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Q.
If $f ( x )= x ^{3}-7 x ^{2}+15$, then the approximate value of $f(5.001)$ is
Application of Derivatives
Solution:
Consider $f ( x )= x ^{3}-7 x ^{2}+15 $
$\Rightarrow f'( x )=3 x ^{2}-14 x$
Let $x =5$ and $\Delta x =0.001$
Also, $f ( x +\Delta x ) \approx f ( x )+\Delta x f'( x )$
Therefore, $f ( x +\Delta x )=\left( x ^{3}-7 x ^{2}+15\right)+\Delta x \left(3 x ^{2}-14 x \right)$
$\Rightarrow f (5.001)=\left(5^{3}-7 \times 5^{2}+15\right)+\left(3 \times 5^{2}-14 \times 5\right)(0.001)$
$($ as $x =5, \Delta x =0.001)$
$=125-175+15+(75-70)(0.001)$
$=-34.995$