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

Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
$f(x+\Delta x)=f(2.01)$
Let $f(x)=2$
$\Delta x=0.01$
$f(2)=4(2)^2+5\times 2 +2$
$\quad\;\;\;=4\times 4+5\times 2+2$
$\quad\;\;\;=16+10+2$
$\quad\;\;\;=28$
Step 2:
$f'(x)=8x+5$[Differentiating with respect to x]
Now $f(x+\Delta x)=f(x)+\Delta f(x)$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=f(x)+f'(x).\Delta x$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+(8x+5).\Delta x$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+(8\times 2+5).\Delta x$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+21\times .0.01$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+0.21$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28.21$
$f(2.01)=28.21$