Common Graphs — Question 10

Question 10 - Solution

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

Step 1: Domain

The denominator is always positive and defined for all real $x$: $$\boxed{\text{Domain: } (-\infty, \infty)}$$

Step 2: Intercepts

Y-intercept: $$f(0) = \frac{0}{\sqrt{0^2 + 1}} = 0$$

X-intercept: $$f(x) = 0 \Rightarrow \frac{x}{\sqrt{x^2 + 1}} = 0 \Rightarrow x = 0$$

Step 3: Symmetry

Check if the function is odd: $$f(-x) = \frac{-x}{\sqrt{(-x)^2 + 1}} = \frac{-x}{\sqrt{x^2 + 1}} = -f(x)$$ So the function is odd, symmetric about the origin.

Step 4: End Behavior

As $x \to \pm\infty$: $$f(x) = \frac{x}{\sqrt{x^2 + 1}} \to \pm 1$$

So horizontal asymptotes: $$\boxed{y = 1} \quad \text{as } x \to \infty \quad\text{and}\quad \boxed{y = -1} \quad \text{as } x \to -\infty$$

Step 5: Range

$$\boxed{\text{Range: } (-1, 1)}$$ The function approaches 1 and -1 but never reaches them.