < f(z) < c$.
A: Because the additional condition of continuity, it is not possible. You can use limits at the only point where the function is non-zero to prove that such a function will not be continuous.
A: I think perhaps your prof used poor wording and meant piecewise continuous, and they are looking for the function given by gebruiker at the top. Had this happen to me in Real Variables and confused me. It is a function and piecewise continuous (over an infinitely large set too!) and if your studying Lebesgue measure/integrals then that is what you want. 
f(x) = 0 when x ≠ xStar, f(x) = c when x = xStar
If you are not studying Lebesgue measure or integrals, and in Reimann then use the intermediate value to show DNE as stated by C. Quilley at the top.
Have a nice day!
A: The Heaviside Step function, a generalized function, performs part of the function initially defined. 
https://en.wikipedia.org/wiki/Heaviside_step_function
This function was created by Heaviside to represent the on and off turning of a switch.
A combination of two of these functions can describe the complete function above.
As noted, the derivative of the Heaviside function is a Dirac delta function.
https://en.wikipedia.org/wiki/Dirac_delta_function
A: Everyone seems to be mistaken about what continuity of a function actually means. The Dirac Delta function answers the question (as others have already suggested). The Dirac delta is a continuous function; as defined by its integrability i.e. the n-dimensional integral of any n-dimensional Dirac delta function is unity (it would be naturally non-integrable if it was discontinuous).
Looking at it from a slightly more intuitive angle; it can be recognised that the Dirac delta is derived via the integration of a complex valued exponential function (which are naturally continuous).
A: Delta Function(impulse function) in x=0 f(0)=inf , x!=0 f(x)=0 this function use in physics as impact function or in electrical engineering for sudden current or voltage and this function is very important.
