To determine where the function $f(x) = \frac{1}{x^2 + 1}$ is increasing, we need to examine the first derivative of the function, $f'(x)$.

Step 1: Differentiate the function

We'll apply the quotient rule to differentiate the function. Recall that the quotient rule states: $\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$ where $u(x) = 1$ and $v(x) = x^2 + 1$.

The derivative of $u(x) = 1$ is $u'(x) = 0$, and the derivative of $v(x) = x^2 + 1$ is $v'(x) = 2x$.

Thus, applying the quotient rule: $f'(x) = \frac{(x^2 + 1)(0) - (1)(2x)}{(x^2 + 1)^2} = \frac{-2x}{(x^2 + 1)^2}$

Step 2: Analyze where the derivative is positive

The function $f(x)$ is increasing where its derivative $f'(x) > 0$. We have: $f'(x) = \frac{-2x}{(x^2 + 1)^2}$ The denominator $(x^2 + 1)^2$ is always positive for all real $x$, so the sign of $f'(x)$ depends on the numerator $-2x$.

• If $x > 0$, then $-2x < 0$, and $f'(x) < 0$.
• If $x < 0$, then $-2x > 0$, and $f'(x) > 0$.

Step 3: Conclusion

Thus, the function $f(x) = \frac{1}{x^2 + 1}$ is increasing on the interval $(-\infty, 0)$.

Answer:

The function is increasing on the open interval $(-\infty, 0)$.

Would you like further details on the solution, or do you have any questions about this process?

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Here are some related questions to explore:

1. How do we find where a function is decreasing?
2. What are the critical points of the function $f(x) = \frac{1}{x^2 + 1}$?
3. How can we use the second derivative test to analyze the concavity of the function?
4. How would the behavior of the function change if we modified the expression $f(x)$ to $\frac{1}{x^2 + 1} + 1$?
5. Can we determine the local maximum or minimum for $f(x) = \frac{1}{x^2 + 1}$?

Tip: Always check the sign of the first derivative to determine intervals of increase or decrease.