# Find the approximate value of

Question:

Find the approximate value of $f(5.001)$, where $f(x)=x^{3}-7 x^{2}+15$.

Solution:

Let $x=5$ and $\Delta x=0.001$. Then, we have:

$f(5.001)=f(x+\Delta x)=(x+\Delta x)^{3}-7(x+\Delta x)^{2}+15$

Now, $\Delta y=f(x+\Delta x)-f(x)$

$\therefore f(x+\Delta x)=f(x)+\Delta y$

$\approx f(x)+f^{\prime}(x) \cdot \Delta x \quad($ as $d x=\Delta x)$

$\begin{array}{rlr}\Rightarrow f(5.001) & \approx\left(x^{3}-7 x^{2}+15\right)+\left(3 x^{2}-14 x\right) \Delta x & \\ & =\left[(5)^{3}-7(5)^{2}+15\right]+\left[3(5)^{2}-14(5)\right](0.001) \quad[x=5, \Delta x=0.001] \\ & =(125-175+15)+(75-70)(0.001) & \\ & =-35+(5)(0.001) & & \\ & =-35+0.005 & & \\ & =-34.995 & \end{array}$

Hence, the approximate value of f (5.001) is −34.995.