Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the approximate value of f (2.01), where f (x) = 4 x ^ 2 + 5 x + 2.

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

Answers (1)

best_answer

Let x = 2 and \Delta x = 0.01
f(x+\Delta x) = 4(x+\Delta x)^2 +5(x+\Delta x)+2
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (8x+5).(0.01) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 4x^2+5x+2 \ and \ \Delta x = 0.01)\\ dy = 0.08x+0.05
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.08x + 0.05 + 4x^2 + 5x +2\\ f(x+\Delta x) = 0.08(2)+0.05+4(2)^2+5(2)+2\\ f(x+\Delta x) = 0.16 + 0.05 + 16 + 10 + 2\\ f(x+\Delta x) = 28.21
Hence, the approximate value of f (2.01), where f (x) = 4x^2 + 5x + 2. is 28.21

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question