Find the approximate value of $f\left(2.01\right)$, where $f\left(x\right) = 4x^2+5x+2$
Let $x=2$, $\Delta x = 0.01$ so that $x + \Delta x = 2.01$
$f\left(x\right) = f\left(2\right) = 4 \times 2^2 + 5 \times 2 + 2 = 28$
$f'\left(x\right) = 8x + 5$
$\begin{aligned}
\text{Now}, f\left(x+\Delta x\right) & = f\left(x\right) + \Delta f \left(x\right) \\
& = f \left(x\right) + f'\left(x\right) \cdot \Delta x \\
& = \left(4x^2 + 5x + 2\right) + \left(8x + 5\right) \times \Delta x
\end{aligned}$
$\begin{aligned}
\therefore \; f\left(2.01\right) & = 28 + \left(8 \times 2 + 5\right) \times 0.01 \\
& = 28 + 0.21 \\
& = 28.21
\end{aligned}$