In this page Lagrange theorem questions solution2 we are going to see solution of the practice questions.
(iii) f (x) = 2 x³ + x² - x - 1 , [0 , 2]
Solution:
If f (x) be a real valued function that satisfies the following conditions.
1. |
f(x) is defined and continuous on the closed interval [0,2] |
2. |
f(x) is differentiable on the open interval (0,2). |
Then there exists at least one point c ∊ (0,2) such that f ' (c) = f (b) - f (a) / (b - a) |
f (x) = 2 x³ + x² - x - 1
f ' (x) = 3 (2x²) + 2 x - 1 - 0
f ' (x) = 6 x² + 2 x - 1
f ' (c) = 6 c² + 2 c - 1
f (0) = 6 (0)² + 2 (0) - 1
= 0 + 0 - 1
f (0) = - 1
f (2) = 2 x³ + x² - x - 1
= 2 (2)³ + 2² - 2 - 1
= 2 (8) + 4 - 3
= 16 + 4 - 3
= 17
f ' (c) = f (b) - f (a) / (b - a)
= [17 - (-1)]/(2 - 0)
= [17 + 1]/2
= 18/2
= 9
6 c² + 2 c - 1 = 9
6 c² + 2 c - 1- 9 = 0
6 c² + 2 c - 10 = 0
now we are going to divide the entire equation by 2
3 c² + c - 5 = 0
We cannot solve this equation by using factorization method. So we are trying to solve this equation by using quadratic formula.
a = 3, b = 1, c = -5
x = - 1 ± √[1² - 4(3)(-5)]/2(3)
x = - 2 ± √[1 + 60]/6
x = - 2 ± √61/6
x = (- 2 + √61)/6 , (- 2 - √61)/6
c = (- 2 + √61)/6 ∈ (0,2) Lagrange theorem questions solution2 Lagrange theorem questions solution2