Can anyone explain this solution (contractive function, interval, first derivative related) 
We are trying to find a positive null $x \geq 0$ of $$f(x) =
\sqrt{x+1}+\frac{\text{sin}(x)}{10}-2$$
Choose $X=\left[0,8\right]$. For $x \in X$ we have that
  $$\frac{1}{15}= \frac{1}{6}-\frac{1}{10} \leq f'(x) =
\frac{1}{2\sqrt{x+1}}+\frac{\text{cos}(x)}{10}\leq
\frac{1}{2}+\frac{1}{10} = \frac{3}{5}$$
(YOU COULD STOP READING HERE BECAUSE I'M RATHER INTERESTING IN JUST UNDERSTANDING THE STEPS TILL HERE.) 
Now try to find a $\beta$ such that $\text{sup} \left|1+\beta
f'(x)\right|<1$, i.e.
$-2 < \beta f'(x) <0 \text{ }\forall  x \in M$. Because of the
  inequality above, we have that $$-\frac{10}{3} < \beta <0$$
We can thus set $\beta = -3$ and $\phi(x) = x-3f(x) =
6+x-3\sqrt{x+1}-\frac{3}{10}\text{sin}(x)$. Thus
  $\left|\phi'(x)\right| \leq \frac{4}{5}= 0.8$
So, we have a contractive mapping.
[... further things that need to be shown so we satisfy all conditions (that there even exists a null). I leave them out to not make this question longer.]

I'm mainly interesting in understanding the beginning where we have $$\frac{1}{15}= \frac{1}{6}-\frac{1}{10} \leq f'(x) =
\frac{1}{2\sqrt{x+1}}+\frac{\text{cos}(x)}{10}\leq
\frac{1}{2}+\frac{1}{10} = \frac{3}{5}$$
Can you tell me how you get that? I understand that in the middle we have the first derivative of $f$. On the right side, we took the beginning of interval zero and put it into the first fraction, so $\frac{1}{2\sqrt{0+1}}= \frac{1}{2}$. Then put zero in the second fraction: $\frac{\text{cos}(0)}{10}=\frac{1}{10}$. This explains why we have $\frac{3}{5}$ on the right side.
But how do you get the left side, especially how you get $-\frac{1}{10}?$ We insert the end of interval on the left, $8$, I think. But why we have $-\frac{1}{10}$ ?

Consider $-1\leq \cos(x)\leq 1$ for all $x\in\mathbb{R}$ yields
$$
-\frac1{10}\leq \frac{\cos(x)}{10}\leq \frac1{10}.
$$
And consider that $x\mapsto\sqrt{x+1}$ is monoton increasing. Therefore you get $$
2\leq2\sqrt{x+1}\leq 6
$$
for $x\in[0,8]$ using the boundary points $x=0$ and $x=8$. This yields
$$
\frac16\leq \frac1{2\sqrt{x+1}}\leq \frac12
$$
for $x\in[0,8]$. Together we get
$$
\frac16-\frac1{10}\leq \frac1{2\sqrt{x+1}}+\frac{\cos(x)}{10}\leq \frac12+\frac1{10}
$$
for $x\in[0,8]$.
