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6 July, 05:53

Let f be the function given by f (x) = 5cos2 (x2) + ln (x+1) - 3. The derivative of f is given by f′ (x) = -5cos (x2) sin (x2) + 1x+1. What value of c satisfies the conclusion of the Mean Value Theorem applied to f on the interval [1,4]?

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  1. 6 July, 07:15
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    The value of c that satisfies the conclusion of the Mean Value Theorem applied to f on the interval [1,4] is approximately 0.1793.

    Step-by-step explanation:

    Given f (x) = 5cos²x² + ln (x + 1) - 3

    The Mean Value Theorem applied to f on an interval [a, b] is given as

    f' (x) = c = [f (b) - f (a) ] / (b - a)

    = [f (4) - f (1) ] / (4 - 1)

    c = [f (4) - f (1) ]/3

    f (4) = 5 (cos4²) ² + ln (5) - 3

    = 3.2296

    f (1) = 5 (cos1²) ² + ln (2) - 3

    = 2.6916.

    c = (3.2296 - 2.6916) / 3

    = 0.538/3

    ≈ 0.1793
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