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I'm looking for a concave down increasing-function, see the image in the right lower corner.

Basically I need a function f(x) which will rise slower as x is increasing. The x will be in range of [0.10 .. 10], so f(2x) < 2*f(x) is true. Also if

I would also like to have some constants which can change the way/speed the function is concaving.

Any suggestion on a formula?

Concave up/down graphs

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    $\begingroup$ $\log(x)$ is a common one $\endgroup$
    – robjohn
    Jan 4, 2017 at 15:21
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    $\begingroup$ What about $\log_a (x)$ ? $\endgroup$
    – VHarisop
    Jan 4, 2017 at 15:22

5 Answers 5

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If you are restricted to a positive $x$, you could use

$$ f(x) = a \sqrt{x}, \quad a > 0 $$

which satisfies

$$ f(2 \cdot x) = a \sqrt{2 \cdot x} = \sqrt{2} \cdot a \sqrt{x} = \sqrt{2} \cdot f(x) < 2 \cdot f(x) $$

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\begin{equation} f(x) = -\frac{a}{x^b} \end{equation} where $a$ is a positive number and $b>0$.

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I think natural logarithm is a good function for your purpose. It's increasing and have low-rate slope, lower than your diagram. Let $f(x)=a+b\ln x$, then $f(2x)<2f(x)$ concludes that $\displaystyle\ln\frac{2}{x}<\frac{a}{b}$. In other hand for $0.1<x<10$ gives us $$-1.6<\ln\frac{2}{x}<3$$ so we must set $\displaystyle3<\frac{a}{b}$ and we have $0<3b<a$.

For example set $a=4$ and $b=1$ which satisfy our condition, so $f(x)=4+\ln x$ has the condition $f(2x)<2f(x)$ with below diagram in $[0.1,10]$:

$f(x)=4+\ln x$

Below graph shows $f(2x)<2f(x)$ in $[0.1,10]$:

enter image description here

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$\log(x)$ comes to mind, but we also need to consider $f(2x)\lt2f(x)$. If we set $f(x)=\log(20x)$, then, since $x\gt0.1$, $$ \begin{align} f(2x) &=\log(40x)\\ &=\log(2)+\log(20x)\\ &=\log(2)+f(x)\\ &\lt f(x)+f(x)\\ &=2f(x) \end{align} $$

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Constant $a$ in

$$ 1- e^{-x/a}$$

changes climb rate as desired. ( Twice the function value more than function with double argument).

2 curves

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