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3 April, 16:23

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.

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Answers (2)
  1. 3 April, 16:32
    0
    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
  2. 3 April, 16:53
    0
    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.
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