Find the approximate value of f (2.01),

Let:

$x=2$

$x+\Delta x=2.01$

$\Rightarrow \Delta x=0.01$

$f(x)=4 x^{2}+5 x+2$

$\Rightarrow f(x=2)=16+10+2=28$

Now, $y=f(x)$

$\Rightarrow \frac{d y}{d x}=8 x+5$

$\therefore d y=\Delta y=\frac{d y}{d x} d x=(8 x+5) \times 0.01=(16+5) \times 0.01=0.21$

$\therefore f(2.01)=y+\Delta y=28.21$