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August 28th, 2018, 06:23 AM   #1
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Question Formula of a decreasing upward curve

Hi!

For a graph of a hypothetical stock price where the price is increasing but at a decreasing rate of change, I need the general formula for the attached graph.

For example, the formula of an increasing upward curve is f(x) = x^n, where n is a value such as 1.5.

Thank you!

Doctor TR
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August 28th, 2018, 09:41 AM   #2
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Do you want the curve to level off as x increases or to have no upper limit?

Whichever way you answer, some further information is needed.
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August 28th, 2018, 11:49 AM   #3
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Quote:
Originally Posted by DoctorTR View Post
For a graph of a hypothetical stock price where the price is increasing but at a decreasing rate of change, I need the general formula for the attached graph.
The classic example is $f(x) = \log x$, where $\log$ is the natural logarithm. It's increasing, but ever more slowly. Its derivative is $\frac{1}{x}$, which goes to zero as $x$ gets large. In other words the graph of $\log x$ keeps increasing, but it gets more and more flat as you move to the right.

There's no general formula, but the pattern is a function that's increasing and whose derivative is decreasing to zero.
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August 28th, 2018, 12:37 PM   #4
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Another general formula is

$\dfrac{ax}{x + b}, \text { where } a \text { and } b \text { are constants such that } a > 0 < b.$

That will increase with x.

$h > 0 \implies \dfrac{a(x + h)}{(x + h) + b} - \dfrac{ax}{x + b} = \dfrac{ax + ah}{x + b + h} - \dfrac{ax }{x + b} =$

$\dfrac{(ax + ah)(x + b) - ax(x + b + h)}{(x + b + h)(x + b} =$

$\dfrac{ax^2 - ax^2 + abx - abx + ahx - ahx + ahb}{(x + b + h)(x + b)} = \dfrac{ahb}{(x + b + h)(x + b)}.$

But a, b, and h are all positive. So if x is not negative

$\dfrac{a(x + h)}{(x + h) + b} - \dfrac{ax}{x + b} > 0 \implies \dfrac{a(x + h)}{(x + h) + b} > \dfrac{ax}{x + b}.$

Unlike the function in post 3, which increases without limit, this function gradually approaches a but never reaches it.
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