logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

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

Can you answer this question?
 
 

1 Answer

0 votes
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 $f(x+\Delta x)=f(x)+\Delta f(x)$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=f(x)+f'(x).\Delta x$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+(8x+5).\Delta x$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+(8\times 2+5).\Delta x$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+21\times .0.01$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28+0.21$
$\qquad\;\;\;\;\;\;\;\;\;\;\;\;\;\;=28.21$
$f(2.01)=28.21$
answered Aug 6, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...