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August 28th, 2018, 06:23 AM  #1 
Newbie Joined: Aug 2018 From: Guttenberg, NJ Posts: 1 Thanks: 0  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 
August 28th, 2018, 09:41 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907 
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. 
August 28th, 2018, 11:49 AM  #3  
Senior Member Joined: Aug 2012 Posts: 2,136 Thanks: 622  Quote:
There's no general formula, but the pattern is a function that's increasing and whose derivative is decreasing to zero.  
August 28th, 2018, 12:37 PM  #4 
Senior Member Joined: May 2016 From: USA Posts: 1,253 Thanks: 519 
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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