[jacks]

A positive real values function satisfied \( \displaystyle \frac{f(x)}{f(y)}\leq 7^{(x-y)^2} \) for all \( x,y\in  \) Domain and \( \displaystyle f\bigg(\frac{3}{2}\bigg)=\frac{\pi}{4} \). Then \( f(x) \) is

[Luis Fuentes]

Hi

A positive real values function satisfied \( \displaystyle \frac{f(x)}{f(y)}\leq 7^{(x-y)^2} \) for all \( x,y\in  \) Domain and \( \displaystyle f\bigg(\frac{3}{2}\bigg)=\frac{\pi}{4} \). Then \( f(x) \) is

Spoiler

Since \( f(x) \) is  a positive function we can take \( g(x)=log_7(f(x)) \), that is, \( f(x)=7^{g(x)}. \)

Thus:

 \( \displaystyle \frac{f(x)}{f(y)}\leq 7^{(x-y)^2}\quad \Leftrightarrow{}\quad 7^{g(x)-g(y)}\leq 7^{(x-y)^2}\quad \Leftrightarrow{}\quad g(x)-g(y)\leq (x-y)^2 \)

and \( g(y)-g(x)\leq (y-x)^2=(x-y)^2 \)

From this \( |g(x)-g(y)|\leq (x-y)^2 \) and we can prove that \( g(x) \) is differentiable with zero derivative:

\( 0\leq \displaystyle\lim_{x \to y}{}\dfrac{|g(x)-g(y)|}{|x-y|}\leq \displaystyle\lim_{x \to y}{}\dfrac{|x-y|^2}{|x-y|}=\displaystyle\lim_{x \to y}|x+y|=0. \)

We deduce that \( g(x) \) and \( f(x)=7^{g(x)} \) are constant. In particular if \( \displaystyle f\bigg(\frac{3}{2}\bigg)=\frac{\pi}{4} \) then:

\( f(x)=\dfrac{\pi}{4}\qquad \forall x\in domain \).

[cerrar]

Best regards.

[jacks]

Thanks Admin.