Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?

Answer 1

Answer:

c =  1

Step-by-step explanation:

Given:-

- The given function f(x) is:

                                f(x) = 3x^2 - 2x + 1 ,   [ 0 , 2 ]

Find:-

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?

f it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem.

Solution:

- The mean value theorem states that if a function f(x) is differentiable over the range [ x1 , x2 ] , then there exist a value c within the range [ x1 , x2 ]. Such that:

                                f'(c) = [ f(x2) - f(x1) ] / [ x2 - x1 ]

- Note: The right hand side of above theorem expresses the " Secant " line.

- We see that the function f(x) is a polynomial function with degree 2 which is continuous and differentiable over the entire interval of real numbers R. For which the differential of the given function is:

                                f'(x) = 6x - 2

- It exist for all real value of x and is continuous ( Linear Line ).

- It satisfies the hypothesis of the mean value theorem. So our function f(x) to be differentiable over the range [ 0 , 2 ]. then there exist a value c within the range [ 0 , 2 ]

                                f'(x) = [ f(2) - f(0) ] / [ 2 - 0 ]

                                f'(x) = [ 3(2)^2 - 2(2) + 1 - 1 ] / [ 2 - 0 ]

                                f'(x) = [ 8 ] / [ 2 ] = 4

                                f'(c) = 6c - 2 = 4

                                c = 6 / 6 = 1

- Hence, the required value of c = 1.

Answer 2

Answer:

7/3

Step-by-step explanation:

f(x) = 3x^2 + 2x + 1, [0, 2]

f(0)=1

f(2)=17

f'(c)=f(2)-f(0)/2-1

f'(c)=17-1/1=16

f'(c)= 6x+2

6x+2=16

6x=16-2

6x=14

x=14/6

x=7/3