Find the approximate value of f (5.001),

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, $y=f(x)$

$\Rightarrow \frac{d y}{d x}=3 x^{2}-14 x$

$\therefore d y=\Delta y=\frac{d y}{d x} d x=\left(3 x^{2}-14 x\right) \times 0.001=(75-70) \times 0.001=0.005$

$\therefore f(5.001)=y+\Delta y=-35+0.005=-34.995$

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