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1).Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = 5x2 − 3x + 2, [0, 2]
Yes, it does not matter if f is continuous or differentiable, every function satifies the Mean Value Theorem.
Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on double-struck R.
No, f is not continuous on [0, 2].
No, f is continuous on [0, 2] but not differentiable on (0, 2).
There is not enough information to verify if this function satifies the Mean Value Theorem.

If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisify the hypotheses, enter DNE).
c =

2).Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = ln x, [1, 7]
Yes, it does not matter if f is continuous or differentiable, every function satisfies the Mean Value Theorem.
Yes, f is continuous on [1, 7] and differentiable on (1, 7).
No, f is not continuous on [1, 7].
No, f is continuous on [1, 7] but not differentiable on (1, 7).
There is not enough information to verify if this function satisfies the Mean Value Theorem.

If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).
c=

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1 answer
  1. #1 The MVT requires that f(x) be continuous and differentiable. Luckily, all polynomials meet these requirements.

    f(x) = 5x^2-3x+2
    f'(x) = 10x-3
    f(0) = 2
    f(2) = 16
    So, we want c where f'(c) = 7
    10x-3 = 6
    x = 0.9
    The tangent line at x=0.9 is
    y-3.35 = 7(x-.9)
    y = 7x-2.95

    http://www.wolframalpha.com/input/?i=plot+y%3D5x^2-3x%2B2%2C+y%3D7x-2.95

    #2
    lnx satisfies the MVT on [1,7]
    f(1) = 0
    f(7) = ln7
    so the slope of the secant is ln7 / 6
    so, we want f'(x) = 1/x = ln7/6
    c = 6/ln7 = 3.083, which is in [1,7]
    f(6/ln7) = 1.126
    So, the tangent line is
    y-1.126 = (ln7/6)(x-3.083)

    http://www.wolframalpha.com/input/?i=plot+y%3Dln+x+%2C+y%3D+%28ln7%2F6%29%28x-3.083%29%2B1.126+for+x%3D1..7

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