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Math Help - 3 question new

  1. #1
    key
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    3 question new

    The other thread wasn't working so here are the problems again . Please help me and my brother this is due towmorrow for him.

    1.
    a) Let f(x) = 2x Find the first differences between the values of f for x= 0,1,2,3,4, and 5. Write a formula that gives the difference between f(x+1) and f(x) in terms of x.
    b) Repeat part (a) but use function g(x)=3x
    c) Repeat part (a) but use the function h(x) = 5x
    d) Make a conjecture for a general formula that gives the difference between f(x+1) and f(x) for the function k(x) = ax , for a > 1.
    2. The average growth rate of exponential function f(x) on an interval a≤x≤b is f(b) f(a)
    b-a
    The instantaneous growth rate at a is the number that this fraction approaches (if any) as b → a. Let f(x)=3x, and let b=a+h. Write the instantaneous growth rate at a as the product of 3a and a function of h (that does not involve a). Use a calculator to estimate the number this second factor approaches as h → 0.

    3. Let f(x) =bx . For a fixed positive integer n, let g(x)= [f(x)]n , and let h(x) = f(f(f((x))))

    a) Express g(x) as f(?)
    b) Express h(x) as f(?)
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  2. #2
    Super Member wingless's Avatar
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    Let's see.
    f(x) = 2x
    Make a table:
    Code:
      _x_      _f(x)_
       0          0          f(0) = 0
       1          2          f(1) = 2
       2          4          f(2) = 4
       3          6          f(3) = 6
       4          8          f(4) = 8
       5          10         f(5) = 10
    So you can find the differences from here,
    f(1) - f(0) = 2 - 0 = 2
    f(2) - f(1) = 4 - 2 = 2
    f(3) - f(2) = 6 - 4 = 2
    ...
    As you see, all differences of f(x)s for consecutive x's are 2. But this is just a basic deduction, it's not proven yet.

    a) In a function (for example, f(x) = 3x + 4) you can replace all x's with anything you want.

    In your function ( f(x) = 2x ) we will replace all x's with x+1 to get a function for f(x+1).
    f(x) = 2x
    f(x+1) = 2(x+1)
    f(x+1) = 2x + 2
    Now we have functions f(x) = 2x and f(x+1) = 2x + 2

    "Write a formula that gives the difference between f(x+1) and f(x) in terms of x"
    So we are looking for f(x+1) - f(x)
    f(x+1) - f(x) = (2x + 2) - (2x)
    f(x+1) - f(x) = 2
    Aha, all differences for consecutive x's are 2, we proved it now.
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  3. #3
    key
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    can someone help with questions 2 and 3
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  4. #4
    Super Member wingless's Avatar
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    "The instantaneous growth rate at a is the number that this fraction approaches (if any) as b → a. Let f(x)=3x, and let b=a+h."

    Is it f(x) = 3x or f(x) = 3^x ?
    I'm asking because it says "The average growth rate of exponential function f(x)..."
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  5. #5
    key
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    it's
    sorry

    Ok i just looked at the problem and everywhere where it says 2x or 3x or bx it's supposed to be or whatever the number is
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  6. #6
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    can someone help please
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  7. #7
    Super Member wingless's Avatar
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    In 2. question, there's the formula for finding instantaneous growth rate. The limit of this formula is called derivative.


    Instantaneous growth rate at the point a is,
    \lim_{h \to 0 } \frac {f(a+h) - f(a)}{h}

    You can't calculate this exactly without knowing how to find limits, but you can get approximate results by giving h approximate values.

    Your function is f(x) = 3^x
     \frac {f(a+h) - f(a)}{h}
     \frac {3^{a+h} - 3^a}{h}
    In here, h is a value near 0 (nearer, more accurate)

    If you want to calculate the instantaneous growth rate at x = a = 1,
     \frac {3^{1+h} - 3^1}{h}
    Let's say h = 0.001
    Then  \frac {3^{1+h} - 3^1}{h}
     \frac {3^{1 + 0.001} - 3^1}{0.001}
    = 3,297..

    Attached Graph:
    f(x) = 3^x
    Attached Thumbnails Attached Thumbnails 3 question new-3-x.jpg  
    Last edited by wingless; December 10th 2007 at 01:44 PM.
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  8. #8
    key
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    thank you and can you please please help me with 3
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  9. #9
    Super Member wingless's Avatar
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    3. Let f(x) =bx . For a fixed positive integer n, let g(x)= [f(x)]n , and let h(x) = f(f(f((x))))

    a) Express g(x) as f(?)
    b) Express h(x) as f(?)

    f(x) = b^x
    g(x) = n f(x)
    h(x) = f(f(f(...x...)))

    (Is it g(x) = n.f(x) or [f(x)]^n? I made it for the first)
    g(x) = n f(x)
    g(x) = n b^x
    g(x) = b^{\log _bn}.b^x
    g(x) = b^{\log _bn + x}
    g(x) = f(\log _bn + x)

    h(x) = f(f(f(...x...)))
    h(x) = f(h(x))
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  10. #10
    key
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    it's for the second thank you so much for your help but yay it's for the second
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