Common Graphs — Question 9

Question 9 - Solution

We are given: $$f(x) = \ln(x^2 + 1)$$

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

Y-intercept: $$f(0) = \ln(0^2 + 1) = \ln(1) = 0$$

X-intercepts: Set $f(x) = 0$: $$\ln(x^2 + 1) = 0 \Rightarrow x^2 + 1 = 1 \Rightarrow x^2 = 0 \Rightarrow x = 0$$ So the only intercept is at the origin.

Symmetry: $$f(-x) = \ln((-x)^2 + 1) = \ln(x^2 + 1) = f(x) \Rightarrow \text{even function}$$

End behavior: As $x \to \pm\infty$, $x^2 + 1 \to \infty \Rightarrow f(x) \to \infty$

Range: Minimum value is at $x = 0$: $\ln(1) = 0$, and it increases without bound. $$\boxed{\text{Range: } [0, \infty)}$$