Question:
Find the approximate value of $f(5.001)$, where $f(x)=x^{3}-7 x^{2}+15$.
Solution:
Let:
$x=5$
$x+\Delta x=5.001$
$\Rightarrow \Delta x=0.001$
$f(x)=x^{3}-7 x^{2}+15$
$\Rightarrow y=f(x=3)=125-175+15=-35$
Now, $y=f(x)$
$\Rightarrow \frac{d y}{d x}=3 x^{2}-14 x$
$\therefore d y=\Delta y=\frac{d y}{d x} d x=\left(3 x^{2}-14 x\right) \times 0.001=(75-70) \times 0.001=0.005$
$\therefore f(5.001)=y+\Delta y=-35+0.005=-34.995$