Question

Show that the derivative of the function f given by
$f\left(x\right)=2x^3-9x^2+12x+9$, at x = 1 and x = 2 are equal.

Answer 1

Given: $f\left(x\right)=2x^3-9x^2+12x+9$

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

$$\begin{gathered} f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h-f(x)}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{2(x+h{)}^3-9(x+h{)}^2+12(x+h)+9-2x^3+9x^2-12x-9}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{2x^3+2h^3+6x^{2}h+6x{h}^2-9x^2-9h^2-18xh+12x+12h+9-2x^3+9x^2-12x-9}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{2h^3+6x^{2}h+6x{h}^2-9h^2-18xh+12h}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{h(h^2+6x^2+6xh-9h-18x+12)}{h} \\ \Rightarrow f\text{'}\left(x\right)=6x^2-18x+12 \end{gathered}$$

So,

$$\begin{gathered} f\text{'}\left(1\right)=6\left(x^2-3x+2\right) \\ =6\times (1-3+2) \\ =0 \\ f\text{'}\left(2\right)=6\left(x^2-3x+2\right) \\ =6\times (4-6+2) \\ =0 \end{gathered}$$

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