In this page Lagrange theorem questions solution3 we are going to see solution of the practice questions.
(iv) f (x) = x ^(2/3) , [-2 , 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 [-2,2] |
2. |
f(x) is not differentiable on the open interval (-2,2). |
Hence Lagrange theorem does not exists. |
(v) f (x) = x³ - 5 x² - 3 x , [1 , 3]
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 [1,3] |
2. |
f(x) is differentiable on the open interval (1,3). |
Then there exists at least one point c ∊ (1,3) such that f ' (c) = f (b) - f (a) / (b - a) |
f (x) = x³ - 5 x² - 3 x
f ' (x) = 3 x² - 5 (2 x) - 3
f ' (x) = 3 x² - 10 x - 3
f ' (c) = 3 c² - 10 c - 3
f (1) = x³ - 5 x² - 3 x
= (1)³ - 5 (1)² - 3 (1)
= 1 - 5 - 3
= - 7
f (1) = - 7
f (3) = x³ - 5 x² - 3 x
= (3)³ - 5 (3)² - 3 (3)
= 27 - 45 - 9
= 27 - 54
= -27
f (3) = - 27
f ' (c) = f (b) - f (a) / (b - a)
= [-27 - (-7)]/(3 - 1)
= [-27 + 7]/2
= -20/2
= -10
3 c² - 10 c - 3 = -10
3 c² - 10 c - 3 + 10 = 0
3 c² - 10 c + 7 = 0
3 c² - 3 c - 7 c - 7 = 0
3 c (c - 1) - 7 (c - 1) = 0
(c - 1) (3 c - 7) = 0
c - 1 = 0 3 c - 7 = 0
c = 1 3 c = 7
c = 7/3 Lagrange theorem questions solution3