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Exponential Growth Functions

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the exponential function f(x)= ab^(x)

    a) Show that (f(x+1))/f(x) = b

    b)Use the results from part (a) to explain why there is no exponential function of the form f(x)= ab^(x) whose graph passes through the points (0,4) (1,4) (2,8) (3,24) and (4,72)

    2. Relevant equations

    -None-

    3. The attempt at a solution

    There is no attempt at part (a) becuase I couldn't find a way to start it, but for part (b) I am thinking it can't be a exponential function becuase the y-values for two of the x-values are the same. That's what I think, except I can't put it togethor the way part (b) wants me to.

    Please Help.
     
  2. jcsd
  3. Jan 5, 2012 #2

    eumyang

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    You are given
    [itex]f(x) = ab^x[/itex]
    Are you familiar with function notation? What goes inside the parentheses next to f can indicate what you're plugging in. For example,
    [itex]f(5) = ab^5[/itex]
    [itex]f(\text{Mickey Mouse}) = ab^\text{Mickey Mouse}[/itex]
    Given that, what is
    [itex]f(x+1) ?[/itex]
    Once you find that, then substitute into
    [tex]\frac{f(x+1)}{f(x)}[/tex]
    Suppose (0, 4) is on the graph of f(x). This is saying that
    [itex]f(0) = 4[/itex]
    Suppose the other points are on the graph of f(x) as well. What is
    [itex]f(1) ?[/itex]
    [itex]f(2) ?[/itex]
    ...
    Can you see how this relates to the expression
    [tex]\frac{f(x+1)}{f(x)} ?[/tex]
     
  4. Jan 5, 2012 #3
    Thanks, I know get part (a), I was misunderstanding the problem before. It has to be like (ab^(x+1))/(ab^(x)) and then you show simplification. :) :)

    Sorry about that whole thing, I understand it now, but how do I take this and put it with part (b)?
     
    Last edited: Jan 5, 2012
  5. Jan 5, 2012 #4

    eumyang

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    If you mean
    [tex]\frac{ab^x + 1}{ab^x}[/tex]
    ... then sorry, that's not right. Is the "+1" really supposed to be separate from abx?
     
  6. Jan 5, 2012 #5

    eumyang

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    In (b), in order for the points to lie on the graph of f(x) = abx, then for any x,
    [itex]\frac{f(x+1)}{f(x)}[/itex]
    has to equal the same base b. So what is
    [itex]\frac{f(1)}{f(0)} ?[/itex]
    [itex]\frac{f(2)}{f(1)} ?[/itex]
    And so on.
     
  7. Jan 5, 2012 #6
    I don't understand what you are saying...

    The question:

    b)Use the results from part (a) to explain why there is no exponential function of the form f(x)= ab^(x) whose graph passes through the points (0,4) (1,4) (2,8) (3,24) and (4,72)

    From the points, which is the base? I understand you take the points and plug it into the equation, but I don't get anything further then that. For example, for the point (0,4). I would do this:

    y=b^x
    4=b^0
    4≠1

    Is that right to show that it is not a function becuase the values are unequal.

    ^^Thats probably wrong but what do I plug into the (f(x+1))/(f(x)) =b ????? <--- Thats where I am getting confused
     
    Last edited: Jan 5, 2012
  8. Jan 5, 2012 #7
    bumping this becuase its the last problem and I am getting irritated :(
     
  9. Jan 6, 2012 #8

    eumyang

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    Why did you plug into y = bx? The function is in the form of y = abx. Anyway, that's not what I would do.

    Let's look at another exponential function: y = 5*2x. Some of the points are these:
    (0, 5), (1, 10), (2, 20), (3, 40), (4, 80)...
    For an exponential function f, and for any x,
    [itex]\frac{f(x+1)}{f(x)} = b[/itex].
    Take the ratio of successive y-values, like thus:
    [itex]\frac{f(1)}{f(0)} = \frac{10}{5} = 2[/itex]
    [itex]\frac{f(2)}{f(1)} = \frac{20}{10} = 2[/itex]
    [itex]\frac{f(3)}{f(2)} = \frac{40}{20} = 2[/itex]
    [itex]\frac{f(4)}{f(3)} = \frac{80}{40} = 2[/itex]
    ...
    Note that they all equal 2, and that is the base of the exponential function y = 5*2x.

    Now look at the points you were given:
    (0,4) (1,4) (2,8) (3,24) and (4,72)
    and find
    [itex]\frac{f(1)}{f(0)} = ?[/itex]
    [itex]\frac{f(2)}{f(1)} = ?[/itex]
    [itex]\frac{f(3)}{f(2)} = ?[/itex]
    [itex]\frac{f(4)}{f(3)} = ?[/itex]
    Do these simplify to the same number?
     
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