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 

             = 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