In this page Lagrange theorem questions solution1 we are going to see solution of the practice questions.
(1) Verify La-grange's law of mean for the following functions:
(i) f (x) = 1 - x², [ 0 , 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 [0,3] |
2. |
f(x) is differentiable on the open interval (0,3). |
Then there exists at least one point c ∊ (0,3) such that f ' (c) = f (b) - f (a) / (b - a) |
f (x) = 1 - x²
f ' (x) = 0 - 2 x
f ' (x) = - 2 x
f ' (c) = - 2 c
f (0) = 1 - 0²
= 1
f (3) = 1 - 3²
= 1 - 9
= - 8
f ' (c) = f (b) - f (a) / (b - a)
= (- 8 - 1)/(3 - 0)
= - 9/3
2 c = -3
c = -3/2 Lagrange theorem questions solution1
(ii) f (x) = 1/x , [ 1 , 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 [1,2] |
2. |
f(x) is differentiable on the open interval (1,2). |
Then there exists at least one point c ∊ (1,2) such that f ' (c) = f (b) - f (a) / (b - a) |
f (x) = 1/x
f (x) = x⁻¹
f ' (x) = (-1) x⁻²
f ' (x) = - 1/x²
f ' (c) = - 1/c²
f (1) = 1/1
= 1
f (2) = 1/2
f ' (c) = f (b) - f (a) / (b - a)
= (1/2) - 1/(2 - 1)
= (-1/2)/1
= (-1/2)
- 1/c² = -1/2
c² = 2
c = ± √2
c = √2 , - √2
- √2 ∉ (1,2) but √2 ∈ (1,2). So the required value of c is √2.