Question

Find the approximate value of $f\left(2.01\right)$, where $f\left(x\right)=4x^2+5x+2$.

Answer 1

Let $x=2$ and $\Delta x=0.01$.
Now, $\Delta y=f(x+\Delta x)-f\left(x\right)$
$f(x+\Delta x)=f\left(x\right)+\Delta y$
$\simeq f\left(x\right)+f^{\prime }\left(x\right)\cdot \Delta x$, as $(dx\simeq \Delta x)$
$\Rightarrow f\left(2.01\right)\approx (4x^2+5x+2)+(8x+5)\Delta x$
$=\left[4\right(2{)}^2+5\left(2\right)+2]+[8\left(2\right)+5\left]\right(0.01)$
$=28+\left(21\right)\left(0.01\right)=28+0.21=28.21$
Hence, the approximate value of $f\left(2.01\right)$ is $\mathrm{28.21.}$