Show that the derivative of the function f given by

Given: $f(x)=2 x^{3}-9 x^{2}+12 x+9$

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by:

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h-f(x)}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2(x+h)^{3}-9(x+h)^{2}+12(x+h)+9-2 x^{3}+9 x^{2}-12 x-9}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2 x^{3}+2 h^{3}+6 x^{2} h+6 x h^{2}-9 x^{2}-9 h^{2}-18 x h+12 x+12 h+9-2 x^{3}+9 x^{2}-12 x-9}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{2 h^{3}+6 x^{2} h+6 x h^{2}-9 h^{2}-18 x h+12 h}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h\left(h^{2}+6 x^{2}+6 x h-9 h-18 x+12\right)}{h}$

$\Rightarrow f^{\prime}(x)=6 x^{2}-18 x+12$

So,

$f^{\prime}(1)=6\left(x^{2}-3 x+2\right)$

$=6 \times(1-3+2)$

$=0$

$f^{\prime}(2)=6\left(x^{2}-3 x+2\right)$

$=6 \times(4-6+2)$

$=0$'

Hence the derivative at $x=1$ and $x=2$ are equal.