Supremum and infimum of Multiplicative inverse Given $$
f(x)=
\begin{cases}
1/x,& 0< x\le 1\\
0,& x=0
\end{cases}
$$ and a partition $P = [0, x_1, x_2,\dots, x_{n-1},1]$ and $I_k = [x_{i-1}, x_i]$, e.g. $I_1 = [0, x_1]$. Is it possible to compute the supremum and infimum of $M_k = \underset{I_k}{\mathbb{sup}} f(x)$ and $m_k = \underset{I_k}{\mathbb{inf}} f(x)$? 
Any help is highly appreciated.
 A: For $k\ge 2$, you have that $I_k$ is a closed interval and $f(x)=1/x$ is a continuous function on it, so it attains its $\min$ and its $\max$ on this interval (which Theorem is it?). Hence, for $k\ge 2$ \begin{align}\sup_{I_k}f(x)=\max_{I_k}f(x)&=f(x_{k-1})\\  \inf_{I_k}f(x)=\min_{I_k}f(x)&=f(x_{k})\end{align} Now, in $I_1=[0,x_1]$, you have that $\lim_{x\to 0} f(x)=+\infty$, so $$\sup_{I_1}f(x)=+\infty$$ On the other hand $f(0)=0$, and $f(x)=1/x>0$ for any other $x$, so $$\inf_{I_1}f(x)=0$$

Edit: To elaborate on $I_1$: $f$ has exactly one value on $x=0$, which is $f(0)=0$. However, $\lim_{x\to 0+} f(x)=+\infty$, which is different than $f(0)$. This does not mean that $f$ has two values for $x=0$. It means that $f$ is discontinuous at $x=0$, since $$0=f(0)\neq \lim_{x\to0+}f(x)=+\infty$$ Now, as $x$ tends to $0$, $f(x)$ increases all the time without bounds. In other words, name a number, then there is an $x$ (very small, meaning very close to $0$) so that $f(x)=1/x$ is bigger than this number you named. This shows that $$\sup_{I_1}f(x)=\lim_{x\to 0}f(x)=+\infty$$ Note: here you have indeed a $\sup$ and not a maximum (since it is attained asymptotically). Now, for the $\inf$ things are simpler. $f$ takes the value $0$ (exactly, not asysmptotically, or anything the like, but exactly), at $x=0$ and is positive for any other value in $I_1$. Hence, $f(0)=0$ is indeed the least value of $f$ in $I_1$, which establishes that $$\inf_{I_1}f(x)=\min_{I_1}f(x)=f(0)=0$$
A: Hint:
$1/x$ is a decreasing function.
Care have to be taken to handle $I_1$.
A: For $i>1$, $f $ is stictly decreasing at $I_i=[x_{i-1},x_i] $ thus
$$M_i=f (x_{i-1})=\frac {1}{x_{i-1}} $$
$$m_i=\frac {1}{x_i} $$
but for $I_1$,
$$m_1=0$$
and
$$M_1=\lim_{0^+}f (x)=+\infty $$
