Is $f(-x) = 1/f(x)$ under these conditions? Let $f\colon \mathbb{R}\to \mathbb{R}$, so that:


for $x\in \mathbb{R}$, $f(x)&gt;0$
for $x,y \in \mathbb{R}$, $f(x+y) = f(x)f(y)$


Prove that $f(-x) = 1/f(x)$.
Using your second condition, you get

$$f(0)=f(x+(-x))=f(x)f(-x)$$
Dividing both sides by $f(x)$
$$\frac{f(0)}{f(x)}=f(-x)$$

Now, what is $f(0)$? Let's use that condition again:

$$f(x)=f(x+0)=f(x)f(0)$$
hence, dividing both sides by $f(x)$, it's evident that $f(0)=1$.

Substituting above, we get

$$\frac{1}{f(x)}=f(-x)$$
If we want to solve this problem in the straightforward way, start by putting $y = -x$, then

$f(0) = f(x)f(-x) =&gt; f(-x) = f(0) / f(x)$

To find out what f(0) would be, we can set $x = y =  0$ in the original equation and see, $f(0) = f(0)^{2} =&gt; f(0) = 0, 1$

So you can see, the only condition we need is, $f(0) \neq 0$

Now, we set y = 0 on the original equation, hence,
$f(x + 0) = f(x)f(0) =&gt; f(x) = f(x) f(0) =&gt; f(x)(1 - f(0)) = 0$

Now, according to condition 1, f(x) can't be zero. Hence, $f(0) = 1$

Now we can conclude, $f(-x) = f(0) / f(x) = 1 / f(x)$

(Q.E.D)

N.B.:
As an addition, if the function is instead defined as $f:R -&gt; R^{+}$, which can be reduced to  Cauchy's Functional Equation.

From this pdf


  All continuous functions $f : R → (0,+∞)$ satisfying $f(x +y) = f(x)f(y)$ are of the form
  $f(x) = a^{x}$ . Namely the function $g(x) = log(f(x))$ is continuous and satisfies the Cauchy equation.


Now, if $g(x) = log(f(x))$, then,

$g(x+y) = log(f(x+y))$

Starting with your equation, taking log at both sides,

$log(f(x+y)) = log(f(x)) + log(f(y)) =&gt; g(x + y) = g(x) + g(y)$

Which is just Cauchy's Functional Equation.