Find the approximate value of


Find the approximate value of $f(2.01)$, where $f(x)=4 x^{2}+5 x+2$



Let $x=2$ and $\Delta x=0.01$. Then, we have:

$f(2.01)=f(x+\Delta x)=4(x+\Delta x)^{2}+5(x+\Delta x)+2$

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 $d x=\Delta x)$

$\Rightarrow f(2.01) \approx\left(4 x^{2}+5 x+2\right)+(8 x+5) \Delta x$

$=\left[4(2)^{2}+5(2)+2\right]+[8(2)+5](0.01) \quad[$ as $x=2, \Delta x=0.01]$





Hence, the approximate value of f (2.01) is 28.21.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now