The function $f\left(x\right)=x^3-6x^2+9x+25$ has

$f\left(x\right)=x^3-6x^2+9x+25$
$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=3x^2-12x+9\\ f^{\prime \prime }\left(x\right)=6x-12\end{array}$
To get the stationary points $\Rightarrow f^{\mathrm{\prime }}\left(x\right)=0$
$\begin{array}{r}\therefore 3x^2-12x+9=0\\ x^2-4x+3=0\\ (x-1)(x-3)=0\\ x=1\text{ or }x=3\end{array}$
To get the maxima and minima of a function
$\begin{array}{l}{f}^{\prime \prime }\left(x\right)=6x-12\\ \text{ At }x=1,f^{\prime \prime }\left(1\right)=6\left(1\right)-12=6-12=-6<0\end{array}$

$\therefore \text{ Maxima of }f\left(x\right)\text{ exists at }x=1\text{. }$,
$\text{ At }x=3,f^{\prime \prime }\left(3\right)=6\left(3\right)-12=18-12=6>0\text{. }$
$\therefore \text{ Minima of }f\left(x\right)\text{ exists at }x=3$