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

f (x) = 3x^2 âˆ' 2x + 1, [0, 2]

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

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.

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