Question 8 (MCQ) - Finding approximate value of function - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Finding approximate value of function
Finding approximate value of function
Last updated at Dec. 16, 2024 by Teachoo
Question 8 If f(x) = 3x2 + 15x + 5, then the approximate value of f (3.02) is (A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66Let x = 3 and ∆x = 0.02 f’(x) = 6x + 15 Now, ∆y = f’(x) ∆x = (6x + 15) 0.02 Also, ∆y = f (x + ∆x) − f(x) f (x + ∆x) = f (x) + ∆y f (3.02) = 3x2 + 15x + 5 + (6x + 15) (0.02) Since x = 3 f (3.02) = 3(3)2 + 15(3) + 5 + (0.02)[6(3)+15] = (27 + 45 + 5) + 33(0.02) = 77 + 0.66 = 77.66 Hence, part (D) is the correct answer.