## Lagrange Theorem Questions Solution2

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