Let there be a function $g:(0, \infty) \to \mathbb{R}$, $g(x) = \frac{1}{x}$, and $S:[1,\infty)\to[0,\infty)$, the area of the domain delimited by $g$'s graph, the $Ox$ axis and the straight lines parallel with $Oy$, which go through the points $A$ and $M$ which are on the graph of $g$ and have the abscissas $1$ and $x$ ($x \ge 1$). Let there be $x_0 \ge 1$, a fixed number.
For $h > 0$, show that $\frac{1}{x_0+h} \le \frac{S(x_0+h)-S(x_0)}{h} \le \frac{1}{x_0}$ if $h$ is the distance between $A$ and $M$.
The solution in the book is the following, but I do not understand it:
$x_0 > 1$, $h > 0$. $\frac{h}{x_0+h} \le S(x_0+h)-S(x_0) \le \frac{h}{x_0}$, which is divided by $h$.
What I tried:
If $g(x)=x^2$, a parabola, then the area of the domain delimited by the graph of $g$, the $Ox$ axis and the straight lines parallel with $Oy$, which go through the points on the graph of $g$ with the abscissas $x_0$ and $x_0+h$, would be $S(x_0+h)-S(x_0)$. The minimum area of this domain, interpreted as a rectangle, is $x_0^2h$ and the maximum area of this domain is $(x_0+h)^2h$. These three expressions can be put in the following relationship:
$$x_0^2h \le S(x_0 + h)-S(x_0) \le (x_0 + h)^2h$$
This relationship can be divided by $h$ and we obtain:
$$x_0^2 \le \frac {S(x_0 + h)-S(x_0)}{(x_0+h)-x_0} \le (x_0 + h)^2$$
Hint:
The figure illustrates the situation.
$S(x_0+h)-S(x_0)$ is the area of the trapezoid under the graph of the function between $x_0$ and $x_0+h$.
$\frac{h}{x_0}=hg(x_0)$ is the area of the rectangle under the green line.
$\frac{h}{x_0+h}=hg(x_0+h)$ is the area of the rectangle under the red line.
Since $y=\frac{1}{x}$ is a monotonic decreasing function for $x>1$ we have
$$ \frac{1}{x_0+h}<\frac{1}{x_0} $$ so $$ \frac{h}{x_0+h}\le S(x_0+h)-S_{x_0}\le \frac{h}{x_0} $$