In this page mean value theorem questions1 we are going to see some practice questions.
(iii) f (x) = |x - 1|, 0 ≤ x ≤ 2
If f(x) be a real valued function that satisfies the following three conditions.
1. |
f(x) is defined and continuous on the closed interval [0,2] |
2. |
f(x) is not differentiable on the open interval (0,2). |
Since the given function is not satisfying all the conditions Rolle's theorem is not admissible. |
(iv) f (x) = 4 x³ - 9 x, -3/2 ≤ x ≤ 3/2
If f(x) be a real valued function that satisfies the following three conditions.
1. |
f(x) is defined and continuous on the closed interval [-3/2,3/2] |
2. |
f(x) is not differentiable on the open interval (-3/2,3/2). |
f (X) = 4 x³ - 9 x f (-3/2) = 4(-3/2)³ - 9 (-3/2) = 4 (-27/8) + 27/2 = -27/2 + 27/2 = 0 f (3/2) = 4(3/2)³ - 9 (3/2) = 4 (27/8) - 27/2 = 27/2 - 27/2 = 0 f (-3/2) = f (3/2) from this we come to know that the given function satisfies all the conditions of Rolle's theorem. c ∈ (-3/2,3/2) we can find the value of c by using the condition f '(c) = 0. f (x) = 4 x³ - 9 x f '(x) = 12 x² - 9 (1) = 12 x² - 9 f '(c) = 12 c² - 9 f '(c) = 0 12 c² - 9 = 0 12 c² = 9 c² = 9/12 c = √(9/12) c = √(3/4) c = ± √3/2 mean value theorem questions1 |
(2) Using Rolle's theorem find the points on the curve y = x² + 1,
- 2 ≤ x ≤ 2 where the tangent is parallel to x - axis.
If f(x) be a real valued function that satisfies the following three 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). |
y = x² + 1 f (-2) = (-2)² + 1 = 4 + 1 = 5 f (2) = (2)² + 1 = 4 + 1 = 5 f (-2) = f (2) from this we come to know that the given function satisfies all the conditions of Rolle's theorem. c ∈ (-2,2) we can find the value of c by using the condition f '(c) = 0. f (x) = x² + 1 f '(x) = 2 x + 0 = 2 x f '(c) = 2 c f '(c) = 0 2 c = 0 c = 0/2 c = 0 f (0) = 0² + 1 = 1 Therefore the required point on the curve is (0,1). mean value theorem questions1 |