## Mean Value Theorem Questions1

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               = 0f (3/2) = 4(3/2)³ - 9 (3/2)             = 4 (27/8) - 27/2             = 27/2 - 27/2               = 0f (-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 - 9f '(c) = 12 c² - 9f '(c) = 012 c² - 9 = 012 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          = 5f (2) = (2)² + 1          = 4  + 1          = 5f (-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 xf '(c) = 2 cf '(c) = 02 c = 0   c = 0/2   c = 0f (0) = 0² + 1        = 1Therefore the required point on the curve is (0,1). mean value theorem questions1