$\mathrm{f}\left(\mathrm{x}\right)=(\mathrm{x}-8{)}^4(\mathrm{x}-9{)}^5,0\le \mathrm{x}\le 10,$monotonically decreases in

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=(\mathrm{x}-8{)}^4(\mathrm{x}-9{)}^5,0\le \mathrm{x}\le 10 \\ \therefore {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=4(\mathrm{x}-8{)}^3(\mathrm{x}-9{)}^5+5(\mathrm{x}-9{)}^4(\mathrm{x}-8{)}^4 \\ =(\mathrm{x}-8{)}^3(\mathrm{x}-9{)}^4\left[4\right(\mathrm{x}-9)+5(\mathrm{x}-8\left)\right] \\ =9(\mathrm{x}-8{)}^3(\mathrm{x}-9{)}^4\left(\mathrm{x}-\frac{76}{9}\right) \end{gathered}$$
Sign scheme of f'(x) is as follows: 
 

${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)<0\text{ if }\mathrm{x}\in \left(8,\frac{76}{9}\right)$,i.e., f(x) decreases if $\mathrm{x}\in \left(8,\frac{76}{9}\right)$.