Find the approximate value of

Let $x=2$ and $\Delta x=0.01$. Then, we have:

$f(2.01)=f(x+\Delta x)=4(x+\Delta x)^{2}+5(x+\Delta x)+2$

Now, $\Delta y=f(x+\Delta x)-f(x)$

$\therefore f(x+\Delta x)=f(x)+\Delta y$

$\approx f(x)+f^{\prime}(x) \cdot \Delta x \quad($ as $d x=\Delta x)$

$\Rightarrow f(2.01) \approx\left(4 x^{2}+5 x+2\right)+(8 x+5) \Delta x$

$=\left[4(2)^{2}+5(2)+2\right]+[8(2)+5](0.01) \quad[$ as $x=2, \Delta x=0.01]$

$=(16+10+2)+(16+5)(0.01)$

$=28+(21)(0.01)$

$=28+0.21$

$=28.21$

Hence, the approximate value of f (2.01) is 28.21.