Are there functions that satisfy all following properties?
Domain: $x>0$
$f'(x)>0$
$f''(x)<\frac{-2f'(x)}{x}$
Could anyone give me an example of such a function?
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Sign up to join this communityAre there functions that satisfy all following properties?
Domain: $x>0$
$f'(x)>0$
$f''(x)<\frac{-2f'(x)}{x}$
Could anyone give me an example of such a function?
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 $f(x)=-\frac 1 {2x^{2}}$
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!