1. **State the problem:** We are given the function $f(x) = \frac{1}{x^2 + 1}$ and want to understand its behavior. 2. **Formula and rules:** The function is a rational function where the denominator is $x^2 + 1$. Since $x^2 + 1 > 0$ for all real $x$, the function is defined everywhere. 3. **Find intercepts:** - **y-intercept:** Evaluate $f(0) = \frac{1}{0^2 + 1} = 1$. - **x-intercepts:** Solve $\frac{1}{x^2 + 1} = 0$. Since the numerator is 1, it never equals zero, so no x-intercepts. 4. **Find extrema:** - Take derivative: $f'(x) = -\frac{2x}{(x^2 + 1)^2}$. - Set $f'(x) = 0$ gives $x=0$. - Second derivative test or analyze sign changes shows $x=0$ is a maximum. 5. **Summary:** The function has a maximum at $(0,1)$, no x-intercepts, and is always positive. **Final answer:** The function $f(x) = \frac{1}{x^2 + 1}$ has a maximum value of 1 at $x=0$, no x-intercepts, and is defined for all real $x$.