We are given the function $f(x) = x^2 + 1$ and we want to determine the interval(s) in which its reciprocal function, $g(x) = \frac{1}{f(x)} = \frac{1}{x^2+1}$, is increasing.

1. Problem Description

We are given the function $f(x) = x^2 + 1$ and we want to determine the interval(s) in which its reciprocal function, $g(x) = \frac{1}{f(x)} = \frac{1}{x^2+1}$, is increasing.

2. Solution Steps

A function is increasing when its derivative is positive. Thus, we need to find the derivative of $g(x)$ and determine when it is greater than

0. First, we find the derivative of $g(x)$ with respect to $x$:

$g(x) = \frac{1}{x^2+1}$

$g'(x) = \frac{d}{dx} (\frac{1}{x^2+1})$

Using the chain rule, we have:

$g'(x) = -\frac{1}{(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1)$

$g'(x) = -\frac{1}{(x^2+1)^2} \cdot (2x)$

$g'(x) = -\frac{2x}{(x^2+1)^2}$

Now, we need to find when $g'(x) > 0$. Since $(x^2+1)^2$ is always positive, we can analyze the sign of $-2x$.

$-\frac{2x}{(x^2+1)^2} > 0$

$-2x > 0$

$2x < 0$

$x < 0$

Thus, the reciprocal function $g(x)$ is increasing when $x < 0$. In interval notation, this is $(-\infty, 0)$.

3. Final Answer

$(-\infty, 0)$