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Are there functions that satisfy all following properties?

1)Domain: x>0

2)f'(x)>0

3) f''(x)<$\frac{-2f'(x)}{x}$

Could anyone give me an example of such a function?

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  • $\begingroup$ You should tell us what you tried and where you are stuck. This is how the site works. $\endgroup$
    – user65203
    Jun 13 '18 at 6:10
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Let $g(x)=f'(x)$ and consider $(x^{2}g(x))'=2xg(x)+x^{2}g'(x)=x\{2g(x)+xg'(x)\}=x\{2f'(x)+xf''(x)\}$. Condition 3) becomes $(x^{2}g(x))'<0$. Equivalently, $x^{2}g(x)$ is a positive strictly decreasing function. There is no other condition, so the answer is $f(x)=c+\int g(x)\, dx=c+\int \frac {h(x)} {x^{2}} \, dx$ where $h$ is an arbitatry strictly decreasing positive function on $(0,\infty)$ and $c$ is a constant. For a specific example take $h(x)=\frac 1 x$ which gives $f(x)=-\frac 1 {2x^{2}}$

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Hint: Try focusing on $g(x)=f'(x)$. You want $g(x)>0$, and $g'(x)<-\frac{2}{x}g(x)$.

How about $g(x)=x^{-k}$ for $k>...$? Then, "backtracking" to obtain $f(x)$ should be easy!

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  • $\begingroup$ I tried this before, but failed to find the way to make it work. $\endgroup$
    – Yhqjlyr
    Jun 13 '18 at 4:52
  • $\begingroup$ @Lyryhqj The hint is now easier! $\endgroup$
    – Anonymous
    Jun 13 '18 at 5:11

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