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Math Help - Mean Value Theorem

  1. #1
    Junior Member
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    Mean Value Theorem

    Hello!

    I need some help please with two revision questions.
    I read my Calculus book and I did not understand it!

    Q1.
    Suppose that 3<= f ' (x) <= 5 for all values of x. Determine the highest and lowest possible values for the expression A = f (8) - f (2).

    Am I meant to use "MVT" ??


    Q2.
    For what values of the numbers a and b does the function f(x) = axe^bx
    have the maximum value f(2) = 1? Give reasons.


    Sorry this I do not know.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    f'(x)={f(x)-f(y)}/{x-y}

    for the second one:

    put (2,1) into f(x).

    this is your first equation with a and b.

    now,
    find f'(x).

    f'(x)=0

    put x=1.

    And there you have second equation.

    solve system of equations finding a and b.
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  3. #3
    MHF Contributor

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    Quote Originally Posted by Jon123 View Post
    Hello!

    I need some help please with two revision questions.
    I read my Calculus book and I did not understand it!

    Q1.
    Suppose that 3<= f ' (x) <= 5 for all values of x. Determine the highest and lowest possible values for the expression A = f (8) - f (2).

    Am I meant to use "MVT" ??
    That seems most reasonable to me (f(8)- f(2))/(8- 2)= (f(8)- f(2))/6 is equal to the derivative of f at some point- but you know that can never be larger than 5 or less than 3.

    Q2.
    For what values of the numbers a and b does the function f(x) = axe^bx
    have the maximum value f(2) = 1? Give reasons.
    At a max or min, the derivative will be 0. f'(x)= ae^{bx}+ abxe^{bx}. Where is that equal to 0? What is the value of f there? What must a and b equal in order that that result be 1?

    Sorry this I do not know.
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