Compute approximate value of a function change
$y=x^3 - x^2$ when $x=2$ and $x=0.01$. Any ideas how to solve it?
Attempt based on feedback in the comments:
$\Delta x = dx$ and $\Delta y= dy$, then $f(x+ \Delta x) = f(x) + f′(x) \cdot \Delta x$. $\Delta x = 2-0,01=1,99$. Then $f(0,01) + f′(0,01) \cdot 1,99 = \ldots$
Where $f'(x) = 3x^2 - 2x$.