Find the approximate value of $f\left( {5.001} \right)$, where $f\left( x \right) = {x^3} - 7{x^2} + 15$.

Given, $f\left( x \right) = {x^3} - 7{x^2} + 15 \& \Rightarrow f'\left( x \right) = 3{x^2} - 14x$

Also, $f\left( {x + \Delta x} \right) \approx f\left( x \right) + \Delta xf'\left( x \right)$

Taking $x = 5$ and $\Delta x = 0.001$, we get
$f\left( {5.001} \right) \approx {5^3} - 7 \cdot {5^2} + 15 + \left( {0.001} \right)\left( {3 \cdot {5^2} - 14 \cdot 5} \right)$

$= 125 - 175 + 15 + \cfrac{1}{{100}}\left( 5 \right) = - 35 + 0.005 = - 34.995$