Bizarre function

Algebra Level 5

Let $f:\mathbb{R}^{+} \to \mathbb{R}$ be a function such that

$f$ is strictly increasing
$f(x) > - \dfrac{1}{x} \quad \forall x>0$
$f(x) f \left(f(x)+\dfrac 1x \right) = 1 \quad \forall x>0$

Find the sum of all possible values of $f(1)$ .

Give your answer to 3 decimal places.

The answer is -0.61803398.

1 solution

Sharky Kesa
Jan 4, 2017

Let $a=f(1)$ . From the second condition, $a > - 1$ .

Setting $x=1$ in (c), we get $a f(a+1)=1$ , so $f(a+1)=\frac{1}{a}$ , with $a>-1, a\neq 0$ . Setting $x=a+1$ in (c), we get $f(a+1) f(f(a+1)+\frac{1}{a+1}) = 1$ , so $f(\frac{1}{a} + \frac{1}{a+1}) = a = f(1)$ . However, $f$ is strictly increasing, so $1 = \frac{1}{a}+\frac{1}{a+1}$ . Solving this, we get $a= \frac{1\pm\sqrt{5}}{2}$ .

Checking, we get if $a=\frac{1 + \sqrt{5}}{2}$ , then $1 < a = f(1) < f(1+a) = \frac{1}{a} < 1$ , which is a contradiction. If $a=\frac{1-\sqrt{5}}{2}$ , then we get that this satisfies. Therefore, $f(1)=\frac{1-\sqrt{5}}{2}=0.61803398...$ .

An example of such a satisfying function is $f(x)=\frac{1-\sqrt{5}}{2x}$ .

In the second paragraph, in order to set $x = a + 1$ , because we require $x > 0$ (restriction on domain), in order for the funcation equation to be valid, you are making the implicit assumptinon that $a > -1$ . Now, of course, this is given by (b), and you should write that in the solution for completeness.

Be careful in such functional equations with a restricted domain for substitution (sometimes implicit). It doesn't always imply that we can make such a substitution. As an explicit example, if we were looking for functional equations that satisfy $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ where $f(x^2) = 0$ , this does not imply that $f(x) = 0$ for all $x$ . We could have $f(-1) = -1000$ .

Calvin Lin Staff - 4 years, 5 months ago

Thanks for showing that such a function does exist :)

Calvin Lin Staff - 4 years, 5 months ago

Excellent question! Just one comment on why $a>-1$ . Note that $a>0\implies 1=f(a+1)a>a^2$ , by monotonicity of $f$ , and similarly, $a<0\implies f(1+a)<a\implies 1=af(1+a)>a^2\implies a\in(-1,1)$ . Also, $a\ne 0$ , since otherwise, $a=0\implies f(1+a)=f(1)=a\implies a^2=1$ , which would be absurd.

Samrat Mukhopadhyay - 4 years, 5 months ago

Ah, nice approach. I obtained it directly from the second condition that $f(1) > - 1$ .

With this, it would seem like the second condition isn't necessary, or at least that it's implied by / necessary for the validity of the domain of c.

Calvin Lin Staff - 4 years, 5 months ago

Yes, actually $a>-1$ simply comes from the second condition,which seems to follow anyway from the domain condition of $f$ but that argument I used reveals that $a<1$ .

Samrat Mukhopadhyay - 4 years, 4 months ago

Bizarre function

The answer is -0.61803398.

1 solution

0 pending reports