Question

Show that the derivative of the function f given by 

\[f\left( x \right) = 2 x^3 - 9 x^2 + 12x + 9\], at x = 1 and x = 2 are equal.

Answer 1

Given: 

\[f(x) = 2 x^3 - 9 x^2 + 12x + 9\]

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of 

\[f\] at 

\[x\]  is given by:

`f'(x) = lim_(h→0)f(x +h -f(x))/h`
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{2(x + h )^3 - 9(x + h )^2 + 12(x + h) + 9 - 2 x^3 + 9 x^2 - 12x - 9}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{2 x^3 + 2 h^3 + 6 x^2 h + 6x h^2 - 9 x^2 - 9 h^2 - 18xh + 12x + 12h + 9 - 2 x^3 + 9 x^2 - 12x - 9}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{2 h^3 + 6 x^2 h + 6x h^2 - 9 h^2 - 18xh + 12h}{h}\]
\[ \Rightarrow f'(x) = \lim_{h \to 0} \frac{h( h^2 + 6 x^2 + 6xh - 9h - 18x + 12)}{h}\]
\[ \Rightarrow f'(x) = 6 x^2 - 18x + 12\]

So,

\[f'(1) = 6\left( x^2 - 3x + 2 \right) \]
\[ = 6 \times (1 - 3 + 2) \]
\[ = 0\]
\[f'(2) = 6\left( x^2 - 3x + 2 \right) \]
\[ = 6 \times (4 - 6 + 2) \]
\[ = 0\]

Hence the derivative at 

\[x = 1\] and 

  \[x = 2\]  are equal.