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Let $x=3$ and $x+\Delta x=2.99$

$\therefore \Delta x=2.99-3=-0.01$

$y=x^{3}+5$      (Given)

Differentiating both sides with respect to x, we get

$\frac{d y}{d x}=3 x^{2}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=3}=3 \times(3)^{2}=27$

$\therefore \Delta y=\left(\frac{d y}{d x}\right) \Delta x$

$\Rightarrow \Delta y=27 \times(-0.01)=-0.27$

Thus, the approximate change in y is −0.27.

If $y=x^{3}+5$ and $x$ changes from 3 to $2.99$, then the approximate change is $y$ is __−0.27___.