Let $f\left(x\right)$ be a differential function for all $x$, If $f\left(1\right)=-2$ and $f^{\text{'}}\left(x\right)>2$ for all $x$ in $[1,6]$, then minimum value of $f(6)=$

$$\begin{gathered} f\left(x\right)=ax+b \\ f\left(1\right)=-2 \\ a+b=-2 \\ f’\left(x\right)=a>2 \\ b=-2-a \\ f\left(x\right)=ax-2-a \\ f\left(x\right)=a(x-1)-2 \\ f\left(6\right)=a(6-1)-2 \\ f\left(6\right)=5a-2 \\ a>2 \end{gathered}$$

$\left(1\right)$

$$\begin{gathered} f\left(6\right)=2 \\ 5a-2=2 \\ 5a=4 \\ a=\frac{4}{5} \end{gathered}$$

$\left(2\right)$

$$\begin{gathered} f\left(6\right)=4 \\ 5a-2=4 \\ 5a=6 \\ a=\frac{6}{5} \end{gathered}$$

$\left(3\right)$

$$\begin{gathered} f\left(6\right)=6 \\ 5a-2=6 \\ 5a=8 \\ a=\frac{8}{5}=1.6 \end{gathered}$$

$\left(4\right)$

$$\begin{gathered} f\left(6\right)=8 \\ 5a-2=8 \\ 5a=10 \\ a=2 \end{gathered}$$

Hence the minimum value of $f\left(6\right)=8$.