Common Graphs — Question 7

Question 7 - Solution

We are given: $$f(x) = \frac{1}{x^2 + 1}$$

Step 1: Domain

Since $x^2 + 1 > 0$ for all real $x$, the domain is: $$\boxed{(-\infty, \infty)}$$

Step 2: Y-intercept

Set $x = 0$: $$f(0) = \frac{1}{0^2 + 1} = 1$$

Step 3: X-intercepts

Set $f(x) = 0$: $$\frac{1}{x^2 + 1} = 0 \Rightarrow \text{no solution}$$ So, no x-intercepts.

Step 4: Range

Since the denominator is always $\geq 1$, the maximum occurs at $x = 0$, and the function decreases on both sides. Therefore: $$\boxed{\text{Range: } (0, 1]}$$

Step 5: Symmetry

$$f(-x) = \frac{1}{(-x)^2 + 1} = \frac{1}{x^2 + 1} = f(x)$$ So the function is even, symmetric about the y-axis.

Step 6: End Behavior

As $x \to \pm \infty$, $f(x) \to 0$