# For the given function find k such that f(x)≠f(x+k) for any value of x

Let $f(x)=\frac{x}{x^2+k}$ then find k such that $f(x)≠f(x+k)$ for any value of x.

In my first attempt, I tried to solve $f(x)=f(x+k)$. But on solving this further, I get a biquadratic(including terms of odd power). Hence, I cannot impose any condition on k from here.

A) Function will be continuous for positive values of k and will be discontinuous for negative values of k.

B) $f(x)=f(x+k)$ means that the function is periodic with period k. If the function is periodic about k, then we can write for integers n that $f(x-nk)=f(x+nk)$ .

But I don't know where to go with this analysis.

How to solve this question?

• Using the IVT for $g(x)=f(x+k)-f(x)$ you can prove that $k\notin (-\infty,-4)\cup (0,\infty)$ and I think it could hold for $k\in[-4,0)$, but I haven't have no proof yet. Your observation $B)$ is wrong, since for no $k$ the function $f$ is periodic. The negation of the statement just say, that there exists $x$ such that $f(x+k)=f(x)$ and not that $f(x+k)=f(x)$ for all $x$ which could imply the periodicity. Commented Aug 7, 2017 at 7:12
• "But on solving this further, I get a biquadratic(including terms of odd power). Hence, I cannot impose any condition on k from here." -- The entire problem is solving a simple quadratic. Commented Aug 7, 2017 at 7:29

$f(x+k)=\dfrac{k+x}{(k+x)^2+1}$

$f(x+k)=f(x)$ if $\dfrac{k+x}{(k+x)^2+k}=\dfrac{x}{k+x^2}$

which is $$\dfrac{k+x}{(k+x)^2+k}-\dfrac{x}{k+x^2}=0$$ Together $$\frac{-kx^2+k^2-k x^2}{\left(k+x^2\right) \left(k^2+2 k x+k+x^2\right)}=0$$ which is verified if the numerator is zero $$k \left(x^2+k x-k\right)=0$$ That is $x=\dfrac{-k\pm\sqrt{k^2+4k}}{2}$

We want NO solutions for this equation, so we put

$k^2+4k<0$ that is $-4<k<0$

for $k\in(-4,\;0)$ doesn't happen that $f(x+k)=f(x)$

Hope this helps

PS

also for $k=-4$ doesn't happen that $f(x-4)=f(x)$

so we can say

for $k\in[-4,\;0)$ doesn't happen that $f(x+k)=f(x)$

for $k=0$ we get the same function so the are identical