Mathematical Analysis: Mean Value Theorem I'm trying to learn how the mean value theorem works by attempting to answer the following but i just dont understand it at all.
Consider the function $f(x)$ $=$ $\sqrt x$ for $x > 0$
i. Show that $f'(x)$ is a decreasing function on $(0, \infty)$
ii. Apply the Mean Value Therorem to $f$ on the interval $[100,102]$ to prove
$10 + \frac{1}{11} \leq \sqrt {102} \leq 10 + \frac{1}{10}$
Any help would be appreciated.

We are given that $f(x) = \sqrt{x}$.
Hence, the first and second derivatives of $f$ are$f'(x) = \frac1{2\sqrt{x}}$ and $f''(x) = -\frac1{4x^{3/2}}$. Note that $f''(x) < 0$ for all $x \in \left(0, \infty\right)$. If the first derivative of a function is always negative, then the function is a decreasing function.
In our case, $f''(x)$ is the first derivative of $f'(x)$ and since $f''(x)<0, \forall x \in \left(0, \infty \right)$, we get that $f'(x)$ is a decreasing function.
For the second part, by mean value theorem, we have that if a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a < b$, and differentiable on the open interval $(a, b)$, then there exists a point $c \in (a, b)$ such that
$$f'(c) = \frac{f(b)-f(a)}{b-a}$$ i.e. $f(b) = f(a) + (b-a) \times f'(c)$. To prove the second part, choose $f(x) = \sqrt{x}$, $a = 100$ and $b=102$. Clearly $f$ satisfies the assumptions in the mean value theorem. Hence, $\exists c \in (a,b)$ such that $\sqrt{b} = \sqrt{a} + (b-a) f'(c) = \sqrt{a} + (b-a) \frac1{2 \sqrt{c}}$. Plugging in the values for $a$ and $b$ we get that $$\sqrt{102} = \sqrt{100} + \frac{2}{2 \sqrt{c}} = 10 + \frac1{\sqrt{c}}$$
Since $c \in (100,102)$, we have that $10 < \sqrt{c} <\sqrt{102} < \sqrt{121} = 11$. Hence, $\frac1{11} < \frac1{\sqrt{c}} < \frac1{10}$. Hence, $10 + \frac1{11} < 10 + \frac1{\sqrt{c}} < 10 + \frac1{10}$ i.e. $10 + \frac1{11} < \sqrt{102} < 10 + \frac1{10}$.
